The temperature is 20 degrees and is expected to fall 2 degrees each hour.
Write a linear equation in slope-intercept form to model this situation.

Yes it is

Step-by-step explanation:

it is a linear function as it is written in the standard form Y=MX+C

Answer:

y = -t

Step-by-step explanation:

Parametric Equation = Type of equation that employs an independent variable called a parameter (often denoted by t) and in which dependent variables are defined as continuous functions of the parameter and are not dependent on another existing variable.

⇒ parametric equation of line passing through a point (a₁, b₁, c₁)

   and parallel to a vector <a, b, c> is given by :

   x =a₁ + at ,   y = b₁ + bt   ,   z = c₁ + ct

now according to question:

given -

point, P(1, 0, -3)

line,   x = −1 + 2t , y = 2−t, and z = 3+3t.

so from the line the vector is=  <2, -1, 3>

now using above formula,

equation of line is =   x = 1 + 2t , y = −t, and z = -3+3t.

we have to solve for 'y' only,

⇒ y = -t    (answer)

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Answer:

-59

Step-by-step explanation:

x = -2

4(-2)³ - 6(-2)² + (-2) - 1

= -32 - 24 - 2 - 1

= - 59

Answer:

$38.50

Step-by-step explanation:

(100-30)% of $55

= 70% of $55

= $38.50

Hope this helps :)

Answer:

-4*(x+1)

Step-by-step explanation:

Simplify

⇒ 8x2 + 8x/-2x

2:  Pulling out like terms

4.1     Pull out like factors :

  8x2 + 8x  =   8x • (x + 1)

Canceling Out :

4.2    Canceling out x as it appears on both sides of the fraction line

Final result :

 -4 • (x + 1)

Answer:

-4(x+1)

Step-by-step explanation:

factor out -2x

-2x(-6x + 2x - 4) / -2x

the -2x in numerator and denominator cancel out

-6x + 2x - 4

-4x - 1

-4(x+1)

Answer:

g

Step-by-step explanation:

jjnnkvigific and it's a boy f a horse y

Answer:

insectistics a category variable also called various validity variable is a variable that can take on a limited as usually fixed number two possible values assigning each into visual or other unit of the vision 2 a particular group aur nominacle category of the basis some qualitative

College students (ages 18-26) tend to make decisions that are tentative (more short-range) and support a desire for autonomy. This depicts more permanent choices.

It should be noted that values are a compass that helps us make decisions and choices.

Choices characterize the stage of life we are in. For example 18-26 ages tend to make tentative choices, later 27-31 ages tend to make permanent choices, and ages 32-42 people make stable choices.

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Check the forward differences of the sequence.

• first-order differences

5 - 1 = 4

12 - 5 = 7

22 - 12 = 10

35 - 22 = 13

• second-order differences (i.e. differences of the first differences)

7 - 4 = 3

10 - 7 = 3

13 - 10 = 3

The second differences are all 3 (as far as we know), so the sequence of first differences is arithmetic/linear, which means the original sequence is quadratic. Let the [tex]n[/tex]-th term be

[tex]x_n = an^2 + bn + c[/tex]

Given that [tex]x_1=1[/tex], [tex]x_2=5[/tex], and [tex]x_3=12[/tex], we have

[tex]\begin{cases} a + b + c = 1 \\ 4a + 2b + c = 5 \\ 9a + 3b + c = 12 \end{cases} \implies a=\dfrac32, b=-\dfrac12, c=0[/tex]

and so the [tex]n[/tex]-th term of the sequence is generated by the rule

[tex]x_n = \dfrac{3n^2 - n}2[/tex]

which most closely resembles the last option,

Square the term number, multiply the result by 3, subtract the term number, and divide the result by 2.

Answer:

7%

Step-by-step explanation:

[tex] \frac{14766 - 13800}{13800} \times 100 = 7 [/tex]

The percent increase in the property's value is 7% if the parcel of land is valued at $13,800. Today, that same parcel of land has a value of $14,766.

It's the ratio of two integers stated as a fraction of a hundred parts. It is a metric for comparing two sets of data, and it is expressed as a percentage using the percent symbol.

We have:

A few years ago, Sarah acquired a parcel of land valued at $13,800. Today, that same parcel of land has a value of $14,766

Percentage increase:

[tex]= \rm \dfrac{14766-13800}{13800}\times 100[/tex]

= 7%

Thus, the percent increase in the property's value is 7% if the parcel of land is valued at $13,800. Today, that same parcel of land has a value of $14,766.

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An inequality relating 100n and n³ is 100n ≥ n³ for n ≤ 10 and 100n ≤ n³ for n ≥ 10.

An inequality is comparison of  two values, showing if one is less than, greater than, or simply not equal to another value.

Since 100n and n³ for n = 1, 2, 3, . . . 9, 10, 11 are 100, 200, 300, . . . 900, 1000, 1100 and 1, 8, 27, . . . 729, 1000, 1331 respectively.

Therefore, an inequality relating 100n and n³ will be 100n ≥ n³ for n ≤ 10 and 100n ≤ n³ for n ≥ 10.

