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GarryVolchara [31]
3 years ago
9

Factor 28+56t+28w28+56t+28w28, plus, 56, t, plus, 28, w to identify the equivalent expressions. Choose 2 answers:

Mathematics
1 answer:
Effectus [21]3 years ago
6 0

Answer: The factorization of  28+56t+28w =28(1+2t+w)

The factors of 28+56t+28w are 28 and 1+2t+w.

Step-by-step explanation:

The given expression : 28+56t+28w

To factorize it, we need to find the common factor.

As 56 can be written as 2 x 28.

So, the above expression would become

28+2\times28t+28w

Now, taking 28 as common from all the terms, we will get

28(1+2t+w)

Thus, the factorization of  28+56t+28w =28(1+2t+w)

And the factors of  28+56t+28w are 28 and 1+2t+w.

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Fred borrows $2710 at 3.8% simple interest per month. When Fred pays the loan back 3 months later, what is the total amount that
Fantom [35]

Step-by-step explanation:

Simple interest =P (1+. R )^n

/100

Where P is the amount

R is rate (%)

n is number of years

So if you are given months your divide ➗ the number of months by 12 years

2710 *(1+ [ 3.8/100])^3/12

5 0
3 years ago
How many ways can 6 people be chosen and arranged in a straight line if there are 8 people to choose from? 720 20,160 40,320 48?
Rudik [331]
The number of permutations of 8 people taken 6 at a time is given by:
8P6=\frac{8!}{(8-6)!}=20,160

5 0
3 years ago
Read 2 more answers
(4y − 3)(2y2 + 3y − 5)
VladimirAG [237]

Answer:

8y^3+6y^2-29y+15

Step-by-step explanation:

the correct expression is

(4y − 3)(2y^2 + 3y − 5)

Given data

We have the expression

(4y − 3)(2y^2 + 3y − 5)

let us open bracket

8y^3+12y^2-20y-6y^2-9y+15

Collect like terms

8y^3+12y^2-6y^2-20y-9y+15

8y^3+6y^2-29y+15

4 0
3 years ago
At 4:00 AM, the temperature is – 12 °F. If the temperature drops 3 °F per hour, what will the temperature be at 6:00 AM?
Scilla [17]
The answer would be -18°F at 6AM
Hope this helps ?(:
3 0
3 years ago
Read 2 more answers
99 POINT QUESTION, PLUS BRAINLIEST!!!
VladimirAG [237]
First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:


y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}

And the derivative of x:

x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}

Now, we can calculate the area of the surface:

A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\

=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\

=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}

Calculate indefinite integral:

\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}

And the area:

A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}

Answer D.
6 0
3 years ago
Read 2 more answers
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