Algebra 2 urgent I need help quickkk 20 points !!!

Answer:

16П or approximately 50.3)

Step-by-step explanation:

The area of a circle is equal to П*r²

Since the circle is inside a rectangle with side that is equal to 8, we know that d = 8

2r = d

r = 8/2

r = 4

A = П*r²

A = П*16

А = 16П ( or 16*3.141592653589793238... = 50.26548... ≈ 50.3)

Answer:

area = 50.27 [tex]cm^2[/tex]

Step-by-step explanation:

In the diagram, we see that the diameter of the circle is 8cm. Therefore the radius of the circle is 8/2 = 4cm.

[tex]area \hspace{0.15cm}of \hspace{0.15cm}circle = $\pi$r^2[/tex]

                        = π [tex](4)^2[/tex]

                        = 50.27 [tex]cm^2[/tex]

Answer:

Given :-

Two legs of a rt. triangle : 8 and 10 respectively.

To find :-

Length of the hypotenuse

Solution :-

Using Pythagorean Theorem;

(hypotenuse)² = (8)² + (10)²

= 64 + 100

hypotenuse = √164

= 12.8 units

Answer:

12.80

Step-by-step explanation:

We have to use the Pythagoras theorem to find the value of the hypotenuse.

Let the value of the hypotenuse be c.

Let us use the below formula to find the length of the hypotenuse.

First, let us replace a and b with 8 and 10 ( the lengths of the legs of the right triangle) with a and b.

a² + b² = c²

8² + 10² = c²

64 + 100 = c²

And now let us solve the above and find the value of c.

164 = c²

√164 = c

12.80 = c

Therefore, the length of the hypotenuse is:

12.80 units

Answer:

3

Step-by-step explanation:

1: Turn the mixed fraction into improper.

2: Multiply 3/5 to each side

3: The you get x = 3

You need to multiply the fraction instead of division because multiplying by the reciprocal is the same as dividing the normal fraction.

[tex]\huge\text{Hey there!}[/tex]

[tex]\mathsf{5 = 1\dfrac{2}{3}x}[/tex]

[tex]\mathsf{1\dfrac{2}{3}x = 5}[/tex]

[tex]\mathsf{\dfrac{1\times3 + 2}{3}x = 5}[/tex]

[tex]\mathsf{\dfrac{3 + 2}{3}x = 5}[/tex]

[tex]\mathsf{\dfrac{5}{3}x = 5}[/tex]

[tex]\large\textbf{MULTIPLY }\rm{\bf \dfrac{3}{5}}\large\textbf{ to BOTH SIDES}[/tex]

[tex]\mathsf{{\dfrac{3}{5}\times\dfrac{5}{3}x= \dfrac{3}{5}\times 5}}[/tex]

[tex]\large\textbf{SIMPLIFY IT!}[/tex]

[tex]\mathsf{x = \dfrac{3}{5}\times5}[/tex]

[tex]\mathsf{x = \dfrac{3}{5}\times\dfrac{5}{1}}[/tex]

[tex]\mathsf{x = \dfrac{3\times5}{5\times1}}[/tex]

[tex]\mathsf{x = \dfrac{15}{5}}[/tex]

[tex]\mathsf{x = 15\div5}[/tex]

[tex]\mathsf{x = 3}[/tex]

[tex]\huge\text{Therefore, your answer should be: \boxed{\mathsf{x = 3}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

68% of the day's sales will be between 87.23 and 96.57 and on 95% of the days, sales will be between 82.56 and 101.24.

If your distribution follows a normal distribution, the standard deviation and mean can tell you where the majority of the data are.

We have:

The mean number of sandwiches sold per day is 91.9 and the standard deviation is 4.67.

