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yulyashka [42]
3 years ago
5

Sarah and amanda each have 2 bags with 4 marbles in each .how many marbles do they have altogether?

Mathematics
2 answers:
STatiana [176]3 years ago
7 0
Sarah has 2 bags with 4 marbles in each, totaling 8 marbles. Amanda has the same amount. Simply add 8 and 8 together to see what they have altogether, which would be 16.
djyliett [7]3 years ago
3 0
Alltogether they have 16 marbles
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How to calculate the diameter of cylinder if you are given volume of 1000Litre and 224cm height ​
Afina-wow [57]

well, let's first notice, all our dimensions or measures must be using the same unit, so could convert the height to liters or the liters to centimeters, well hmm let's convert the volume of 1000 litres to cubic centimeters, keeping in mind that there are 1000 cm³ in 1 litre.

well, 1000 * 1000 = 1,000,000 cm³, so that's 1000 litres.

\textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ V=1000000~cm^3\\ h=224~cm \end{cases}\implies \stackrel{cm^3}{1000000}=\pi r^2(\stackrel{cm}{224}) \\\\\\ \cfrac{1000000}{224\pi }=r^2\implies \sqrt{\cfrac{1000000}{224\pi }}=r\implies \cfrac{1000}{\sqrt{224\pi }}=r\implies \stackrel{cm}{37.7}\approx r

now, we could have included the "cm³ and cm" units for the volume as well as the height in the calculations, and their simplication will have been just the "cm" anyway.

6 0
1 year ago
Monica measures 91 milliliters of water into 9 tiny beakers. she measures an equal amount of water into the first 8 beakers. she
UkoKoshka [18]
To determine the amount of water in the first 8 beakers, you will subtract the 19 ml that are in the 9th beaker from the total to see how much water was actually put into the first 8 beakers.

91 ml - 19 ml = 72 ml.

The 72 ml are divided evenly between 8 beakers.

72/8 = 9 ml

There would be 9 ml of water in each of the first 8 beakers.
8 0
3 years ago
I need help. this is the problem x^2/3+10=7x^1/3
Arlecino [84]

\it x^{\frac{2}{3}}+10=7x^{\frac{1}{3}} \Leftrightarrow  (x^{\frac{1}{3}})^2-7x^{\frac{1}
  {3}} +10=0 
\\\;\\
We\ note\ x^{\frac{1}{3}} =t \ \ and \ the \ equation \ will \ be:
\\\;\\
t^2-7t+10 = 0 \Leftrightarrow t^2 -2t-5t+10=0 \Leftrightarrow t(t-2) -5(t-2)=0

\it \Leftrightarrow (t-2)(t-5)=0 
\\\;\\
t-2=0 \Rightarrow t=2 \Rightarrow x^{\frac{1}{3}}=2 \Rightarrow (x^{\frac{1}{3}})^3 =2^3 \Rightarrow x = 8
\\\;\\
t-5=0 \Rightarrow t=5 \Rightarrow x^{\frac{1}{3}}= 5  \Rightarrow (x^{\frac{1}{3}})^3 = 5^3 \Rightarrow x = 125

S = {8; 125}


7 0
3 years ago
point X is located at (2,-6), and point Z is located at (0,5). find the value for the point Y that is located 1/5 the distance f
HACTEHA [7]
Let point is(x,y)
x=(2-0)/5=2/5
y=(-6-5)/5=-11/5
(2/5,-11/5)
8 0
3 years ago
What is 100 ÷ 5 × 4 + 43 <br> A. 69<br> B. 144<br> C. 0.3<br> D. 1.2
inn [45]
123 - Use PEMDAS but since you have both division and multiplication you can just go from left to right
4 0
2 years ago
Read 2 more answers
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