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3 years ago

Help please i dont want to get 75%

Mathematics

2 answers:

tankabanditka [31]3 years ago

6 0

Stamps with animal 48/3=16
stamps with birds=16\4*3=12
the response is 12

KiRa [710]3 years ago

5 0

48/3 = 16

so 16 stamps have pictures of animals

3/4 out of 16 stamps have pictures of animals are pictures of birds
so
16 x 3/4 = 12.

anwer
12 stamps with pictures of birds on them

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the volume of a cylinder is given by the formula v=3.14r^2h, where r is the radius of the cylinder and h is the height. suppose

photoshop1234 [79]

we know that

The volume of a cylinder is given by the formula

$V=3.14 r^{2} h$

where

r is the radius of the cylinder

h is the height of the cylinder

in this problem

$r=x+8\\h=2x+3$

Substitute the values in the formula above

$V=3.14*(x+8)^{2}*(2x+3)$

$V=3.14*(x^{2}+16x+64)*(2x+3) \\ V=3.14 *(2x^{3} +3x^{2} +32x^{2} +48x+128x+192)\\ V=3.14*(2x^{3} +35x^{2} +176x+192)$

therefore

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5 0

3 years ago

7. The planet Abydos experiences drastic changes in temperature over the course of the day. On one day it reached a minimum temp

Delicious77 [7]

Answer:

The temprature change was +190 degrees farenheit

Step-by-step explanation:

subtract 290-100 using absolute value

6 0

3 years ago

-24x -15 = -27. Step by step explanation on how to solve this problem.

Jet001 [13]

Answer:

x = $\frac{1}{2}$

Step-by-step explanation:

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7 0

3 years ago

Read 2 more answers

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

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2 years ago

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alina1380 [7]

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3 years ago