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3 years ago

The function h(x) is quadratic and h(3) = h(-10) = 0. Which could represent h()?

Mathematics

2 answers:

lilavasa [31]3 years ago

6 0

Answer:

h(x) = 2x^2+14x - 60

Step-by-step explanation:

user100 [1]3 years ago

4 0

Answer:

the answer is the last option

Step-by-step explanation:

took the test

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X/6 + 12 = 6 Please help me

otez555 [7]

Answer:

$x=-36$

Step-by-step explanation:

$\frac{x}{6} +12=6$

$\frac{-36}{6} +12=6$

$-6+12=6$

Hence, x = -36

8 0

3 years ago

Read 2 more answers

Aaron is planning to attend a public university after he graduates from high school in 4 years. He is devising a plan to save mo

rjkz [21]

Answer:

b, 418.57

Step-by-step explanation:

subtract the total his family is contribution from tuition then divide by total months remaining

please rank brainliest

6 0

3 years ago

"What is the value of-3|15-s|+2^3 when s-=3"?

Serggg [28]

You probably made a typing error in writing s = -3 and wrote s - = 3 instead.

The given expression is:

$-3|15-s|+ 2^{3} \\ \\ 
-3|15-s|+8
$

Substituting the given value of s in the previous equation, we get:

$-3|15-(-3)|+8 \\ \\ 
=-3|15+3|+8 \\ \\ 
=-3|18|+8 \\ \\ 
=-54+8 = -46$

Thus the value of given expression becomes -46 for s = -3

3 0

3 years ago

Please help me to prove this!

Sophie [7]

Answer: see proof below

Step-by-step explanation:

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

Proof LHS → RHS:

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$