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Elina [12.6K]
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)
Mathematics
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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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.

Learn more about rectangular box on:

brainly.com/question/13103197

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2 years ago
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Gemiola [76]

Step-by-step explanation:

length = (2x + 3) units

width = (x - 7) units

Area = 2( length × width)

Area = 2( (2x + 3) × (x - 7) )

Area = 2( 2x² - 14x + 3x - 21)

Area = 2 ( 2x² -11x - 21)

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Answer:

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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.

Function B is a rational function. x cannot be 0, since that would result in an undefined number. This also means that B does not have an x-intercept.

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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.

5 0
3 years ago
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