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3 years ago

Seventh grade

Mathematics

1 answer:

victus00 [196]3 years ago

7 0

Answer:

Maybe it could be six or 30

Step-by-step explanation:

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Can someone help me with this math question.

Veronika [31]

Answer:

which math question

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Step-by-step explanation:

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4 years ago

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Find the slope of the line y = 5x + 4.

avanturin [10]

Answer:

x=1/5y+−4/5

Step-by-step explanation:

Let's solve for x.

y=5x+4

Step 1: Flip the equation.

5x+4=y

Step 2: Add -4 to both sides.

5x+4+−4=y+−4/5x=y−4

Step 3: Divide both sides by 5.

5x/5=y−4/5x=1'5y+−4'5

Answer:

x=1/5y+−4/5

( I hope this was helpful) >;D

6 0

3 years ago

The data from the U.S. Census Bureau for 1982−2003 shows that the surface area of the United States that is covered by rural lan

Wewaii [24]

If we plug in 0 for x, we can receive the data for the year 1982. Doing so shows us that the Rural area was 1,417.4 million acres and the total area was 1,839.4 million acres. Therefore, we can conclude that the majority of the area of the United States was rural in 1982.

5 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,

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

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3 years ago

2(2x + 7) = 2 + 2x<br>A. One Solution <br>B. No Solution <br>C. Infinite Solutions

Rudik [331]

B. I am so sorry if it is not right

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3 years ago

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