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2 years ago

10

How would you explain to someone the rules of a horizontal translation where we are subtracting from the inputs(within the paren

thesis) before evaluating the function

1 answer:

chubhunter [2.5K]2 years ago

5 0

Answer:

Ok, we know that we can write a horizontal translation as:

y' = f(x - A)

where if A is positive, this moves the graph of f(x) A units to the right.

Why is this?

Ok, let's compare:

y = f(x)

and

y' = f(x - A)

in y, when x = 0 we have f(0).

While to have this same point in y', we need to evaluate in x = A.

f(A - A) = f(0).

Then the value f(0) is now at x = A, this means that the point moved A units to the right.

And you can do this for all the values, so you will find that the entire graph of f(x) has ben moved A units to the right.

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The school bus Evie rides is scheduled to arrive at her stop at 8:20 a.m. each day. The table below shows the actual arrival tim

saveliy_v [14]

Im not sure what the answer is but dont open the link its a virus

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2 years ago

Read 2 more answers

X/2-1/5=x/3+1/4<br>solve this equation

Alekssandra [29.7K]

Answer:

$x=\frac{27}{10}$

Step-by-step explanation:

$\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}$

$\frac{x}{2}-\frac{1}{5}-\frac{x}{3}=\frac{x}{3}+\frac{1}{4}-\frac{x}{3}$

$-\frac{1}{5}+\frac{x}{6}=\frac{1}{4}$

$-\frac{1}{5}+\frac{x}{6}+\frac{1}{5}=\frac{1}{4}+\frac{1}{5}$

$\frac{x}{6}=\frac{9}{20}$

$\frac{6x}{6}=\frac{9\times \:6}{20}$

$x=\frac{27}{10}$

3 0

3 years ago

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Louisa ran at an average speed of five miles per hour along an entire circular park path. Calvin ran along the same path in the

docker41 [41]

Answer:

15 miles

Step-by-step explanation:

Let $x$ be the miles in the circular park path, $t_{L}$ the time Louisa takes to finish and $t_{C}$ the time Calvin takes to finish both in hours.

Then $x$ , the longitude is equal to the velocity times the time used to finish. So

$x=5t_{L}$

$x=6t_{C}$

And the difference between Louisa's time and Calvin' time is 30 minutes, half an hour. So:

$t_{C}=t_{L}-0.5$

Three equations, three unknowns, the system can be solved.

Equalizing the equation with x :

$5t_{L}=6t_{C}$

In this last equation replace $t_{C}$ with the other equation and solve:

$5t_{L}=6(t_{L}-0.5)\\ 5t_{L}=6t_{L}-3\\ 3=6t_{L}-5t_{L}\\ 3=t_{L}\\ t_{L}=3$

With Louisa's time find x:

$x=5t_{L}\\ x=5(3)\\ x=15$

7 0

2 years ago

In the given figure △ABC ≅△DEC. Which of the following relations can be proven using CPCTC ?

Serhud [2]

Option B:

$\overline{A B}=\overline{D E}$

Solution:

In the given figure $\triangle A B C \cong \triangle D E C$ .

If two triangles are similar, then their corresponding sides and angles are equal.

By CPCTC, in $\triangle A B C \ \text{and}\ \triangle D E C$ ,

$\overline{AB }=\overline{DE}$ – – – – (1)

$\overline{B C}=\overline{EC}$ – – – – (2)

$\overline{ CA}=\overline{CD}$ – – – – (3)

$\angle ACB=\angle DCE$ – – – – (4)

$\angle ABC=\angle DEC$ – – – – (5)

$\angle BAC=\angle EDC$ – – – – (6)

Option A: $\overline{B C}=\overline{D C}$

By CPCTC proved in equation (2) $\overline{B C}=\overline{EC}$ .

Therefore $\overline{B C}\neq \overline{D C}$ . Option A is false.

Option B: $\overline{A B}=\overline{D E}$

By CPCTC proved in equation (1) $\overline{AB }=\overline{DE}$ .

Therefore Option B is true.

Option C: $\angle A C B=\angle D E C$

By CPCTC proved in equation (4) $\angle ACB=\angle DCE$ .

Therefore $\angle A C B\neq \angle D E C$ . Option C is false.

Option D: $\angle A B C=\angle E D C$

By CPCTC proved in equation (5) $\angle ABC=\angle DEC$ .

Therefore $\angle A B C\neq \angle E D C$ . Option D is false.

Hence Option B is the correct answer.

$\Rightarrow\overline{A B}=\overline{D E}$

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2 years ago

What is the equation of the line perpendicular to the line y=-x+2 through the point (3,5)?

mina [271]

A. Y=x+2

I thinks that’s right hope this helps :)

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2 years ago