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irina [24]
3 years ago
8

PRECal please help right answer only

Mathematics
1 answer:
Svetlanka [38]3 years ago
8 0

What does x² look like?   <em>goes to +∞ on the left and the right</em>

Well, x⁴ is an even exponent so it is has the same end behavior as x².

Answer: +∞


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3 years ago
Find the product.
4vir4ik [10]
Answer :

6{y}^{4} {z}^{2} + 12 {y}^{3} {z}^{2}-3 {y}^{3} {z}+3 {y}^{2} {z}^{2}

Step-by-step explanation :

To find the product of

3 {y}^{2} z(2 {y}^{2} z + 4yz - y + z)

First we expand the bracket ,

it implies that, we use the expression outside the bracket to multiply individual expressions inside the bracket.

Hence

3 {y}^{2} z(2 {y}^{2} z + 4yz - y + z)

= 3 {y}^{2} z(2 {y}^{2} z) + 3 {y}^{2} z(4yz) - 3 {y}^{2} z(y )+3 {y}^{2} z( z)

we now apply the law of indices

{a}^{m} \times {a}^{n} = {a}^{m+n}

meaning, when you are multiplying two expressions with the same bases , repeat one of the bases and add the exponents.

Then, simplify to obtain

= 6{y}^{4} {z}^{2} + 12 {y}^{3} {z}^{2}-3{y}^{3} {z}+3 {y}^{2} {z}^{2}
3 0
3 years ago
What Is The Prime Exponent Form Of 192​
Mandarinka [93]

Answer:

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Step-by-step explanation:

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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.

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Answer:678

Step-by-step explanation:

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3 years ago
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