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3 years ago

B=0

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

4 0

Answer:

Step-by-step explanation:

-(-B) = -(-0) =0

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I need help please

ASHA 777 [7]

Answer:

the answer would be 81

Step-by-step explanation:

6 to the second power would be 36 then you would multiply 36x2 and it would equal 72. then you would do 2x6 which is 12. then 72+12-3= 81

5 0

3 years ago

Membership in an art club is expected to grow by about 4% per month. Currently, there are 350 members in the club. In about how

igomit [66]

500=350(1+0.04/12)^12x
Solve for x
500/350=(1+0.04/12)^12x
X=(Log(500/350)/log(1+0.04/12))/12
X=9 months

6 0

3 years ago

Read 2 more answers

camela is trying to find the equation of a line that passes through the points (-1, 16) and (5, 88). does the equation y= 12

lutik1710 [3]

I'm assuming you meant does y=12x pass through those coordinates, and if that's the case, then the answer is no.

Hope that helps :)

6 0

3 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

3 years ago

Match the vocab word with the definition:

love history [14]

A:8

B:14

C:3

D:11

E:7

F:4

G:1

H:12

J:9

K:10

L:2

M:6

N:5

O:13

4 0

3 years ago