helpp me please
Find the missing term.
The roots of x2 − ( ) + 34 are 5 ± 3i.

Answer:

True, True, False

Step-by-step explanation:

Answer:

sure how can I help you? please tell me

Answer:

68.5% seats filled

76% points earned

Step-by-step explanation:

Percentages are formed when one finds a ratio of two related quantities, usually comparing the first partial quantity to the amount that "should" be there.

[tex]\text{ratio}=\dfrac {\text{the "part"}}{\text{the whole}}[/tex]

For instance, if you have a pie, and you eat half of the pie, you're in effect imagining the original pie (the whole pie) cut into two equal pieces, and you ate one of them (the "part" of a pie that you ate).  To find the ratio of pie that you ate compared to the whole pie, we compare the part and the whole:

[tex]\text{ratio}=\dfrac {\text{the number of "parts" eaten}}{\text{the number of parts of the whole pie}}[/tex]

[tex]\text{ratio}=\dfrac {1}{2}[/tex]

If you had instead eaten three-quarters of the pie, you're in effect imagining the original pie cut into 4 equal pieces, and you ate 3 of them.

[tex]\text{ratio}=\dfrac {\text{the number of "parts" eaten}}{\text{the number of parts of the whole pie}}[/tex]

[tex]\text{ratio}=\dfrac {3}{4}[/tex]

There can be cases where the "part" is bigger than the whole.  Suppose that you are baking pies and we want to find the ratio of the pies baked to the number that were needed, the number of pies you baked is the "part", and the number of pies needed is the whole.  This could be thought of as the ratio of project completion.

If we need to bake 100 pies, and so far you have only baked 75, then our ratio is:

[tex]\text{ratio}=\dfrac {\text{the number of "parts" made}}{\text{the number of parts of the whole order}}[/tex]

[tex]\text{ratio}=\dfrac {75}{100}[/tex]

But, suppose you keep baking pies and later you have accidentally made more than the 100 total pies.... you've actually made 125 pies.  Even though it's the bigger number, the number of pies you baked is still the "part" (even though it's bigger), and the number of pies needed is the whole.

[tex]\text{ratio}=\dfrac {\text{the number of "parts" made}}{\text{the number of parts of the whole order}}[/tex]

[tex]\text{ratio}=\dfrac {125}{100}[/tex]

To find a percentage from a ratio, there are two small steps:

Going back to the pies:

When you ate half of the pie, your ratio of pie eaten was [tex]\frac{1}{2}[/tex]

Dividing the two numbers, the result is [tex]0.5[/tex]

Multiplying by 100 gives 50.  So, the percentage of pie that you ate (if you ate half of the pie) is 50%

When you ate three-quarters of the pie, the ratio was [tex]\frac{3}{4}[/tex]

Dividing the two numbers, the result is 0.75

Multiplying by 100 gives 75.  So, the percentage of pie that you ate (if you ate three-quarters of the pie) is 75%.

When you were making pies, and 100 pies were needed, but so far you'd only baked 75 pies, the ratio was [tex]\frac{75}{100}[/tex]

Dividing the two numbers, the result is 0.75

Multiplying by 100 gives 75.  So, the percentage of the project that you've completed at that point is 75%.

Later, when you had made 125 pies, but only 100 pies were needed, the ratio was [tex]\frac{125}{100}[/tex]

Dividing the two numbers, the result is 1.25

Multiplying by 100 gives 125%.  So, the percentage of pies you've made to complete the project at that point is 125%.... the number of pies that you've made is more than what you needed, so the baking project is more than 100% complete.

1.   27400 spectators n a 40000 seat stadium percentage.

Here, it seems that the question is asking what percentage of the stadium is full, so the whole is the 40000 seats available, and the "part" is the 27400 spectators that have come to fill those seats.

[tex]\text{ratio}=\dfrac {\text{the number of spectators filling seats}}{\text{the total number of seats in the stadium}}[/tex]

[tex]\text{ratio}=\dfrac {27400}{40000}[/tex]

Dividing gives 0.685.  Multiplying by 100 gives 68.5.  So, 68.5% of the seats have been filled.

