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Vikki [24]
4 years ago
11

QUESTION 9

Mathematics
1 answer:
Ratling [72]4 years ago
8 0
The first one is (v+6)(v-6)
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What is the minimum value for h(x)=x2−16x+60?
Mila [183]
Minimum value is equal to x=8, y=-4First find the derivative of the original equation which equals= d/dx(x^2-16x+60) = 2x - 16at x=8, f'(x), the derivative of x equals zero, so therefore, at point x = 8, we have a minimum value.Just plug in 8 to the original equation to find the answer for the minimum value. 

7 0
4 years ago
Giving brainlist to whoever answers
stealth61 [152]

Answer:

62.4 cm³

Step-by-step explanation:

Hope this helps :)

7 0
3 years ago
Fine the product mentally. [(x+1) (x-1)^2
Murljashka [212]
If the product is: [(x+1) (x-1)]^2 , the answer is:
[(x+1) (x-1)]^2 = [x^2-1]^2 = x^4-2x^2+1
5 0
3 years ago
Read 2 more answers
In which quadrant would point (-4.5, -4.5) be located
Leviafan [203]
It’s the 3rd one.. . i’m
7 0
3 years ago
NEED ANSWERED ASAP
fomenos
The answer is 10 2/3, 5

Proof:

Solve the following system:
{x/2 + y/3 = 7 | (equation 1)
{x/4 + (2 y)/3 = 6 | (equation 2)

Subtract 1/2 × (equation 1) from equation 2:
{x/2 + y/3 = 7 | (equation 1)
{0 x+y/2 = 5/2 | (equation 2)

Multiply equation 1 by 6:
{3 x + 2 y = 42 | (equation 1)
{0 x+y/2 = 5/2 | (equation 2)

Multiply equation 2 by 2:
{3 x + 2 y = 42 | (equation 1)
{0 x+y = 5 | (equation 2)

Subtract 2 × (equation 2) from equation 1:
{3 x+0 y = 32 | (equation 1)
{0 x+y = 5 | (equation 2)

Divide equation 1 by 3:
{x+0 y = 32/3 | (equation 1)
{0 x+y = 5 | (equation 2)
Collect results:
Answer: {x = 10 2/3, y = 5
8 0
4 years ago
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