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2 years ago

Simplify 4 to the power of 2 and 4 to the power of 8

Mathematics

1 answer:

4vir4ik [10]2 years ago

8 0

Answer:

The second option.

Step-by-step explanation:

4^2 × 4^8

(4 × 4)(4 × 4 × 4 × 4 × 4 × 4 × 4 × 4) = 4^10

4^2 × 4^8

=4^2+8

4^10

When solve an exponential with the same constant, you ga no problem. Just take one of those and and the exponent, you are good to go.

Please mark as brainliest.

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Polly planted a magnolia in her front yard. When she planted the tree it was 8 feet tall. Every 2 year. She measured the height

kompoz [17]

Answer:

Y = 2.5X + 8

Y: height (in feet)

X: time (in years)

Step-by-step explanation:

Height = 8 + 5(t/2)

= 8 + 5t/2

Height = 8 + 2.5t

Y = 2.5X + 8

6 0

2 years ago

Calculate the area of a rectangle with an altitude of 30cm and a diagnol of 50cm

romanna [79]

Answer:

2000cm²

Step-by-step explanation:

find the other side using pythagorean theorem

50²=30²+x²

x=40

Area = 40 * 50 = 2000cm²

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2 years ago

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Makovka662 [10]

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2 years ago

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Alexandra [31]

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2 years ago

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How do I find the integral<br> ∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

defon

$\int\frac{10}{(x-1)(x^2+9)}\ dx=(*)\\\\\frac{10}{(x-1)(x^2+9)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+9}=\frac{A(x^2+9)+(Bx+C)(x-1)}{(x-1)(x^2+9)}\\\\=\frac{Ax^2+9A+Bx^2-Bx+Cx-C}{(x-1)(x^2+9)}=\frac{(A+B)x^2+(-B+C)x+(9A-C)}{(x-1)(x^2+9)}\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1$

$(*)=\int\left(\frac{1}{x-1}+\frac{-x-1}{x^2+9}\right)\ dx=\int\left(\frac{1}{x-1}-\frac{x+1}{x^2+9}\right)\ dx\\\\=\int\frac{1}{x-1}\ dx-\int\frac{x+1}{x^2+9}\ dx=\int\frac{1}{x-1}-\int\frac{x}{x^2+9}\ dx-\int\frac{1}{x^2+9}\ dx=(**)\\\\\#1\ \int\frac{1}{x-1}\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int\frac{1}{t}\ dt=lnt+C_1=ln(x-1)+C_1$

$\#2\ \int\frac{x}{x^2+9}\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=\frac{1}{2}\ du\end{array}\right|\Rightarrow\int\left(\frac{1}{2}\cdot\frac{1}{u}\right)\ du=\frac{1}{2}\int\frac{1}{u}\ du\\\\\\=\frac{1}{2}ln(u)+C_2=\frac{1}{2}ln(x^2+9)+C_2$

$\#3\ \int\frac{1}{x^2+9}\ dx=\int\frac{1}{x^2+3^2}\ dx=\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-\frac{1}{2}ln(x^2+9)+C_2-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C_3$

$\boxed{=ln(x-1)-\frac{1}{2}ln(x^2+9)-\frac{1}{3}tan^{-1}\left(\frac{x}{3}\right)+C}$

4 0

3 years ago