As we know, from the empirical rule, 68% of the data will lie in one standard deviation of the mean, i.e., in the interval with the endpoints x±s

for samples and with endpoints u±б for the population.

x = 91.9

s = 4.67

x - s = 91.9 - 4.67 = 87.23

x + s = 91.9 + 4.67 = 96.57

68% of the day's sales will be between 87.23 and 96.57

95% of the data will lie within two standard deviations of the mean, which means in the interval with the endpoints  x±2s for samples and with endpoints  u±б population.

x - 2s = 82.56

x + 2s = 101.24

On 95% of the days, sales will be between 82.56 and 101.24

Thus, 68% of the day's sales will be between 87.23 and 96.57 and on 95% of the days, sales will be between 82.56 and 101.24.

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The inductive step is assuming that the statement is true for some  n = k , where  k  is one of the values we want to prove the statement for, it shows that it is also true for  n = k+1.

Suppose we have the statement 3n > 4n for all positive integers  n.

Now, the statement above is false when n = 1 , since 3 is less than 4. However, it will be true for all other positive integers.

In that case, we can still prove this statement by induction for all integers  n ≥ 2 .

Thus, the inductive step is assuming that the statement is true for some  n = k , where  k  is one of the values we want to prove the statement for, it shows that it is also true for  n = k+1.

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O is parallel to q

let's take the angle opposite to angle 4 is alpha

angle 4 = angle alpha ( vertically opposite angles)

angle alpha = angle 3 ( as angle 4 = angle 3).

hence o is parallel to q as both the angles are equal

The amount of water will remain in the bottle after Cory rides for 2 1/2 hours will be;

⇒ y = 12 ounces

What is Equation of line?

The equation of line in point-slope form passing through the points (x₁ , y₁) and (x₂, y₂) with slope m is defined as;

⇒ y - y₁ = m (x - x₁)

Where, m = (y₂ - y₁) / (x₂ - x₁)

Given that;

The average amount of water, in ounces, in his water bottle can be represented by the equation,

⇒ y = - 8x + 32

Where, y is the amount of water remaining after x hours.

Now,

Since, The average amount of water, in ounces, in his water bottle can be represented by the equation,

⇒ y = - 8x + 32

Hence, The amount of water will remain in the bottle after Cory rides for 2 1/2 hours is,

Substitute x = 2 1/2 in above equation,

⇒ y = - 8 × 2 1/2 + 32

⇒ y = - 8 × 5/2 + 32

⇒ y = - 20 + 32

⇒ y = 12 ounces

Thus, The amount of water will remain in the bottle after Cory rides for 2 1/2 hours is,

⇒ y = 12 ounces

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The question is incomplete. The complete question is

A teacher promised a movie day to the class that did better, on average, on their test. The box plot shown in the below-mentioned figure shows the results of the test.

Which class should get the reward, and why?

- The 2nd-period class should get the reward. They have the highest score, a perfect 100.

- The 2nd-period class should get the reward. They have a higher median.

- The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

- The 4th-period class should get the reward. Their lowest score is an outlier and should be thrown out.

The box plot shown in the question provides the information that
In 2nd period the minimum of the data is 77, first quartile is 78, median of the data is 89, third quartile is 92 and the maximum of the data is 100.
Hence, the mean of the results of 2nd period is

[tex]\frac{77+78+89+92+100}{5}=87.2[/tex]

In 2nd period the minimum of the data is 72, first quartile is 83, median of the data is 89, third quartile is 96 and the maximum of the data is 98.

Hence, the mean of the results of the 4th period is

[tex]\frac{72+83+89+96+98}{5}=87.6[/tex]

Hence, obviously, the 4th period is having more mean.

Moreover, we can observe that if more of the marks are on the higher side the average will automatically be more. Hence, as the first and the third quartiles are more on the 4th-period, so the average marks for the 4th period must be better.

So, just by observing the box plot, we can give the statement that  "The 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class. "

Therefore, the 4th-period class should get the reward. Though the medians are the same, the first and third quartiles are higher, so the students did better on average than in the 2nd-period class.

