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3 years ago

Write the equation of a line that is parallel to y = -5/4x+ + 7 and that passes through the point (-4,1).

Mathematics

1 answer:

k0ka [10]3 years ago

5 0

Answer: y- 1 = -5/4(x+4)\\

Step-by-step explanation:

y = mx + b, where m is the slope, and in the equation, the slope is -5/4.

For two-line that are parallel, the slope is the same, -5/4.

$y - y_{1}= m( x - x_{1})\\y- 1 = -5/4(x-(-4))\\y- 1 = -5/4(x+4)\\$

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olga55 [171]

Answer:

f(x) = -(x-3)^2 + 4

Step-by-step explanation:

(h; k) are the coordinates of the vertex.

On the graph are (3, 4), therefore we have:

f(x) = a(x-3)^2 + 4

We have the x- intercept (1,0) -> x=1; y=0.

Substitute them into the equation:

0 = a(1 - 3)^2 + 4

0 = a(-2)^2 + 4

4a + 4 = 0 | -4

4a = -4 |-4

a = -1

So, we have the answer:

f(x) = -(x-3)^2 + 4

5 0

3 years ago

What is the correct answer?

Tasya [4]

Given:

The function is

$f(x)=(x+3)^2(x-5)^6$

To find:

The zeros of the given function.

Solution:

The general form of polynomial is

$P(x)=a(x-c_1)^{m_1}(x-c_2)^{m_2}...(x-c_n)^{m_n}$ ...(i)

where, a is a constant, $c_1,c_2,...,c_n$ are zeros of respective multiplicities $m_1,m_2,...,m_n$ .

We have,

$f(x)=(x+3)^2(x-5)^6$

On comparing this with (i), we get

$c_1=-3,m_1=2$

$c_2=5,m_2=6$

It means, -3 is a zero with multiplicity 2 and 5 is a zero with multiplicity 6.

Therefore, the correct option is B.

5 0

3 years ago

Hi, Could anyone help me with my homework

tatiyna

Answer:

$\hookrightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

solving steps:

$\rightarrow \sf (x + 4)^6$

$\bold{rewrite \ the \ following}$

$\rightarrow \sf (x + 4)^2 (x + 4)^2 (x + 4)^2$

$\bold {formula \ used : \sf (x+a)^2 = (x^2 + 2xa + a^2)}$

$\rightarrow \sf (x^2 + 8x+16) (x^2 + 8x+16) (x^2 + 8x+16)$

$\bold{simplify \ by \ removing \ parenthesis}$

$\rightarrow \sf (x^4 +8x^3 + 16x^2 + 8x^3 +64x^2 + 128x+16x^2+128x+256 ) (x^2 + 8x+16)$

$\bold{basic \ addition \ of \ integers }$

$\rightarrow \sf (x^4+16x^3+96x^2+256x+256) (x^2 + 8x+16)$

$\bold{remove \ parenthesis}$

$\rightarrow \sf (x^6 + 16x^5 + 96x^4 + 256x^3 + 256x^2 + 8x^6 + 128x^4 + 768x^3 + 2048x^2 + 2048 + 16x^4 + 256x^3 + 1536x^2 + 4096x + 4096)$

$\bold {final \ answer:}$

$\rightarrow \sf x^6+24x^5+240x^4+1280x^3+3840x^2+6144x+4096$

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2 years ago