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Llana [10]
1 year ago
10

2x^{2} + 20x + 48Completely factor the given polynomial, if possible. If the polynomial cannot be factored write "not factorable

".
Mathematics
1 answer:
astra-53 [7]1 year ago
4 0

Given:

2x^2+20x+48

To find the factors:

The given equation can be written as,

\begin{gathered} 2x^2+20x+48=2x^2+8x+12x+48 \\ =2x(x+4)+12(x+4)_{} \\ =(x+4)(2x+12)_{} \\ =2(x+4)(x+6) \end{gathered}

Hence, the factors are 2,(x+4), and (x+6).

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What is the slope for 10,15 and 15,20
makvit [3.9K]
So it will be y1-y2, so 15-20 is -5

Then x1-x2, so 10-15 is -5.

-5/-5 = 1

There’s ya slope : 1
6 0
3 years ago
Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab
Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx

If f(x)=x^2, then

f'(x)=2x

and

b=0\\ \\a=2

Therefore,

SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx

Apply substitution

x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du

Then

SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du

Now

\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))

Hence,

SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}

3 0
3 years ago
What is 1231221 ÷ 272727÷0
xz_007 [3.2K]
??? what type of question is that??
7 0
2 years ago
Someone please help me with this question please
Eddi Din [679]

Answer: False

Step-by-step explanation:

6 0
3 years ago
The total length of pencils A, B and C is 29 cm. Pencil a is 11 cm shorter then pencil B, and pencil B is twice as long a pencil
satela [25.4K]

The length of pencil A is 5 cm

<em><u>Solution:</u></em>

Let the length of pencil A be "x"

Let the length of pencil B be "y"

Let the length of pencil C be "z"

<em><u>The total length of pencils A, B and C is 29 cm</u></em>

Therefore,

length of pencil A + length of pencil B + length of pencil C = 29

x + y + z = 29 ------------ eqn 1

<em><u>Pencil A is 11 cm shorter then pencil B</u></em>

x = y - 11 ------- eqn 2

<em><u>Pencil B is twice as long a pencil C</u></em>

y = 2z

z = \frac{y}{2} ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

y - 11 + y + \frac{y}{2} = 29\\\\2y + \frac{y}{2} = 29 + 11\\\\\frac{4y+y}{2} = 40\\\\5y = 80\\\\y = 16

<em><u>Substitute y = 16 in eqn 2</u></em>

x = 16 - 11

x = 5

Thus length of pencil A is 5 cm

5 0
3 years ago
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