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Xelga [282]
2 years ago
9

It takes peter 3/5 of an hour to paint one room. How long will it take for him to paint three rooms?

Mathematics
1 answer:
rosijanka [135]2 years ago
6 0

Answer:

1 4/5 hours

Step-by-step explanation:

3/5 x 3 = 9/5 or 1 4/5

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Aleta's puppy gained 3/8 pound each week for 4 weeks. Altogether, how much weight did the puppy gain during the 4 weeks?
kirill [66]
It would be 1 1/2 or 1.5 pounds in total

3/8 as a decimal is .375 (divide 3 by 8)

Then you multiply .375 by 4 then you get 1.5 or 1/ 1/2.
6 0
3 years ago
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PLEASE HELP !
scZoUnD [109]
It will be 33/74
77-41 = 33 and 33/74 is the correct fraction for the cars that are NOT boxcars
5 0
3 years ago
What is 2x + 11y when x = 4 and y = 2?<br><br> A. 30<br><br> B. 38<br><br> C. 48<br><br> D. 136
Alla [95]
The correct answer is A
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3 years ago
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A+0=0+a=0 is true except for all teal numbers except 0 true or false
jolli1 [7]

Try it with some random number, like a=3.

       Is "3+0 = 0+3 = 0" true?    No.

If you were doing multiplication, then it would be true, but not with addition.

6 0
3 years ago
T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as
Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is


K(x)= \frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}


The given curve is y=7e^{x}


{y}''=7e^{x}\\ {y}'=7e^{x}


k(x)=\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}


{k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}

For Maxima or Minima

{k(x)}'=0

7(e^x)(1+49e^{2x})(98e^{2x}-1)=0

→e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0

e^{x}=0  ,  ∧ 1+49e^{2x}=0   [not possible ∵there exists no value of x satisfying these equation]

→98e^{2x}-1=0

Solving this we get

x= -\frac{1}{2}\ln{98}

As you will evaluate {k(x})}''<0 at x=-\frac{1}{2}\ln98

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[-\frac{1}{2}\ln98,1/√2]

k(x)=\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}

k(x)=\frac{7}{\infty}

k(x)=0







5 0
3 years ago
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