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Elanso [62]
3 years ago
11

What numbers go into 12 and 32 that can multiply and add?

Mathematics
1 answer:
Ilya [14]3 years ago
5 0
I think the answer is 4. Hope this helps.
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The formula for the area of a trapezoid is a=1/2h(b1+b2) are the are the length of the bases, and is the height, to answer the q
tatuchka [14]
A=1/2*75*(125+81)
a=1/2*75*206
a=77.25 sq ft
8 0
4 years ago
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Determine the period of the function g(x) = (cos ;)
Romashka [77]

Answer:

6\pi

Step-by-step explanation:

The period of a sinusoidal function in the form a * cos (b (x + h)) + m is \frac{2\pi }{|b|}. Therefore, we can see that "b" in g(x) is 1/3 and 1/3 substituted into \frac{2\pi }{|b|} is 2\pi/1/3 which is 6\pi.

5 0
3 years ago
Please help me with #4 <br> (10 points )
myrzilka [38]

Answer:

40 feet

Step-by-step explanation:

We are given a right isosceles triangle having lengths of two sides 12 feet and 16 feet.

<em>Since, the triangle is an isosceles triangle i.e. two sides of the triangle are equal.</em>

That is, the three sides of the triangle are 12 feet, 12 feet and 16 feet.

We know that, Perimeter of a triangle = Sum of the sides

Thus, Perimeter of the given triangle = 12 + 12 + 16 = 24 + 16 = 40 feet.

Hence, the total length of the fencing needed is 40 feet.

8 0
4 years ago
Find all numbers p such that p and p^2 are both prime. prove that you found them all
Anuta_ua [19.1K]

0 and 1 are neither prime nor composite. A prime is any number greater than 1 that has just 1 and itself as factors. Primes can only start at x > 1


When that happens (when you start with numbers greater than one) p^2 is a composite consisting of 2 primes, so any composite will obey the law that he number will have at least 3 factors making it up -- in this case p p^2 and 1.


So the answer to the question by definition is that 0 numbers can have the property of both p and p^2 to be prime.

5 0
4 years ago
Read 2 more answers
Who can help me d e f thanks​
12345 [234]

d)

y = (2ax^2 + c)^2 (bx^2 - cx)^{-1}

Product rule:

y' = \bigg((2ax^2+c)^2\bigg)' (bx^2-cx)^{-1} + (2ax^2+c)^2 \bigg((bx^2-cx)^{-1}\bigg)'

Chain and power rules:

y' = 2(2ax^2+c)\bigg(2ax^2+c\bigg)' (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} \bigg(bx^2-cx\bigg)'

Power rule:

y' = 2(2ax^2+c)(4ax) (bx^2-cx)^{-1} - (2ax^2+c)^2 (bx^2-cx)^{-2} (2bx - c)

Now simplify.

y' = \dfrac{8ax (2ax^2+c)}{bx^2 - cx} - \dfrac{(2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

y' = \dfrac{8ax (2ax^2+c) (bx^2 - cx) - (2ax^2+c)^2 (2bx-c)}{(bx^2-cx)^2}

e)

y = \dfrac{3bx + ac}{\sqrt{ax}}

Quotient rule:

y' = \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{\left(\sqrt{ax}\right)^2}

y'= \dfrac{\bigg(3bx+ac\bigg)' \sqrt{ax} - (3bx+ac) \bigg(\sqrt{ax}\bigg)'}{ax}

Power rule:

y' = \dfrac{3b \sqrt{ax} - (3bx+ac) \left(-\frac12 \sqrt a \, x^{-1/2}\right)}{ax}

Now simplify.

y' = \dfrac{3b \sqrt a \, x^{1/2} + \frac{\sqrt a}2 (3bx+ac) x^{-1/2}}{ax}

y' = \dfrac{6bx + 3bx+ac}{2\sqrt a\, x^{3/2}}

y' = \dfrac{9bx+ac}{2\sqrt a\, x^{3/2}}

f)

y = \sin^2(ax+b)

Chain rule:

y' = 2 \sin(ax+b) \bigg(\sin(ax+b)\bigg)'

y' = 2 \sin(ax+b) \cos(ax+b) \bigg(ax+b\bigg)'

y' = 2a \sin(ax+b) \cos(ax+b)

We can further simplify this to

y' = a \sin(2(ax+b))

using the double angle identity for sine.

7 0
2 years ago
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