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3 years ago

I need help with #3-7 algebra is hard.

Mathematics

1 answer:

Hunter-Best [27]3 years ago

7 0

I believe this is the answer to #3.
A. The value of x is five
B. Perimeter of square = 68
Perimeter of triangle = 18

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Strike441 [17]

Answer:

$\sqrt{2} - \sqrt{2} = 0$

0 is real number, rational number, integers, whole numbers,

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3 years ago

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olga nikolaevna [1]

Answer:

Step-by-step explanation:

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3 years ago

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Help me with my homework please ;-;

cluponka [151]

Answer:

For problem 31: 25 counters in the 5th figure, 81 counters in the 9th figure, and 121 counters in the 11th figure.

Step-by-step explanation:

Since in the examples it shows 1 exponent 2, 2 exponent 2, and 3 exponent 2, that means all the rest will have 2 as an exponent. So the fifth figure would be 5 exponent 2, which is 5 times 5 = 25. The 9th figure would be 9 exponent 2, which is 9 times 9 = 81. The 11th figure would be 11 exponent 2, which is 11 times 11 = 121.

Is that all, or do you need answers for #'s 32, 33, and 34?

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3 years ago

The selling price of an item is $ 455 After 6 months of not selling, it is marked down by 20%. After another 6 months of not

olga_2 [115]

It would be 443 bc you would subtract by 6 2x

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4 years ago

What is the antiderivative of sin^2(x)cos^2(x)?

marin [14]

Answer:

$\frac{x}{8}-\frac{\sin(4x)}{32}+C$

Step-by-step explanation:

[Most of the work here comes from manipulating the trig to make the term (integrand) integrable.]

Recall that we can express the squared trig functions in terms of cos(2x). That is,

$\cos(2x)=2\cos^2x-1 \\ \cos(2x)=1 - 2\sin^2x.$

And so inverting these,

$\cos^2x=\frac{1}{2} (1+\cos2x) \\ \sin^2x=\frac{1}{2} (1-\cos2x)$ .

Multiply them together to obtain an equivalent expression for sin^2(x)cos^2(x) in terms of cos(2x).

$\sin^2x \cdot \cos^2x =\frac{1}{2} (1-\cos2x) \cdot \frac{1}{2} (1+\cos2x) = \frac{1}{4}(1-\cos^2(2x)).$

Notice we have cos^2(2x) in the integrand now. We've made it worse! Let's try plugging back in to the first identity for cos^2(2x).

$\cos(2x)=2\cos^2x-1 \Rightarrow \cos(4x)=2\cos^2(2x)-1 \Rightarrow \cos^2(2x) = \frac{1}{2}(1+\cos(4x))$

So then,

$\sin^2x \cdot \cos^2x = \frac{1}{4}(1-\cos^2(2x)) = \frac{1}{4}(1-\frac{1}{2}(1+\cos(4x))) = \frac{1}{4}(1-\frac{1}{2}-\frac{1}{2}\cos(4x))=\frac{1}{8}(1-\cos(4x)).$

This is now integrable (phew),

$\int \sin^2x\cos^2x \ dx = \int \frac{1}{8}(1-\cos(4x)) \ dx = \frac{1}{8} \int (1-\cos(4x)) \ dx = \frac{1}{8}(x-\frac{1}{4}\sin(4x))+C.$

7 0

3 years ago