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3 years ago

If 9 oz make 10 servings how many oz do I need to make 13 servings

Mathematics

1 answer:

Luden [163]3 years ago

4 0

9/10 = x/13. Solve for x. X = 11.7.
Your answer would be 11.7oz. Hope this helps

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If 6:2=8:x, then<br> a. 6x=16<br> b. 6x=10<br> c.2x=48<br> d. 2x=14

otez555 [7]

Answer:

a)6x = 16

Step-by-step explanation:

<h2><u>Finding the value</u></h2>

the ratio can be solved by <u>cross multiplication :</u>

6 : 2

8 : x

6*x = 8*2

6x = 16

answer is : <u>a) 6x = 16</u>

4 0

3 years ago

Read 2 more answers

What is the difference between constructing a square and constructing a regular hexagon?

dexar [7]

Answer:

The diameter of the circle is used for the square construction, but the radius of the circle is used for the regular hexagon construction.Step-by-step explanation:

5 0

2 years ago

What is thee answer?

Vanyuwa [196]

I think that it is B

4 0

3 years ago

Translate 5x+7 into a verbal sentence.

algol13

B im pretty sure....

7 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

4 years ago