2.   an archer scores 95 points out of a possible 125 points percentage

Here, it seems that the question is asking what percentage of the points possible were earned, so the whole is the 125 points possible, and the "part" is the 95 points that were earned.

[tex]\text{ratio}=\dfrac {\text{the number of points earned}}{\text{the total number of points possible}}[/tex]

[tex]\text{ratio}=\dfrac {95}{125}[/tex]

Dividing gives 0.76.  Multiplying by 100 gives 76.  So, 76% of points possible were earned.

The correct answer is option A which is the number of cars sold for the price range of $4000 to $4999 will be 15.

A histogram is a graph for the representation of the data on the plot of the rectangular boxes. It has the data sets on the horizontal and the vertical axes.

As we can see from the histogram data we will conclude that the price range of $4000 to $4999 is shown by the third block and this third block reaches the height of 15.

Therefore the correct answer is option A which is the number of cars sold in the price range of $4000 to $4999 will be 15.

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Answer: A. -3.65

Step-by-step explanation: The best way to figure this out is by using a calculator. One specific rule to remember is the positive/negative rule. When it applies to multiplication, it also applies to division.

[tex](+)*(+)=+\\(+)*(-)=-\\(-)*(+)=-\\(-)*(-)=+\\[/tex]

Knowing this, we can input it into the calculator and find the answer's sign.

[tex]-27.375/7.5=-3.65[/tex]

Therefore, the answer is A. -3.65.

I hope this helps! Pls mark brainliest!! :)

Answer:

It appears to already be solved. What do you need help with?

Step-by-step explanation:

Answer:

radius ≈ 15.5

Step-by-step explanation:

the radius is RS

the angle between a tangent and the radius at the point of contact is 90°

then Δ RST is a right triangle

using Pythagoras' identity in the right triangle.

the square on the hypotenuse is equal to the sum of the squares on the other 2 sides , then

RS² + ST² = RT² ( substitute values )

RS² + 7² = 17²

RS² + 49 = 289 ( subtract 49 from both sides )

RS² = 240 ( take square root of both sides )

RS = [tex]\sqrt{240}[/tex] ≈ 15.5 ( to 1 dec. place )

Answer:

f=- 6, x=-1

f = - 6, x=-3

Step-by-step explanation:

Solve for the first variable in one of the equations, then substitute the result into the other equation.

The  values of x and y would make the following expression represent a real number is 4 and -5 respectively

The standard form of writing a complex number is given asl

z= x + iy

where

x is the real part

y is the imaginary part

Given the expression below;

(4 +5i)(x + yi)

Expand

4x + 4yi + 5ix + 5y(-1)

4x + 4yi + 5ix - 5y

Hence the  values of x and y would make the following expression represent a real number is 4 and -5 respectively

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The question is asking you to understand that a population which is raised by a common ratio over a certain time period follows an exponential pattern. See:

After 1 hour, the population is 3*10

After 2 hours, the population is 3*3*10

After 3 hours, the population is 3*3*3*10

.....

This can be generalised as a function of t, the time in hours:

f(t) = (3^t) * 10

Since the function determines the population at a given time, it is more prudent to replace f(t) with P, the population after time t:

P = (3^t) * 10

Since 20 minutes is equal to (1/3) hours, t can be substituted for (1/3) in order to calculate the population size after 20 minutes:

P = (3^(1/3)) * 10 = 14.4224957031 ≈ 14

Therefore the population after 20 minutes is 14.

Jaira is completing the construction of a regular hexagon inscribed in a circle, as shown below: The next step in her construction should be:

A closed polygon with six equal sides and six equal angles is called a regular hexagon.

Thus, the steps highlighted above will help Jaira complete her construction.

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I'll use the method of transformations.