Induction hypothesis:

Suppose 100n ≤ k³ for some positive integer k ≥ 10.

We need to show that 100( k + 1 ) ≤ ( k + 1 )³ = k³ + 3k² +3k + 1.

Note 100( k + 1 ) = 100k + 100 ≤ k³ + 100

                                                 ≤ k³ + 3k²    (∵ k ≥ 10 )

                                                 ≤ k³ + 3k² + 3k

                                                 ≤ k³ + 3k²+3k + 1

So 100( k + 1 ) ≤ ( k + 1 )³, which is true.

Hence by the principle of mathematical induction, 100n ≤ k³ for every integer  k ≥ 10.

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Answer:

27i

Step-by-step explanation:

Since we define i as [tex]\sqrt{-1}[/tex], [tex]\sqrt{-27} = 27i[/tex]

Answer:

Step-by-step explanation:

i [tex]i \sqrt{-27} =i\sqrt{-1 \times 27} =i\sqrt{27i^2} =i \times i\sqrt{27} =i^2\sqrt{9 \times 3} \\=-1\times 3\sqrt{3} \\=-3\sqrt{3}[/tex]

Answer:

t = 2

Step-by-step explanation:

[tex]2t + 4 = 14 - 3t[/tex]

[tex]5t = 10[/tex]

[tex]t = 2[/tex]

Checking:

[tex] log(2(2) + 4) = log(8) [/tex]

[tex] log(14 - 3(2)) = log(8) [/tex]

The range of y will be all real numbers such that 0≤y≤40

The complete question is:

Raj’s bathtub is clogged and is draining at a rate of 1.5 gallons of water per minute. The table shows that the amount of water remaining in the bathtub, y, is a function of the time in minutes, x, that it has been draining. What is the range of this function? all real numbers such that y ≤ 40 all real numbers such that y ≥ 0 all real numbers such that 0 ≤ y ≤ 40 all real numbers such that 37.75 ≤ y ≤ 40

The relation between two expressions that are not equal, employing a sign such as ≠ ‘not equal to’, > ‘greater than, or < ‘less than.

Raj’s bathtub is clogged and is draining at a rate of 1.5 gallons of water per minute.

The amount of water remaining in the bathtub = y

The function of  time in minutes, that it has been draining = x

At 0 minutes the amount of water is 40 gallons.

The highest volume of water is 40 which is decreasing at the rate of 1.5 gallons per minute.

The given function is a linear function  

y = 0

However, the volume of water can be 0 but cannot ever be negative.

Therefore the range of y will be all real numbers such that 0≤y≤40

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An equivalent ratio to the tape diagram shown is 1 to 2

From the tape diagram, we have:

Red = 2

Blue = 4

Express as a ratio

Red : Blue = 2 : 4

Divide by 2

Red : Blue = 1 : 2

Hence, an equivalent ratio to the tape diagram shown is 1 to 2

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Answer: 19683/64

Step-by-step explanation:

[tex]a_{n}=8\left(\frac{3}{2} \right)^{n-1}\\\\\implies a_{10}=8 \left(\frac{3}{2} \right)^{10-1}=\boxed{\frac{19683}{64}}[/tex]

The weight of each pile yesterday will be 6 and 8 pounds.

The unitary method is a method for solving a problem by the first value of a single unit and then finding the value by multiplying the single value.

It is Given that Yesterday the ratio of seeds in these piles was 3:4.

Let 3x and 4x represent the seeds.

If the 2 pounds of seeds are added in the bigger pile, then

4x + 2

He also ate 1/4 pound from the smaller pile,

3x - 1/4

The quantities of seeds in those piles is in the ratio of 5:16.

So, 4x + 2 = 5x

3x - 1/4 = 16x

Solve;

So, 4x + 2 = 5x

2 = 5x - 4x

x = 2

The weight of each pie will be

3x = 6 pound

4x = 8 pound

Hence, The weight of each pile yesterday would be 6 and 8 pounds.

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..................................

Answer:

please provide the diagram

Step-by-step explanation:

for the solution figure is necessary so provides us to solve this problem

Answer:

Sonu is 15years and Monu is 21 years

Using the probability concept, we have that:

a) It would not be unusual to observe one component​ fail, since the probability that one component​ fails is greater than 0.05.

b) It would be unusual to observe two components​ fail, since the probability that two components​ fail is less than 0.05.

A probability is given by the number of desired outcomes divided by the number of total outcomes. If a probability is less than 0.05, the event is considered unusual.

In this problem, the probabilities are given as follows:

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the end behavior is:

If we have a polynomial of even degree, then the end behavior in both ends is the same one.

Particularly, in these cases (even degree) we only need to look at the leading coefficient. If it is positive, then as x tends to infinity and negative infinity, the function tends to infinity.

In this case, the polynomial is:

y = 9x⁴ + 8x - 2

Notice that the degree is 4, even, and the leading coefficient is 9 (positive).

Then the end behavior is:

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[tex]{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}[/tex]

★ Curved surface area of right circular cylinder is 4.4 m².

