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Shkiper50 [21]
2 years ago
14

How do I solve for d in the following equation? c(ad)+x=bd+ad(yz)

Mathematics
1 answer:
Klio2033 [76]2 years ago
8 0
Easy
remember associative property
a(bc)=(ab)c
so theerfor we can move dem around
also rembmer distributive property and how it can be expanded
ab+ac=a(b+c) therefor
ab+ac+ad=a(b+c+d)

(ca)d+x=bd+d(ayz)
minus (ca)d from both sides
x=db+d(ayz)-d(ca)
undistribute d
x=d(b+ayz-ca)
divide both sides by (b+ayz-ca)
\frac{x}{b+ayz-ca}=d
d=\frac{x}{b+ayz-ca}
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Given:

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To find:

The statement that best compares the graph of g(x) with the graph of f(x).

Solution:

The horizontal stretch is defined as:

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If 0, the function f(x) is horizontally stretched by factor \dfrac{1}{k}.

If k>1, the function f(x) is horizontally compressed by factor \dfrac{1}{k}.

We have,

f(x)=x^2

g(x)=\left(\dfrac{1}{3}x\right)^2

Using these functions, we get

g(x)=f\left(\dfrac{1}{3}x\right)         ...(ii)

On comparing (i) and (ii), we get

k=\dfrac{1}{3}

Since 0, the function f(x) is horizontally stretched by factor \dfrac{1}{\frac{1}{3}}=3.

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2 years ago
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