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2 years ago

What are the length and width of the rectangle? (picture included)

Mathematics

1 answer:

Marianna [84]2 years ago

6 0

Answer:

a or c

Step-by-step explanation:

I don’t know ask for help sorry don’t blam this on me

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Jacob distributed a survey to his fellow students asking them how many hours they'd spent playing sports in the past day. He als

Salsk061 [2.6K]

Answer: y=1.5x+5 estimate:8.75

7 0

2 years ago

NEED HELLP ASAP !!!

ss7ja [257]

Answer:

m<x = 48

m<y = 132

Step-by-step explanation:

Since x and y are supplementary angles, that means their sum is 180 degrees.

And we are given that y is 12 less than 3 times the measure of x. With this information we can make a system of linear equations.

x + y = 180

y = 3x - 12

So now plug in the value of y into the first equation.

x + y = 180

x + (3x - 12) = 180

x + 3x = 192

4x = 192

x = 48

And then plug this computed value of x back into the first equation to find the measure of y.

x + y = 180

48 + y = 180

y = 180 - 48

y = 132

So the measures of the angles are as follows:

m<x = 48

m<y = 132

Cheers.

7 0

3 years ago

I only need to know is it 360 or 180?

Sveta_85 [38]

Answer:

So we know that all angles add up to 360. We first subtract it by 81

Now we have 279 as an answer. But since 2w + 81 cant be 360 because it doesn't add up because you have to spaces in between them. It adds up to 180 not 360

<h2><u>From your question its 180</u></h2>

3 0

2 years ago

Read 2 more answers

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

Which statement about this figure is true?

Arisa [49]

Answer: I'm thinking it's reflectional symmetry.

Step-by-step explanation:

3 0

3 years ago