Please I need help!!!!!!

Mathematics

1 answer:

8090 [49]3 years ago

8 0

Answer:

What? I'm in middle school ;-;

Step-by-step explanation:

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What is the volume of the cylinder below radius 7 height 4

Veseljchak [2.6K]

V = pi * r^2 *h

where r is the radius and h is the height of the cylinder

V = pi * 7^2 *4

V = pi* 49*4

pi is approximated by 3.14 or 22/7

V = 22/7 * 49 * 4

V = 616 units ^3

6 0

4 years ago

Read 2 more answers

Help me with 16,17,18,and 19 please simpl

mixas84 [53]

Answer:

Ques 16)

We have to simplify the expression:

$\dfrac{t^2}{t^2+3t-18}-(\dfrac{5t}{t^2+3t-18}-\dfrac{t-3}{t^2+3t-18})\\ \\=\dfrac{t^2}{t^2+3t-18}-(\dfrac{4t+3}{t^2+3t-18})\\ \\=\dfrac{t^2-4t-3}{t^2+3t+18}$

Ques 17)

$\dfrac{3w^2+7w-7}{w^2+8w+15}+\dfrac{2w^2-9w+4}{(2w^2+9w-5)(w^2-w-12)}\\ \\=\dfrac{3w^2+7w-7}{w^2+8w+15}+\dfrac{(2w-1)(w-4)}{(2w-1)(w+5)(w+3)(w-4)}\\\\=\dfrac{3w^2+7w-7}{(w+3)(w+5)}+\dfrac{1}{(w+3)(w+5)}\\\\=\dfrac{3w^2+7w-7+1}{(w+3)(w+5)}\\\\=\dfrac{(3w-2)(w+3)}{(w+5)(w+3)}\\\\=\dfrac{3w-2}{w+5}$

Ques 18)

Let the blank space be denoted by the quantity 'x'.

$\dfrac{x}{12a^2+8a}+\dfrac{15a^2}{12a^2+8a}=\dfrac{7a}{3a+2}\\ \\\dfrac{x+15a^2}{12a^2+8a}=\dfrac{7a}{3a+2}\\\\=\dfrac{x+15a^2}{4a(3a+2)}=\dfrac{7a}{3a+2}\\\\=\dfrac{x+15a^2}{4a}=7a\\\\x+15a^2=28a^2\\\\x=28a^2-15a^2\\\\x=13a^2$

Ques 19)

Let the missing quantity be denoted by 'x'.

$\dfrac{p^2+7p+2}{p^2+5p-14}-\dfrac{x}{p^2+5p-14}=\dfrac{p-1}{p-2}\\ \\\dfrac{p^2+7p+2-x}{p^2+5p-14}=\dfrac{p-1}{p-2}\\\\\dfrac{p^2+7p+2-x}{(p-2)(p+7)}=\dfrac{p-1}{p-2}\\\\p^2+7p+2-x=(p-1)(p+7)\\\\p^2+7p+2-x=p^2+6p-7\\\\x=p+9$