If [tex]f_X(x)[/tex] denotes the PDF of [tex]X[/tex], and [tex]y=g(x)=x^3 \iff x=g^{-1}(y) = y^{1/3}[/tex], we have

[tex]f_Y(y) = f_X\left(g^{-1}(y)\right) \left|\dfrac{dg^{-1}}{dy}\right|[/tex]

[tex]\dfrac{dg^{-1}}{dy} = \dfrac13 y^{-2/3}[/tex]

[tex]\implies f_Y(y) = f_X\left(y^{1/3}\right) \left|\dfrac13 y^{-2/3}\right| = \boxed{\begin{cases} \dfrac53 y^{2/3} & \text{if } 0 \le y \le 1 \\ 0 & \text{otherwise} \end{cases}}[/tex]

Answer:

6.871 decimeters

Step-by-step explanation:

687.1 square centimeters = 6.871 square decimeters

1 dm² = 100 cm²

Answer:

6.871

Step-by-step explanation:

calculator

Answer:

it's 1,1 is the correct ans

Step-by-step explanation:

Because my teacher told its absolutely correct answer I got the same question in exam

Answer:

the answer should be the 2 one

Step-by-step explanation:

I got it right just had it.

Only Set B  is a partition of set N of natural numbers .

Numbers starting from 1 to infinity comes under Natural Numbers.

It is asked in the question that to determine whether or not each of the following is a partition of set N of natural numbers

A  ) {{n | n > 5}, {n | n < 5}]

It is not a set of natural numbers as it does not include 5.

B  )  [{n | n ≥ 6}, {1, 3,5} , {2, 4}]

{1,3,5} {2,4}and {n|n≥6} include all the natural numbers

Therefore , it forms a partition of N.

C ) {{n | n^2 > 11}, {n | n^2 < 11}

Every natural number n must satisfy either n^2 > 11 or n^2 < 11

This cannot be possible , therefore it is not the partition of Natural Numbers.

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The solution to all the answers are given below.

The complete question includes

Grades on a standardized test are known to have a mean of 1,000 for students in the Delhi. 453

randomly selected Delhi students take the test, yielding sample mean of 1,013 and sample standard

deviation (s) of 108.

It is given by

Confidence Interval for 95% confidence Interval is given by

[tex]\rm Z = X \pm 1.96 \dfrac{\sigma}{\sqrt{n}}[/tex]

(a) Construct a 95% confidence interval for the mean test score for Delhi students.

The confidence interval is given by

[tex]\rm 1,013 \pm 1.96 \dfrac{\sigma}{\sqrt{n}}[/tex]

[tex]\rm 1,013 \pm 1.96 \dfrac{108}{\sqrt{453}}[/tex]

1013 ± 5.07

so the interval is  [1003.06,1022.94]

(b)Yes, since the null of no difference is rejected at the 5% significance level (interval excludes Delhi sample mean of 1,013)

(c) Another 503 Delhi students are randomly selected to take a 3-hour prep course and then give the test. Their average score is 1,019 with a standard deviation of 95.

The standard deviation now is

[tex]\rm \sqrt{\dfrac{95^2}{453} + \dfrac{108^2}{503}}[/tex]

= 6.61

The interval is given by

[tex](1,019-1,013) \pm 1.96 \dfrac{\sigma}{\sqrt{n}}[/tex]

= [-7,+19]

(d) No, the interval includes 0, the null difference between the two populations

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We can make 64 different equations using the power of ten.

The power of a number identifies how many times that particular number is multiplied by itself.

Here, let us assume that the different equations = x

Using a power of 10, we have 10x making a total of 640.

Divide both sides by 10

10x/10 = 640/10

x = 64

Therefore, we can conclude that we can make 64 different equations using the power of ten.

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Answer:

ANSWER IS B

Step-by-step explanation:

NEXT TIME USE any graph website online

Answer: (1+2a)^2 or (1+2a)(1+2a)

Step-by-step explanation:

Answer:

[tex](2a +1)( 2a + 1)[/tex]

Step-by-step explanation:

Answer:

Step-by-step explanation:

It represents a linear function because there is a constant rate of change.