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Answer:

A = 5, B = 9, C = 6

Step-by-step explanation:

let me know if you want an explanation :))

All the equations have been solved

A quadratic equation is what can be represented by

ax² +bx+c = f(x)

It is given to solve the given quadratic equations

1.  x² = 25

x = ± 5

2. 16x² = 25

x² = 25/16

x = ± 5/4

3. (3x+2)²= 25

9x² +4+12x = 25

9x² +12x -21 = 0

3x(3x+7) -3(3x+7) = 0

x =1

x = -7/3

4. x² = 11x

x² -11x = 0

x (x-11) = 0

x = 0, 11

5. 3x²= 14x

3x² -14x = 0

x (3x - 14) = 0

x = 0 , 14/3

6. x²+2x = 3

x²+2x -3 = 0

x (x+3) -1(x+3) = 0

x = 1 ,-3

7.x²+2x -5 = 0

determined from the formula for the roots

x=−1±√6

x=1.44949

x=−3.44949

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Answer:

  y ≈ 15272.9(1.06306^x) . . . . from regression tool

  y = 15300(1.0624^x) . . . . calculated "by hand"

Step-by-step explanation:

An exponential model is one that uses the exponential function ...

  f(x) = a·b^x

to model the data. The value of 'a' is the function value when x=0, the initial value. The value of 'b' is the growth factor, the multiplier between function values for successive values of x.

__

An exponential regression tool makes short work of this. Such tools are available on some calculators, apps, and spreadsheets. All that is required is entering the data into a table and invoking the tool. The one shown in the attachment tells you the model is approximately ...

  y ≈ 15272.9(1.06306^x)

__

We can average the growth rate factors a couple of different ways. When we think of "averaging," we usually think of the arithmetic mean, the sum of values divided by their number. If we average the growth rates in this way, we get ...

  average rate of growth =

  ((161/153) +(173/161) +(184/173) +(196/184) +(207/196) +(220/207))/6 ≈ 1.0624

Using the first table value as the initial value, the model formed this way is ...

  y = 15300(1.0624^x)

__

For exponential growth it may make more sense to compute the geometric mean. This is the n-th root of the product of n numbers. In this case, it reduces to the 6-th root of the ratio of the last to the first:

  average rate of growth = (220/153)^(1/6) ≈ 1.0624

This gives the same growth factor as above, hence the same model.

The second attachment shows the application of this model to the domain used in the table. We see there are some minor differences when rounding the function value to the nearest hundred.

The differences in our two models seem to arise from rounding the data values in the table to the nearest hundred.

__

The values given in the table can be used in different ways to arrive at two similar, but different exponential models of the table data:

Interestingly, the function calculated "by hand" has a very slightly better correlation with the table data. The difference in correlation coefficients is in the 6th decimal place.

Using proportions, the expression for the price per person is:

[tex]P = \frac{850}{p}[/tex]

Jordan's expression is wrong because the price per person is inverse proportional to the number of people in the party, not direct proportional.

A proportion is a fraction of a total amount, and the measures are related using a rule of three.

The cost of renting a bowling alley for a party is $850, hence the cost per person is given by:

[tex]P = \frac{850}{p}[/tex]

Jordan's expression is wrong because the price per person is inverse proportional(that is, the amount is divided by p) to the number of people in the party, not direct proportional.

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The given differential equation is linear with constant coefficients,

[tex]\dfrac{d^2y}{dx^2} - 49y = 0[/tex]

with characteristic equation

[tex]r^2 - 49 = 0[/tex]

and hence characteristic roots [tex]r=\pm7[/tex]. This means the general solution to the ODE is

[tex]y = C_1 e^{7x} + C_2 e^{-7x}[/tex]

In fact, you're given the solution already,

[tex]y = C_1 e^{kx} + C_2 e^{-kx}[/tex]

and you've determined that

[tex]\dfrac{d^2y}{dx^2} = k^2 (C_1 e^{kx} + C_2 e^{-kx}) = k^2 y[/tex]

Comparing this to the given ODE, it's obvious that [tex]k=7[/tex], so you can just replace [tex]k[/tex] with 7 in the given template solution.

