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choli [55]
4 years ago
6

Evaluate 1−4×(−3)+8×(−3)

Mathematics
2 answers:
Ivan4 years ago
5 0
The Answer to this equation is -11
Svetradugi [14.3K]4 years ago
4 0

Answer:

-11

Step-by-step explanation:

I believe this is it

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A party rental company has chairs and tables for rent. The total cost to rent
Tema [17]
2c+5t=43

c=(43-5t)/2

8c+3t=36 and using c found above in this equation gives you:

4(43-5t)+3t=36

172-20t+3t=36

-17t=-136

t=8, and since c=(43-5t)/2

c=1.5

So a table costs $8.00 to rent and a chair costs $1.50
5 0
3 years ago
Which is greater than, -9/4 or -3?
Firdavs [7]

-9/4 is greater than -3 because -9/4 is also -2.25 and the less negative number is always greater than the more negative number.

Hope this helps!


7 0
3 years ago
Read 2 more answers
Can someone help me!?!?
amm1812

It's the first step.4x-x-2+6=6x+16     ##substract 2 rather an add 2

Check it carefully next time.!!

5 0
3 years ago
Christina describes her shape. She says it has 3 equal sides that are each 4 centimeters in length. It has no right angles. Do u
Salsk061 [2.6K]


draw an equilateral triangle that has a side length of all 4cm


4 0
3 years ago
Read 2 more answers
The core melt- down and explosions at the nuclear reactor in Chernobyl in 1986 released large amounts of strontium-90, which dec
elena-14-01-66 [18.8K]

If S(t) is the amount of strontium-90 present in the area in year t, and it decays at a rate of 2.5% per year, then

S(t+1)=(1-0.025)S(t)=0.975S(t)

Let S(0)=s be the starting amount immediately after the nuclear reactor explodes. Then

S(t+1)=0.975S(t)=0.975^2S(t-1)=0.975^3S(t-2)=\cdots=0.975^{t+1}S(0)

or simply

S(t)=0.975^ts

So that after 50 years, the amount of strontium-90 that remains is approximately

S(50)=0.975^{50}s\approx0.282s

or about 28% of the original amount.

We can confirm this another way; recall the exponential decay formula,

S(t)=se^{kt}

where t is measured in years. We're told that 2.5% of the starting amount s decays after 1 year, so that

0.975s=se^k\implies k=\ln0.975

Then after 50 years, we have

S(50)=se^{50k}\approx0.282s

5 0
3 years ago
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