All categories

2 years ago

The graph of fx) = 7^X is stretched vertically by a factor of five.

Mathematics

1 answer:

Strike441 [17]2 years ago

4 0

Answer:

g(x) = 5*7ˣ

Step-by-step explanation:

Given function:

f(x) = 7ˣ

The vertical stretch is the function with:

g(x) = a*f(x), where a is the stretch factor

With a = 5 the function becomes:

g(x) = 5 f(x) = 5*7ˣ

You might be interested in

Use the number line to find (One-fourth) + (negative StartFraction 21 over 8 EndFraction) Negative StartFraction 19 over 4 EndFr

Shalnov [3]

Answer:

- $\frac{19}{8}$

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

When a figure is translated, its orientation remains the same or changes ?

Leviafan [203]

No it does not change it is a imaginary like used to move a figure. Nor does it move when it is rotated

6 0

3 years ago

What value of n makes the equation true?

otez555 [7]

2/3 n = -12 Write equation

(3/2) 2/3n = -12/1 (3/2) Isolate n

n = -36/2 Simplify

n = -18 Answer

4 0

3 years ago

Read 2 more answers

ABCD∼EFGH AD=45 in. , EH=75 in. , and AB=30 in. What is EF ?

Marrrta [24]

EF = 50 in.

In similar figures, corresponding sides are proportional. We use the similarity statement ABCD~EFGH to write a proportion:

$\frac{AB}{EF}=\frac{AD}{EH}
\\
\\ \frac{30}{x}=\frac{45}{75}$

Cross multiply the proportion:
45*x = 75*30
45x = 2250

Divide both sides by 45:
45x/45 = 2250/45
x = 50

3 0

3 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

3 years ago