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Vikki [24]
3 years ago
13

A brand new filled can of chicken broth is 10 cm tall and has a radius of 10 cm. Mishka uses some of the broth to cook and now t

he broth left in the can is 6 cm high. How much broth did Mishka use? Round only your final answer to the nearest whole number.
Mathematics
2 answers:
Maurinko [17]3 years ago
8 0

we have chicken broth in cylindrical can.

we have given the height radius of cylindrical can.

h=10cm r=10 cm .

and she have 6cm broth height in the tin that means she had used 4 cm of height of broth .

now we need to find volume of two cylinder with h=10 cm and 6cm

and we need to find the difference between the volumes .

volume1=\pi *r*h_{1} =\pi *(10)^2*10=3.14*1000=3140 cm^3

volume 2=\pi *(10)^2*6=1884cm^3

volume used =volume 1-volume2=3140-1884=1256cm^3 is used

alekssr [168]3 years ago
4 0

Answer:

The Answer fam is................ 1257 cm³

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Genrish500 [490]

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I'm pretty sure the answer is D

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3 years ago
Can someone help me on 64 67 68 and if you do you will get a lot of points
andre [41]

Answer: #64, A. #67, A. #68, C


Step-by-step explanation:

#64

0.96/6= 0.16

2.4/16= 0.15

32= 0.18

16 oz is the least


#67

100/20=5

3*5=15

SO, it's 15%.


#68

Divide

$430/43= $10

$594/54= $11

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8 0
3 years ago
Y = 2x - 8 3x - 2y = 13​
SSSSS [86.1K]

Answer:

{x,y} = {3,-2}

Step-by-step explanation:

// Solve equation [1] for the variable  y

 [1]    y = 2x - 8

// Plug this in for variable  y  in equation [2]

  [2]    3x - 2•(2x-8) = 13

  [2]    -x = -3

// Solve equation [2] for the variable  x

  [2]    x = 3

// By now we know this much :

   x = 3

   y = 2x-8

// Use the  x  value to solve for  y

   y = 2(3)-8 = -2

Solution :

{x,y} = {3,-2}

8 0
3 years ago
Please help . I will mark brainliest
Step2247 [10]

Answer:

c: car b was 10 miles east of car a

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
I’m Really lost if I could get a answer it would be greatly appreciated
Nataly [62]
<h3>Answers:</h3>

f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}

================================================

Work Shown:

Part 1

f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

------------------

Part 2

g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out \left(\sqrt{x-1}+5\right)^2

6 0
3 years ago
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