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3 years ago

Which point in on the graph of

Mathematics

1 answer:

maria [59]3 years ago

8 0

Answer:

(-2, 9)

Because it is (-2,9)

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Sarah tried to evaluate 21 / 7 * 8 step by step.

vladimir2022 [97]

Answer:

24.

Step-by-step explanation:

First, use the Order of Operations:

PEMDAS.

First, divide, then multiply to get your solution.

21 / 7 = 3.

3 · 8 = 24.

Therefore, your answer is 24.

4 0

4 years ago

While on vacation in Mexico Jeremiah reads a distance marker that indicates he is 89 kilometers from Juarez. If 1 mile is approx

Sedbober [7]

Answer:

Distance between Jeremiah and Juarez is 55.3 miles

Step-by-step explanation:

We are given

1 mile is approximately 1.61 kilometers

so,

1.61 km =1 mile

we can divide both sides by 1.61

$1km=\frac{1}{1.61}mile$

now, to find for 89 km

we can multiply both sides by 89

$89\times 1km=89\times\frac{1}{1.61}mile$

now, we can simplify it

and we get

89 km =55.2795 mile

so, nearest tenth is

89 km =55.3 mile

So,

Distance between Jeremiah and Juarez is 55.3 miles

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

The line segment AB with endpoints A (-3, 6) and B (9, 12) is dilated with a scale

Gwar [14]

Answer:

C) (-2, 4), (6,8) is the correct answer.

Step-by-step explanation:

Given that line segment AB:

A (-3, 6) and B (9, 12) is dilated with a scale factor 2/3 about the origin.

First of all, let us calculate the distance AB using the distance formula:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here,

$x_2=9\\x_1=-3\\y_2=12\\y_1=6$

Putting all the values and finding AB:

$AB = \sqrt{(9-(-3))^2+(12-6)^2}\\\Rightarrow AB = \sqrt{(12)^2+(6)^2}\\\Rightarrow AB = \sqrt{144+36}\\\Rightarrow AB = \sqrt{180}\\\Rightarrow AB = 6\sqrt{5}\ units$

It is given that AB is dilated with a scale factor of $\frac{2}{3}$ .

$x_2'=\dfrac{2}{3}\times x_2=\dfrac{2}{3}\times9=6\\x_1'=\dfrac{2}{3}\times x_1=\dfrac{2}{3}\times-3=-2\\y_2'=\dfrac{2}{3}\times y_2=\dfrac{2}{3}\times 12=8\\y_1'=\dfrac{2}{3}\times y_1=\dfrac{2}{3}\times 6=4$

So, the new coordinates are A'(-2,4) and B'(6,8).

Verifying this by calculating the distance A'B':

$A'B' = \sqrt{(6-(-2))^2+(8-4)^2}\\\Rightarrow A'B' = \sqrt{(8)^2+(4)^2}\\\Rightarrow A'B' = \sqrt{64+16}\\\Rightarrow A'B' = \sqrt{80}\\\Rightarrow A'B' = 4\sqrt{5}\ units = \dfrac{2}{3}\times AB$

So, option C) (-2, 4), (6,8) is the correct answer.

5 0

3 years ago

A middle school is having a fundraiser. Edwin has $25.00 to pay for his purchases

antiseptic1488 [7]

Answer:

He needs $1.64 he has to pay $26.64

Step-by-step explanation:

$25 is what Edwin has

5.95 x 2 = $11.19

$2.50 x 3 = $7.50

11.19 + 7.50 + 7.95 = $26.64

6 0

3 years ago

Read 2 more answers