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2 years ago

Identify the vertex and axis of symmetry. f(x) = 3(x+7)²-5

Mathematics

2 answers:

Verizon [17]2 years ago

5 0

Answer:

haiii i can only solve for the vertex...

Step-by-step explanation:

so to find the vertex you have to make it to

y - h = a(x - k)^2

i think

(you could also just make it y = a(x - k)^2 + h but i like the second one better)

soo

y + 5 = 3(x + 7)^2

the vertex is

(-7, -5)

and ummm i think the axis for the axis of symmetry .ss

x = 2a/b but im not sure

masha68 [24]2 years ago

5 0

(-7,-5) i am pretty sure if not i’m sorry

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The diameter of a semicircle is 2 feet. What is the semicircle's perimeter?

Mice21 [21]

Answer:

11/7 feet

Step-by-step explanation:

the perimeter of a circle is πr²

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so we derived at 11/7

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2 years ago

Does 514, 684, 855 make a right triangle

Naily [24]

Answer:

No, it does not.

Step-by-step explanation:

This does not make a right triangle because in order to have a right triangle you must have 2 of the same numbers.

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900, 800, 900 makes a right triangle

900, 800, 700 does not.

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2 years ago

What is the variance? What is the standard deviation, rounded to the nearest whole number?

Dmitry [639]

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3 years ago

Nine hundred people who attended a movie were asked whether they enjoyed it. The table shows the results.

andrew11 [14]

Answer:

184

Step-by-step explanation:

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6 0

3 years ago

Read 2 more answers

Please prove this........

Crazy boy [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B + C = π → C = π - (A + B)

→ sin C = sin(π - (A + B)) cos C = sin(π - (A + B))

→ sin C = sin (A + B) cos C = - cos(A + B)

Use the following Sum to Product Identity:

sin A + sin B = 2 cos[(A + B)/2] · sin [(A - B)/2]

cos A + cos B = 2 cos[(A + B)/2] · cos [(A - B)/2]

Use the following Double Angle Identity:

sin 2A = 2 sin A · cos A

Proof LHS → RHS

LHS: (sin 2A + sin 2B) + sin 2C

$\text{Sum to Product:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-\sin 2C$

$\text{Double Angle:}\qquad 2\sin\bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A - 2B}{2}\bigg)-2\sin C\cdot \cos C$

$\text{Simplify:}\qquad \qquad 2\sin (A + B)\cdot \cos (A - B)-2\sin C\cdot \cos C$

$\text{Given:}\qquad \qquad \quad 2\sin C\cdot \cos (A - B)+2\sin C\cdot \cos (A+B)$

$\text{Factor:}\qquad \qquad \qquad 2\sin C\cdot [\cos (A-B)+\cos (A+B)]$

$\text{Sum to Product:}\qquad 2\sin C\cdot 2\cos A\cdot \cos B$

$\text{Simplify:}\qquad \qquad 4\cos A\cdot \cos B \cdot \sin C$

LHS = RHS: 4 cos A · cos B · sin C = 4 cos A · cos B · sin C $\checkmark$

7 0

3 years ago