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4 years ago

Which equation in standard form has a graph that passes through the point (-4,-3) and has a slope of 3/8?

Mathematics

1 answer:

Sidana [21]4 years ago

5 0

Answer:

y = 3/8x - 1.5

Step-by-step explanation:

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Can someone help solve for number 3a and 3b

Arada [10]

3a should be 9/16.

3b should be 56.25%

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3 years ago

What is the solution of the system? Use any method.

Bad White [126]

Answer:

Step-by-step explanation:

-3,2

thats the answer your welcome

6 0

3 years ago

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Two-fifths of a number is decreased by three. Find the value of the expression if the number is twenty-five.

Sindrei [870]

Let's take the number as x.

two-fifths of a number = 2x / 5
decreased by 3 = (2x/5) - 3

(2x/5) - 3
(2*25/5) - 3
(2*5) - 3
10 - 3
7

8 0

3 years ago

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20 points pls help quick

masha68 [24]

Answer:

$\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\ f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2$

Step-by-step explanation:

We are given the two functions:

$\displaystyle f(x) = 3x^2 + mx +5 \text{ and } g(x) = nx^2 - 4x -2$

And that:

$\displaystyle h(x) = f(x)\cdot g(x)$

With the given conditions that (1, -40) and (-1, 24) satisfy the new function, we want to determine functions f and g.

First, find h:

$\displaystyle \begin{aligned} h(x) & = f(x)\cdot g(x) \\ \\ & = (3x^2 + mx +5)(nx^2 - 4x -2) \end{aligned}$

Because (1, -40) and (-1, 24) are points on the graph of h, we have that h(-1) = 40 and h(-1) = 24. In other words:

$\displaystyle \begin{aligned} h(1) = -40 & = (3(1)^2 + m(1) +5)(n(1)^2 - 4(1) -2) \\ \\ & = (3 + m +5)(n-4 -2) \\ \\ & = (m+8)(n-6) \\ \\ -40 &= mn-6m+8n-48 \end{aligned}$

And:

$\displaystyle \begin{aligned} h(-1) = 24 & = (3(-1)^2 + m(-1) +5)(n(-1)^2 -4(-1) -2) \\ \\ & = (3 - m +5)(n + 4 -2) \\ \\ & = (-m+8)(n+2) \\ \\ 24 & = -mn -2m + 8n +16 \end{aligned}$

Solve the system of equations. Adding the two equations together yield:

$\displaystyle -16 = -8m+16n - 32$

Solve for either m or n:

$\displaystyle \begin{aligned} -16 & = -8m + 16n - 32 \\ \\ 16 & = -8m + 16n \\ \\ 8m & = 16n - 16 \\ \\ m & = 2n -2\end{aligned}$

Substitute this into one of the two equations above and solve:

$\displaystyle \begin{aligned} -40 & = mn - 6m + 8n - 48 \\ \\ 0 & = (2n-2)n -6 (2n-2) + 8n -8 \\ \\ &= (2n^2 - 2n) + (-12n + 12) +8 n - 8 \\ \\ & = 2n^2 -6n + 4 \\ \\ & = n^2 - 3n + 2 \\ \\ &= (n-2)(n-1) \\ \\ & \end{aligned}$

Therefore:

$\displaystyle n = 2 \text{ or } n = 1$

Solve for m:

$\displaystyle \begin{aligned}m &= 2n-2 & \text{ or } m & = 2n-2 \\ \\ & = 2(2) - 1 &\text{ or } & =2(1) -2 \\ \\ &= 2 &\text{ or } & = 0 \end{aligned}$

Hence, the values of n and m are either: 2 and 2, respectively; or 1 and 0, respectively.

In conclusion, functions f and g are:

$\displaystyle f(x) = 3x^2 + 2x + 5\text{ and } g(x) =2x^2 - 4x -2\text{ or } \\ \\ f(x) = 3x^2 + 5 \text{ and } g(x) = x^2 - 4x -2$

7 0

2 years ago

Which operation would you do second in the following problem?

eimsori [14]

it's addition i looked it up

7 0

3 years ago

Read 2 more answers