★ Radius of base of the cylinder is 0.7 m.

★ [tex]\tt \pi = \dfrac{22}{7}[/tex]

[tex]{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}[/tex]

★ The height of cylinder.

[tex] {\large{\textsf{\textbf{\underline{\underline{Using \: Formula :}}}}}}[/tex]

[tex] \star \: \tt C.S.A \: of \: cylinder = \boxed{ \tt \pink{{ 2πrh}}}[/tex]

[tex] {\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}[/tex]

Let,

❍ The height of circular cylinder be [tex]h[/tex]

❍ Radius [r] of base of cylinder be 0.7 m

We know,

[tex] \star \: \tt C.S.A \: of \: cylinder = 2πrh[/tex]

Putting,

☆ [tex]\tt \pi = \dfrac{22}{7}[/tex]

☆ r as 0.7

[tex] \longrightarrow \tt 4.4 {m}^{2} = 2πrh[/tex]

[tex] \longrightarrow \tt 4.4 {m}^{2} = \bigg( 2 \times \dfrac{22}{7} \times 0.07 \times h \bigg)m[/tex]

[tex] \longrightarrow \tt \dfrac{44}{10} {m}^{2} = \bigg( {2 }\times \dfrac{22}{ \cancel{7}} \times \dfrac{ \cancel{7}}{10} \times h \bigg)m[/tex]

[tex] \longrightarrow \tt \dfrac{44}{10} {m}^{2} = \bigg( 44 \times \dfrac{ {1}}{10} \times h \bigg)m[/tex]

[tex] \longrightarrow \tt \dfrac{44}{ \cancel{10}} \times \cancel{10} \: m = \bigg( 44 \times 1 \times h \bigg)[/tex]

[tex] \longrightarrow \tt 44 m = 44h[/tex]

[tex]\longrightarrow \tt \dfrac{44}{44} m = h[/tex]

[tex]\longrightarrow \tt \red{ 1 m } = h[/tex]

Therefore, the height of cylinder = 1 m.

[tex]\begin{gathered} {\underline{\rule{290pt}{2pt}}} \end{gathered}[/tex]

The solution set for the given inequality is 3<x<5.

The given inequality is 6<x+3<8.

In mathematics, a solution set is the set of values that satisfy a given set of equations or inequalities.

Now, solve the given inequality:

6<x+3<8⇒6<x+3 and x+3<8

6-3<x ⇒3<x

x+3<8⇒x<5

Thus, 3<x<5.

Therefore, the solution set is 3<x<5.

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Inequality and graph the solution set and a number line, express the Solution set in interval notation is 3 < x < 5

6<x+3<8

[tex]6 < x+3\quad \mathrm{and}\quad \:x+3 < 8[/tex]

[tex]\mathrm{If}\:a < u < b\:\mathrm{then}\:a < u\quad \mathrm{and}\quad \:u < b[/tex]

[tex]\mathrm{Combine\:the\:intervals}[/tex]

[tex]x > 3\quad \mathrm{and}\quad \:x < 5[/tex]

[tex]3 < x < 5[/tex]

Therefore the Inequality and graph of the solution set and a number line, express the Solution set in interval notation as 3 < x < 5.

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Answer:

1.9x10^8

hope I helped

Answer:

0.0000000019= 1.9*10^ -9

Answer:

Day 5 ................

Answer:

See proof below

Step-by-step explanation:

Important points

Onto

For a function with a given co-domain to be "onto," every element of the co-domain must be an element of the range.

However, the co-domain here is suggested to be [tex]\mathbb R[/tex], whereas the range of f is not [tex]\mathbb R[/tex] (proof below).

Proof (contradiction)

Suppose that f is onto [tex]\mathbb R[/tex].

Consider the output 7 (a specific element of [tex]\mathbb R[/tex]).

Since f is onto [tex]\mathbb R[/tex], there must exist some input from the domain [tex]\mathbb R[/tex], "p", such that f(p) = 7.

Substitute and solve to find values for "p".

[tex]f(x)=-3x^2+4\\f(p)=-3(p)^2+4\\7=-3p^2+4\\3=-3p^2\\-1=p^2[/tex]

Next, apply the square root property:

[tex]\pm \sqrt{-1} =\sqrt{p^2}[/tex]

By definition, [tex]\sqrt{-1} =i[/tex], so

[tex]i=p \text{ or } -i =p[/tex]

By the Fundamental Theorem of Algebra, any polynomial of degree n with complex coefficients, has exactly n complex roots.  Since the degree of f is 2, there are exactly 2 roots, and we've found them both, so we've found all of them.

However, neither [tex]i[/tex]  nor [tex]-i[/tex] are in [tex]\mathbb R[/tex], so there are zero values of p in [tex]\mathbb R[/tex] for which f(p)= 7, which is a contradiction.

Therefore, the contradiction supposition must be false, proving that f is not onto [tex]\mathbb R[/tex]

answer:

w = (-m+y)/y

Step-by-step explanation:

y-wy = m

-wy = m-y

y = -(m-y)/w

wy= -m + y

w = (-m+y)/y

.