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3 years ago

9

34 students in Mr. Jones math class passed the final exam. There are 40 students in his class total. What percent of students pa

ssed.
help please

1 answer:

kati45 [8]3 years ago

8 0

Answer 6 you do 40-34=6

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Solve For X. Please add an explanation.

Vikki [24]

Answer:

Your answer is 2.5

Hope that this is helpful. Vote and like it

5 0

2 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

2 years ago

What are the intercepts of this line?

Fiesta28 [93]

Answer:

x-intercept: -1

y-intercept: 0.5

Step-by-step explanation:

The intercept for each is where the other coordinate is at 0.

7 0

1 year ago

Find the equation of all tangent lines having slope of -1 that are tangent to the curve y=(9)/(x+1)

fredd [130]

Answer:

$f(x)=\frac9{x+1}\\ f'(x)=-\frac9{(x+1)^2}\\ f'(x)=-1\ \iff\ -\frac9{(x+1)^2}=-1\ \to \ \frac9{(x+1)^2}=1\ \to \ (x+1)^2=9\\ |x+1|=3\ \to \ x+1=3\ \vee\ x+1=-3\\ x_1=2\ \vee\ x_2=-4\\ f(x_1)=f(2)=\frac9{2+1}=3\\ f(x_2)=f(-4)=\frac9{-4+1}=-3$

First tangent line:

$y=f'(x_1)\cdot (x-x_1)+f(x_1)\ \to \ y=-1(x-2)+3\ \to \ y=-x+5$

Second tangent line:

$y=f'(x_2)\cdot (x-x_2)+f(x_2)\ \to \ y=-1(x+4)-3\ \to \ y=-x-7$

Notice: slope of -1 means that both $f'(x_1), \ f'(x_2)$ are equal to -1, so $f'(x_1)=-1 \ and \ f'(x_2)=-1$

6 0

2 years ago

Read 2 more answers

The value of 82 is between which two integers?

hoa [83]

Hey there!

Let's look at the squares of all of our answer options. We will compare them to eighty two to see which it belongs in.

A. 36 and 49.

B. 49 and 64.

C. 64 and 81.

D. 81 and 100

As you can see, 82 is in between 81 and 100, so the answer is D. 9 and 10.

Also, the square root of 82 is about 9.05, and this fits our answer.

Have a wonderful day!

8 0

3 years ago