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3 years ago

The graph of g(x) = (1/2)x is the graph of f(x) = 2x reflected over the y-axis. Which graph represents g(x)?

2 answers:

4 0

Reflect across y axis means that you replace x with -x

f(-x)=2(-x)=-2x
g(x)=-2x
de graph is the one that passes through (0,0) and (1,-2) and -1,2)

loris [4]3 years ago

4 0

Answer:

The answer is option two on the quiz. The graph is asking us to reflect the given graph over the y-axis. Therefore, the only one that matches is the second option. The graph that passes through (0,1), (-1,2), and (0,5)

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Y intercept of this graph

Aleksandr [31]

Answer:

The y intercept is - 2

Step-by-step explanation:

Because in the y axis the line crosses - 2

8 0

3 years ago

Round to the nearest hundredth? A 4 ? 35° с B

DaniilM [7]

The length of the unknown side length, |AB|, is 6.97

<h3>Trigonometry</h3>

From the question, we are to determine the value of the unknown side length

The unknown side length is the hypotenuse

Using SOH CAH TOA

We can write that

sin 35° = 4/|AB|

∴ |AB| = 4 /sin35°

|AB| = 4 /sin35°

|AB| = 6.97

Hence, the length of the unknown side length, |AB|, is 6.97

Learn more on Trigonometry here: brainly.com/question/22099203

#SPJ1

3 0

1 year ago

How many seconds will light leaving Los Angeles take to reach the following locations (a) San Francisco (about 500km), (b) Londo

kondor19780726 [428]

Answer:

a) It takes 0.0017s for the light to reach San Francisco.

b) It takes 0.033s for the light to reach London.

c) It takes 1.334s for the light to reach Mars.

d) It takes 149.7s for the light to reach Venus.

Step-by-step explanation:

Here we can solve this problem by using this following formula:

$s = \frac{d}{t}$

In which s is the speed(in km/s), d is the distance(in km) and t is the time(in s).

The light speed is 299 792 458 m / s = 299,792.458 km/s, so $s = 299,792.458$

(a) San Francisco (about 500km)

Find t when $d = 500$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{500}{t}$

$299,792.458t = 500$

$t = \frac{500}{299,792.458}$

$t = 0.0017s$

It takes 0.0017s for the light to reach San Francisco.

b) London(about 10,000km)

Find t when $d = 10,000$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{10,000}{t}$

$299,792.458t = 10,000$

$t = \frac{10,000}{299,792.458}$

$t = 0.033s$

It takes 0.033s for the light to reach London.

(c) the Moon (400,000km)

Find t when $d = 400,000$ . So

$s = \frac{d}{t}$

$299,792.458 = \frac{400,000}{t}$

$299,792.458t = 400,000$

$t = \frac{400,000}{299,792.458}$

$t = 1.334s$

It takes 1.334s for the light to reach Mars.

(d) Venus (0.3 A.U. from Earth at its closest approach).

Each A.U. has 149,59,7871 km.

So 0.3 A.U. = 0.3*(149,597,871) = 44,879,361.3. This means that $d = 44,879,361.3$ . So:

$s = \frac{d}{t}$

$299,792.458 = \frac{44,879,361.3}{t}$

$299,792.458t = 44,879,361.3$

$t = \frac{44,879,361.3}{299,792.458}$

$t = 149.7s$

It takes 149.7s for the light to reach Venus.

3 0

3 years ago

What is 1 4/5 + 1 1/5??

Rudik [331]

$1\dfrac{4}{5}+1\dfrac{1}{5}=(1+1)+\left(\dfrac{4}{5}+\dfrac{1}{5}\right)=2+\dfrac{1+4}{5}=2+\dfrac{5}{5}=2+1=3$

8 0

3 years ago

Solve for equation x. 3In(x)+2In(4)=In(128)

Serggg [28]

$Domain:\ x > 0\\\\3\ln(x)+2\ln(4)=\ln(128)\qquad\text{use}\ \lag_ab^n=n\log_ab\\\\\ln(x^3)+\ln(4^2)=\ln(128)\qquad\text{use}\ \log_ab+\ln_ac=\ln_a(bc)\\\\\ln(16x^3)=\ln(128)\iff16x^3=128\qquad\text{divide both sides by 16}\\\\x^3=8\to x=\sqrt[3]8\\\\\boxed{x=2}\in D$

3 0

3 years ago