She should spend 1 hour working on problems , 3 hour reading , Option B is the right answer.

The gain of the desired effect with respect to the previous gain is called margin gain

The marginal gain from 8:00 am to 9:00 am is

(40-0) = 40 problems

The marginal gain from 9:00 am to 10:00 am is

= 70-40 = 30 problems

The marginal gain from 10:00 am to 11:00 am is

(90-70) = 20 problems

As if she does more than one hour of problem solving her productivity decreases to 30 which is less than 1 hour of reading textbook

Therefore she should not spend more than 1 hour on problem solving.

Therefore Option B is the right answer.

1 hour working on problems , 3 hour reading

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Considering the number of subsets of the set of options, it is found that there are 16 options for purchasing a house in this community.

The number of subsets is given by:

[tex]N_s = 2^n [/tex]

The set of options is given by:

O = { pool , alarm , lake view , balcony }

Which has cardinality n = 4.

The number of options for the purchased home is the number of subsets of set O, hence:

[tex]N_s = 2^4 = 16[/tex]

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Answer:

22,22,29 in

Step-by-step explanation:

x + x     +    x+7 = 73

3x = 66

x = 22      then lengths = 22, 22, 29 in

The values are x= -2 and x =4 of the given function f (x) =-5|x+1| +3

Function is an expression, rule, or law that defines a relationship between one variable (the independent variable) and another variable (the dependent variable).

According to the question,

The given function is

[tex]f(x) = -5|x + 1| + 3[/tex]

Here f(x) = -12

Substitute f(x) and solve for x

[tex]- 12 = -5|x + 1| + 3[/tex]

By subtracting 3 on both sides

[tex]- 12 - 3 = -5|x + 1| + 3 - 3[/tex]

[tex]- 15 = -5|x + 1|[/tex]

By dividing -5 on both sides

[tex]3 = |x + 1|[/tex]

We get absolute values for the function.

[tex]x + 1 = 3 , x + 1 = -3[/tex]

By subtracting 1 on both sides

[tex]x + 1 - 1 = 3 - 1 = 2\\\\x + 1 - 1 = - 3 - 1 = -4\\\\x = 2 , x = - 4[/tex]

Therefore, the values of x are 2 and - 4 of the given function f (x) =-5|x+1| +3

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If [tex]m=-4[/tex]

and a point is [tex](2,3)[/tex]

[tex]y=mx+b[/tex]

[tex]3=(-4)(2)+b[/tex] using the point and the slope

[tex]3=-8+b[/tex]

[tex]b=8+3[/tex]

[tex]b=11[/tex]

Therefore, the equation of the line will be [tex]y=-4x+11[/tex]

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:f(4)= -1 [/tex]

____________________________________

[tex] \large \tt Solution \: : [/tex]

If a point point (x , y) satifys a function, then it can be written as :

[tex]\qquad \tt \rightarrow \: f(x) = y[/tex]

Here, point (x , y) is replaced by (4 , -1)

[tex]\qquad \tt \rightarrow \: f(4) = - 1[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

Answer:

Step-by-step explanation:

Part A: Since 400 is being multiplied 1.06^t, for it to be equal to 399 the value 1.06^t must be less than 1. The lowest value for t given the domain restriction is 0 which would only result in 1 which will given the initial amount of 400. t would need to be less than 0 for 1.06^t to start going below the value 1 but since there is a domain restriction since t represents time, that value cannot go below 400 to reach 399.

Answer:

here you go

Step-by-step explanation:

no of times he failed = -3

points he losed = -3 x 4 = -12

no of times he need to hit the target to get 48 points = 6x

equation

-12 + 6x = 48

6x = 60

x = 10

no of points he  will  gain = 10 x 6 = 60

Answer:

  8 cylinders

Step-by-step explanation:

The surface area of a cylinder is given by the formula ...

