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3 years ago

Given the parent function f(x)=x^2 describe the graph of y=(0.2x)^2

1 answer:

8 0

We have the following function:
f (x) = x ^ 2
We have the following transformation:
Expansions and horizontal compressions
The graph of y = f (bx):
If 0 <b <1, the graph of y = f (x) expands horizontally by the factor of 1 / b.
Applying the transformation:
y = (0.2x) ^ 2
The factor is:
1 / b = 1 / 0.2 = 5
Answer:
b. expanded horizontally by a factor of 5

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worty [1.4K]

$$x=\frac{\ln \left(\frac{3}{2}\right)}{4 \ln (1.0125)}$

Solution:

Given expression:

$$(1.0125)^{4 x}=\frac{3}{2}$

To solve the expression:

$$(1.0125)^{4 x}=\frac{3}{2}$

If f(x) = g(x) then ln(f(x)) = ln(g(x)).

Using the above condition, we can write

$$\ln \left(1.0125^{4 x}\right)=\ln \left(\frac{3}{2}\right)$

Apply log rule: $\log _{a}\left(x^{b}\right)=b \cdot \log _{a}(x)$

$$4 x \ln (1.0125)=\ln \left(\frac{3}{2}\right)$

Divide both side of the equation by $4 \ln (1.0125)$ .

$$\frac{4 x \ln (1.0125)}{4 \ln (1.0125)}=\frac{\ln \left(\frac{3}{2}\right)}{4 \ln (1.0125)}$

$$x=\frac{\ln \left(\frac{3}{2}\right)}{4 \ln (1.0125)}$

The answer is $x=\frac{\ln \left(\frac{3}{2}\right)}{4 \ln (1.0125)}$ .

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3 years ago

Find the image of c under the translation described by the translation rule T<4,5> (C)

iVinArrow [24]

Answer:

A. Point E

Step-by-step explanation:

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4 years ago

Evaluate the integral using the indicated trigonometric substitution. (use c for the constant of integration.) x^3 / sqrt x^2 +

slava [35]

$\displaystyle\int\frac{x^3}{\sqrt{x^2+49}}\,\mathrm dx$

Taking $x=7\tan\theta$ gives $\mathrm dx=7\sec^2\theta\,\mathrm d\theta$ , so that the integral becomes

$\displaystyle\int\frac{(7\tan\theta)^3}{\sqrt{(7\tan\theta)^2+49}}(7\sec^2\theta)\,\mathrm d\theta$
$=\displaystyle7^4\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{49\tan^2\theta+49}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\tan^2\theta+1}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sqrt{\sec^2\theta}}\,\mathrm d\theta$
$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{|\sec\theta|}\,\mathrm d\theta$

When $\sec\theta>0$ , we have

$=\displaystyle7^3\int\frac{\tan^3\theta\sec^3\theta}{\sec\theta}\,\mathrm d\theta$
$=\displaystyle7^3\int\tan^3\theta\sec^2\theta\,\mathrm d\theta$

and from here we can substitute $u=\tan\theta$ to proceed from here.

Quick note: When we set $x=7\tan\theta$ , we are implicitly enforcing $-\dfrac\pi2$ just so that the substitution can be undone later via $\theta=\tan^{-1}\dfrac x7$ . But note that over this domain, we automatically guarantee that $\sec\theta>0$ , so the absolute value bars can be dropped immediately.

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Csqrt%28%7B3%5Csqrt%7B3%7D%20%29%5E%7B2%7D%20%2B5%5E%7B2%7D" id="TexFormula1" title="\sqrt({

Triss [41]

Answer:

2√13

Step-by-step explanation:

⇒ √(3√3)² + 5²

⇒ √27 + 25

⇒ √52

⇒ 2√13

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2 years ago

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Ghella [55]

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4 years ago