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Anna35 [415]
3 years ago
9

Which equation can be used to calculate the output

Mathematics
1 answer:
ioda3 years ago
5 0
The answer is a. trust me
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Ayudaaáaaaaaaaa porfaaaa​
mrs_skeptik [129]

Answer:

a) 4

b) 3

c) 5

d) 5

e) 2

Step-by-step explanation:

a) 4+4=8, 8+4=12, 12+4=16, 16+4=20

b) 6+3=9, 9+3=12, 12+3=15, 15+3=18

c) 8+5=13, 13+5=18, 18+5=23, 23+5=28

d) -10+5= -5, -5+5=0, 0+5=5, 5+5=10

e) 3+2=5, 5+2=7, 7+2=9, 9+2= 11

3 0
3 years ago
What correctly classifies triangle STU?
Brums [2.3K]

Check the picture below.

notice, all three sides in the triangle are of different lengths, thus is an scalene triangle.

6 0
3 years ago
The middle 5 in 0.555 is 1/10 the value of the 5 to its
Marta_Voda [28]
Great, you know that. no need to ask more.
3 0
3 years ago
A ball has a volume of 128 cubic inches. Find the diameter of the ball.
Natalija [7]
\bf \textit{volume of a sphere}\\\\
V=\cfrac{4\pi r^3}{3}~~
\begin{cases}
r=radius\\
-----\\
V=128
\end{cases}\implies 128=\cfrac{4\pi r^3}{3}\implies 128(3)=4\pi r^3

\bf \cfrac{128(3)}{4\pi }=r^3\implies \sqrt[3]{\cfrac{128(3)}{4\pi }}=r^3\implies \sqrt[3]{\cfrac{96}{\pi }}=r\implies \sqrt[3]{\cfrac{8\cdot 12}{\pi }}=r
\\\\\\
\sqrt[3]{\cfrac{2^3\cdot 12}{\pi }}=r\implies 2\sqrt[3]{\cfrac{12}{\pi }}=r\\\\
-------------------------------\\\\
\textit{and since the diameter is \underline{twice as the radius}}\qquad d=4\sqrt[3]{\cfrac{12}{\pi }}
7 0
4 years ago
Lisa spends part of her year as a member of the gym. She then finds a better deal at another gym, so she cancels her membership
stellarik [79]

Answer:

Number of month in first gym = 7

Number of month in second gym = 5

Step-by-step explanation:

Given:

First gym member ship cost = $75 per month

Second gym member ship cost = $50 per month

Total spending = $775

Computation:

Number of month in first gym = a

Number of month in second gym = b

So

a + b = 12....eq1

75a + 50b = 775 ....eq2

From eq 1 and 2

a = 7 month

b = 5 month

Number of month in first gym = 7

Number of month in second gym = 5

5 0
3 years ago
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