  A = 2πr(r +l) . . . . . for radius r and length l

__

The cylinders Rufus has each have an area of ...

  A = 2π(5 cm)(5 cm +4 cm) = 90π cm² ≈ 282.74 cm²

That means 2500 cm² of paint will cover ...

  2500 cm²/(282.74 cm²/cylinder) ≈ 8.84 cylinders

Rufus has enough paint to cover 8 cylinders completely.

Answer:

[tex]\boxed {1)log_{b}(75) = 4.317}[/tex]

[tex]\boxed {2)ln(20) = 2.9957}[/tex]

Step-by-step explanation:

[tex]\textsf {Question l :}[/tex]

[tex]\longrightarrow \mathsf {log_{b}(3) = 1.099}[/tex]

[tex]\longrightarrow \mathsf {log_{b}(5) = 1.609}[/tex]

[tex]\textsf {Identities applied :}[/tex]

[tex]\boxed {log(ab) = loga + logb}[/tex]

[tex]\boxed {log(a)^{x} = xloga}[/tex]

[tex]\textsf {We can rewrite the problem as :}[/tex]

[tex]\longrightarrow \mathsf {log_{b}(75)}[/tex]

[tex]\longrightarrow \mathsf {log_{b}(25 \times 3)}[/tex]

[tex]\longrightarrow \mathsf {log_{b}(5^{2} \times 3)}[/tex]

[tex]\longrightarrow \mathsf {log_{b}(5)^{2} + log_{b}(3)}[/tex]

[tex]\longrightarrow \mathsf {2log_{b}(5) + log_{b}(3)}[/tex]

[tex]\textsf {Now, substitute the values :}[/tex]

[tex]\longrightarrow \mathsf {2(1.609) + (1.099)}[/tex]

[tex]\longrightarrow \mathsf {3.218 + 1.099}[/tex]

[tex]\longrightarrow \mathsf {4.317}[/tex]

[tex]\boxed {log_{b}(75) = 4.317}[/tex]

[tex]\textsf {Question ll :}[/tex]

[tex]\longrightarrow \mathsf {ln(4) = 1.3863}[/tex]

[tex]\longrightarrow \mathsf {ln(5) = 1.6094}[/tex]

[tex]\textsf {Rewriting the problem :}[/tex]

[tex]\longrightarrow \mathsf {ln(20)}[/tex]

[tex]\longrightarrow \mathsf {ln(4 \times 5)}[/tex]

[tex]\longrightarrow \mathsf {ln(4) + ln(5)}[/tex]

[tex]\longrightarrow \mathsf {1.3863 + 1.6094}[/tex]

[tex]\longrightarrow \mathsf {2.9957}[/tex]

[tex]\boxed {ln(20) = 2.9957}[/tex]

Answer:

[tex]\sf \log_b(75)=4.317[/tex]

[tex]\sf \ln (20)=2.9957[/tex]

Step-by-step explanation:

Question 1

Given:

  [tex]\sf \log_b(3)=1.099[/tex]

  [tex]\sf \log_b(5)=1.609[/tex]

To evaluate [tex]\sf \log_b(75)[/tex],  replace 75 with (5 × 5 × 3):

[tex]\implies \sf \log_b(5 \cdot 5 \cdot 3)[/tex]

[tex]\textsf{Apply the Product log law}: \quad \log_axy=\log_ax + \log_ay[/tex]

[tex]\implies \sf \log_b5+\log_b5+\log_b3[/tex]

Substitute the given values to solve:

[tex]\implies \sf 1.609 + 1.609 + 1.099=4.317[/tex]

Question 2

Given:

  [tex]\sf \ln(4)=1.3863[/tex]

  [tex]\sf \ln(5)=1.6094[/tex]

