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3 years ago

7

Use the method of symmetry to find the extreme value of each quadratic function and the value of x for which it occurs h(x)=1/2(

3-x)(2-x)

1 answer:

sertanlavr [38]3 years ago

7 0

Answer:

x=2, and x=3

Step-by-step explanation:

0=1/2*(3-x)(2-x)

x=2, and x=3

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yanalaym [24]

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3 years ago

Demarcus has to wrap a gift for his friend's birthday party. the gift is in a rectangular box with the dimensions shown below. h

Leya [2.2K]

The gift wrap needed by Damarcus = total surface area of a rectangular box = 1,048 in.².

<h3>What is the Total Surface Area of a Rectangular Box?</h3>

Total surface area (TSA) = 2(wl+hl+hw), where:

l = length
w = width
h = height of the box.

The amount of gift wrap needed to cover the whole box = total surface area of the rectangular box

l = 20 in.

w = 8 in.

h = 13 in.

Plug in the values into the formula for total surface area of a rectangular box:

TSA = 2(8×20 + 13×20 + 13×8)

TSA = 1,048 in.²

Therefore, the gift wrap needed by Damarcus = total surface area of a rectangular box = 1,048 in.².

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2 years ago

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Gemiola [76]

Step-by-step explanation:

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width = (x - 7) units

Area = 2( length × width)

Area = 2( (2x + 3) × (x - 7) )

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4 years ago

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Please help I'm Timed Will Name Brainliest if Correct.

Reptile [31]

Answer:

A

Step-by-step explanation:

We can see that Function A's y coordinate doubles every time. The function A = f(x) = 5(2)^x. It is an exponential growth function, and therefore y can never be 0. This means that A does not have an x-intercept.

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3 years ago

Trig proofs with Pythagorean Identities.

lorasvet [3.4K]

To prove:

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Solution:

$$LHS = \frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}$

Multiply first term by $\frac{1+cos x}{1+cos x}$ and second term by $\frac{1-cos x}{1-cos x}$ .

$$= \frac{1(1+\cos x)}{(1-\cos x)(1+\cos x)}-\frac{\cos x(1-\cos x)}{(1+\cos x)(1-\cos x)}$

Using the identity: $(a-b)(a+b)=(a^2-b^2)$

$$= \frac{1+\cos x}{(1^2-\cos^2 x)}-\frac{\cos x-\cos^2 x}{(1^2-\cos^2 x)}$

Denominators are same, you can subtract the fractions.

$$= \frac{1+\cos x-\cos x+\cos^2 x}{(1^2-\cos^2 x)}$

Using the identity: $1-\cos ^{2}(x)=\sin ^{2}(x)$

$$= \frac{1+\cos^2 x}{\sin^2x}$

Using the identity: $1=\cos ^{2}(x)+\sin ^{2}(x)$

$$=\frac{\cos ^{2}x+\cos ^{2}x+\sin ^{2}x}{\sin ^{2}x}$

$$=\frac{\sin ^{2}x+2 \cos ^{2}x}{\sin ^{2}x}$ ------------ (1)

$RHS=2 \cot ^{2} x+1$

Using the identity: $\cot (x)=\frac{\cos (x)}{\sin (x)}$

$$=1+2\left(\frac{\cos x}{\sin x}\right)^{2}$

$$=1+2\frac{\cos^{2} x}{\sin^{2} x}$

$$=\frac{\sin^2 x + 2\cos^{2} x}{\sin^2 x}$ ------------ (2)

Equation (1) = Equation (2)

LHS = RHS

$$\frac{1}{1-\cos x}-\frac{\cos x}{1+\cos x}=2 \cot ^{2} x+1$

Hence proved.

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3 years ago