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Georgia [21]
3 years ago
6

A family has a half a pizza left. Of each whole pizza has 8 pieces and the four children want to share the half pizza, what part

of a whole pizza does each one get?
Mathematics
1 answer:
KiRa [710]3 years ago
3 0

Answer:

\frac{1}{8}

Step-by-step explanation:

Since in the question it is mentioned that

The whole pizza contained eight pieces

But as if now the half pizza is left that means the half of eight peices is 4 pieces that represent the half pizza

Now there are the four children so the each one get is

= \frac{1}{4} \times \frac{1}{2}

= \frac{1}{8}

Hence the each one get 1 by 8th part of the whole pizza and the same is to be considered

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If side AB is parallel to side DC, how are side A'B' and side D'C' related? Explain.
anastassius [24]
Side A’B’ and D’C’ are related because they are both parallel which I think means that they would be the same length
7 0
2 years ago
Three sets of a sum of a number and four are added to the sum of 7 times the same number and 13
Reika [66]
Let
x--------------> a number

we know that
<span>three sets of a sum of a number and four----------> 3(x+4)
t</span><span>he sum of 7 times the same number and 13--------> 7x+13

therefore

</span>(Three sets of a sum of a number and four) are added (to the sum of 7 times the same number and 13) ----------> [3(x+4)] + [7x+3]

[3(x+4)] + [7x+3]------> [3x+12] + [7x+3]=10x+15

the answer is 10x+15


5 0
2 years ago
Find the height of this triangle.
Bogdan [553]

Answer:

\sqrt{3}

Step-by-step explanation:

x^2 + 1 = 4

x^2 = 3

\sqrt{3}

7 0
3 years ago
a) Estimate the volume of the solid that lies below the surface z = 7x + 5y2 and above the rectangle R = [0, 2]⨯[0, 4]. Use a Ri
SSSSS [86.1K]

In the x direction we consider the m=2 subintervals [0, 1] and [1, 2] (each with length 1), while in the y direction we consider the n=2 subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of R are (1, 0), (2, 0), (1, 2), (2, 2).

Let f(x,y)=7x+5y^2. The volume of the solid is approximately

\displaystyle\iint_Rf(x,y)\,\mathrm dx\,\mathrm dy\approx f(1,0)\cdot1\cdot2+f(2,0)\cdot1\cdot2+f(1,2)\cdot1\cdot2+f(2,2)\cdot1\cdot2=\boxed{164}

###

More generally, the lower-right-corner Riemann sum over m=\mu and n=\nu subintervals would be

\displaystyle\sum_{m=1}^\mu\sum_{n=1}^\nu\left(7\frac{2m}\mu+5\left(\frac{4n-4}\nu\right)^2\right)\frac{2-0}\mu\frac{4-0}\nu=\frac83\left(101+\frac{21}\mu+\frac{40}{\nu^2}-\frac{120}\nu\right)

Then taking the limits as \mu\to\infty and \nu\to\infty leaves us with an exact volume of \dfrac{808}3.

7 0
3 years ago
1/2(4x−5)+52=10
Sav [38]
In the first one the options they give you are wrong, in fact you can prove it yourself, none of the options will give you as a result 10
I hope I helped you out
greetings!

7 0
3 years ago
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