To evaluate ln(20) replace 20 with (4 × 5):

[tex]\implies \sf \ln (4 \cdot 5)[/tex]

[tex]\textsf{Apply the Product log law}: \quad \ln xy=\ln x + \ln y[/tex]

[tex]\implies \sf \ln (4)+\ln (5)[/tex]

Substitute the given values to solve:

[tex]\implies \sf 1.3863+1.6094=2.9957[/tex]

Answer:

y = C/h - 2

Step-by-step explanation:

Equation

C = (2 + y)*h

Solution

C = (2 + y)*h                          Divide by h

C/h = (2 + y)*h/h

C/h = 2 + y                            Subtract 2 from both sides

C/h-2 = 2-2+ y                      Combine

C/h - 2 = y

Answer:

[tex]360xy^2|xz^3|[/tex]

Step-by-step explanation:

Given expression:

[tex]12x \sqrt{900x^2y^4z^6}[/tex]

[tex]\textsf{Apply radical rule} \quad \sqrt{ab}=\sqrt{a}\sqrt{b}:[/tex]

[tex]\implies 12x \sqrt{900}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}[/tex]

Replace 900 with 30² :

[tex]\implies 12x \sqrt{30^2}\sqrt{x^2}\sqrt{y^4}\sqrt{z^6}[/tex]

[tex]\textsf{Apply radical rule} \quad \sqrt{a^2}=a, \quad a \geq 0:[/tex]

[tex]\implies 12x \cdot 30|x|\sqrt{y^4}\sqrt{z^6}[/tex]

[tex]\implies 360x|x|\sqrt{y^4}\sqrt{z^6}[/tex]

(We need to use the absolute value of √x² since the x term was originally to the power of 2, which means the value of x² is always positive since the exponent is even).

[tex]\textsf{Apply exponent rule} \quad \sqrt{a^m}=a^{\frac{m}{2}}:[/tex]

[tex]\implies 360x|x|\cdot y^{\frac{4}{2}}\cdot z^{\frac{6}{2}}[/tex]

Simplify:

[tex]\implies 360xy^2|xz^3|[/tex]

(We need to use the absolute value of z³ since the z term was original to the power of 6, which means the value of z⁶ is always positive since the exponent is even).

Answer:

B

Step-by-step explanation:

when x is 0 y=4(0)+12 y=12

y is 0 y=4(-3) +12 = 12=12

Using the permutation formula, it is found that there are 57,120 ways to fill the 4 positions.

There are different roles, hence the order is important, which means that the permutation formula is used.

The number of possible permutations of x elements from a set of n elements is given by:

[tex]P_{(n,x)} = \frac{n!}{(n-x)!}[/tex]

In this problem, 4 roles are chosen from a set of 17, hence the number of ways is given by:

[tex]P_{17,4} = \frac{17!}{13!} = 57120[/tex].

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Answer:

g(x) = x³ - 2

(option B)

Step-by-step explanation:

because the shape of the graph has not changed, we know that there was no direct change to x

(this would have looked like 2x³ instead of x³)

and because the graph was not shifted along the x-axis (left or right), we know that the change did not happen inside the parenthesis

(this would have looked like (x³ - 3) instead of (x³) )

we know that the only change that has happened to this graph was a downward shift along the y-axis  which can be expressed as

g(x) = x³ - 2

The equation of the red graph will be g(x) = x³ - 2.Option B is correct.

A mathematical statement consisting of an equal symbol between two algebraic expressions with the same value is known as an equation.

We know that there was no immediate change to x since the graph's form has not altered. Instead of x³, this would have appeared as 2x³.

Thus we can be confident that the change did not occur inside the parenthesis since the graph was not moved down the x-axis.

We may infer that this graph's sole modification was a downward shift along the y-axis, which is written as;

g(x) = x³ - 2

The graph is also attached in the attachment.

The equation of the red graph will be g(x) = x³ - 2.

Hence option B is correct.

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