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sergij07 [2.7K]
3 years ago
10

What is the volume of this shape

Mathematics
2 answers:
elena55 [62]3 years ago
7 0
Its 25 hope it helps
gtnhenbr [62]3 years ago
3 0
The answer is 25. Hope it helps
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PQRS is a parallelogram. PR and QS are diagonals. Match each element to its value. Tiles value of length of value of length of
ss7ja [257]

Since it is a parallelogram cente at point T, then the measure of PT is equal to the measure of TR. And the measure of QT is equal to the measure of TS.

 

PT = TR

a + 4 = 2a

4 = 2a -a

a = 4

 

PT = TR = 8 units

 

QT = TS

b = 2b -3

3 = 2b – b

b = 3

 

<span>QT = TS = 3 units</span>

5 0
3 years ago
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Bill has $25 to buy fruit from the store, He buys 10 apples and has $15 left over. What is the price for 1 apple?
ddd [48]

Answer:

1 apple = $1

Step-by-step explanation:

7 0
3 years ago
-1/4x – 2/3x = -6 + 28
77julia77 [94]

Answer:

x= -24

Step-by-step explanation:

-1/4x-2/3x= -6+28

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-1/4x-2/3x=22

multiply all by 12

-3x-8x=264

add -3x to -8x

-11x=264

divide both sides by -11

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3 0
3 years ago
Expreessions represent The sum of 3 and n
weeeeeb [17]

Answer:

3+n

Step-by-step explanation:

Simple questions but it can be technical atimes, according to the question we are to solve in terms of n, simple...

The sum of 3 and n can be written as 3+n

This cannot be solved further because there are two different variables and it is impossible to add them together

Therefore the final answer is 3+n

But in case a value is given for n we can then substitute and solve further

Hope this will help you

6 0
3 years ago
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Given the function f(x)= 14x^2 + 5x - 1 answer the question if you could change the 14 to any other number, but the rest has to
polet [3.4K]
This is a polynomial of degree 2, and remember that complex roots always come in pairs; therefore, in a second degree polynomial you can only have 0 real roots and 2 complex roots, or 2 real roots and 0 complex roots. There is no way you can have 1 real root and 1 complex root because, as stated before, complex (non-real) roots always come in pairs; so the only way of having 1 real root is dumping the polynomial of degree 2 (14 x^{2}) by changing the 14 to 0. That way we'll left with f(x)=5x-1 which only has one real root:
0=5x-1
5x=1
x= \frac{1}{5}

7 0
3 years ago
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