Step-by-step explanation:

Since, if the rate of change for y ( output value ) with respect to x ( input value ) remains constant for a function then it is called linear function.

Here, number of hours represents input value and minutes represents the output value,

Now, By the given table,

The function is passing through the points (1,60), (2,120), (3,180), (4,240) and (5,300),

Also,

⇒  The rate of change output value with respect to input value remains constant,

Hence, It represents a linear function because there is a constant rate of change.

b. Total fixed cost are $49885

Sometimes production of a product may increase or decrease, but the cost which do not change with the change of output is called fixed cost of the product.

Highest monthly total cost = $83800

Highest monthly production units = 3500 desks

Lowest monthly total cost = $61900

Lowset monthly production units = 1240 desks

So, variable cost per unit = (Highest cost-lower cost)/(highest production units-lowest production units)

= ($83800-$61900)/(3500-1240)

= $21900/2260

= $9.7

∴ Total fixed costs=Highest monthly total cost-(highest production units×variable cost per unit)

= $83800-(3500×$9.7)

= $83800-$33950

= $49850

Nearest value is $49885, whch is the correct one.

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The result follows from the fundamental theorem of calculus.

[tex]\displaystyle \int_a^b f(x) \, dx = F(b) - F(a) = 2022 - 10 = \boxed{2012}[/tex]

Answer: B. (2,2)

Step-by-step explanation:

The vertex of the parabola in the form [tex](x-h)^2 = 4p(y-k)[/tex] is (h, k).

Answer:

See below

Step-by-step explanation:

Answer:

The chain of the length 7 inches weighs 112 ounces.

Step-by-step explanation:

As we know there are 12 inches in a foot and 16 ounces in a pound

That is 1 foot = 12 inches.

and 1 pound = 16 ounces.

Given that the weight of the chain that is 1 foot long = 12 pounds

So weight of the chain per inch is = 12/12

which is equal to 1 pound

and according to the formula 1 pound = 16 ounces

So weight of the chain per inch is =  16 ounces

therefore weight of the chain that is 7 inch long = 7 × 16

that is 112 ounces.

The chain of the length 7 inches weighs 112 ounces.

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The inverse of a function is n(a) = (a+30)/3 option (A) is correct by using the concept of the inverse of a function.

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

We have a function:

a(n) =  3n - 30

To find make subject n and solve

a(n) + 30 = 3n

[tex]\rm n = \dfrac{a(n) + 30}{3}[/tex]

Plug n = n(a) and a(n) = a

[tex]\rm n(a) = \dfrac{a + 30}{3}[/tex]

Thus, the inverse of a function is n(a) = (a+30)/3 option (A) is correct by using the concept of the inverse of a function.

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Notice that the difference in the absolute values of consecutive coefficients is constant:

|-7| - 1 = 6

13 - |-7| = 6

|-19| - 13 = 6

and so on. This means the coefficients in the given series

[tex]\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots[/tex]

occur in arithmetic progression; in particular, we have first value [tex]c_1 = 1[/tex] and for [tex]n>1[/tex], [tex]|c_i|=|c_{i-1}|+6[/tex]. Solving this recurrence, we end up with

[tex]|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5[/tex]

So, the sum to [tex]n[/tex] terms of this series is

[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S[/tex]

The second sum [tex]S[/tex] is a standard geometric series, which is easy to compute:

[tex]S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}[/tex]

Multiply both sides by [tex]-a[/tex] :

[tex]-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n[/tex]

Subtract this from [tex]S[/tex] to eliminate the intermediate terms to end up with

[tex]S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}[/tex]

The first sum [tex]S'[/tex] can be handled with simple algebraic manipulation.

[tex]S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}[/tex]

[tex]\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i[/tex]

[tex]\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i[/tex]

[tex]\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}[/tex]

[tex]\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S[/tex]

[tex]\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S[/tex]

[tex]\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}[/tex]

[tex]\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}[/tex]

[tex]\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}[/tex]

Putting everything together, we have

[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S[/tex]

[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}[/tex]

[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}[/tex]