All categories

2 years ago

Helpppppppppp.......

Mathematics

2 answers:

ruslelena [56]2 years ago

5 0

Answer:

36 centimeters in the poster

Step-by-step explanation:

Let us plan out our steps, and solve for each:

1. Given the information, let us create a proportionality as such:

3 = 1 ⇒ x - centimeters in the poster

x 12

2. Now let us cross multiply, and solve through simple algebra for x:

3 * 12 = x,

x = 36 centimeters in the poster

Andrew [12]2 years ago

4 0

Answer:

36cm

Step-by-step explanation:

The scale is 3 to 1, so the poster size is 3 times larger than the actual thing.

You might be interested in

Explain how to use the combine place values strategy to find 223-119

Vikentia [17]

So, the subtraction looks like this:

   223
-119
_______

The reason why combining place values here is necessary is that 3 is less than 9 (so I can't just substract in the one's place without going to negative numbers)

so we combine the ones and tens places and substract the whole: 23-19 is four!
   223
-119
_______
   04

and then we continue
   223
-119
_______
    204

and that's the result!

3 0

2 years ago

Read 2 more answers

Please help I only have 2 more trys.

andrezito [222]

Answer:

Question 1: -2xy + 7x

Question 2: 8t - 4v and -4v + 8t

Question 3: -e + 10k and 10k -e

Question 4: x = -3

8 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

2 years ago

What is the measure of arc AC?

8_murik_8 [283]

If Im not mistaken, the answer should be d because if you add all the degrees up, then you would get 222. And if you subtract 360 from 222 you get 138. I hope this helps. :)

7 0

3 years ago

Write three ratios that are equal to 60/180

Fudgin [204]

1/3 or 3/9 or 9/27
I hope this helps

4 0

2 years ago