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Ber [7]
3 years ago
7

Round 630 to the nearest ten

Mathematics
2 answers:
Sergio [31]3 years ago
8 0
Rounding it to the nearest tens place . . . here are your options:
 . . .
620
630
640
 . . .

obviously 630 rounds to 630 since it is equal to it

but the rule is, if the rounded value (in this case, the value in the ones place) is equal to or greater than five (5), then you round up . . . if the rounded value is less than five (5), then you round down . . . since our value in the ones place is 0 (which is less than 5), we round down to the next multiple of ten, which happens to be 630 itself.
rjkz [21]3 years ago
3 0
630 rounded to the nearest tenth is 600.4 and under let it rest 5 and up wake it it up
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Find the slope and y-intercept of the line through the points (-3,-4) and (0,-1)​
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Slope=1, y-intercept=-1.

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(-1-(-4))/(0-(-3))

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3 years ago
Julia and carl buy 2 sandwiches,1 salad,1piece of fruit and 2 drinks for lunch.They give the cashier 20.03.What coins and bills
Sergeeva-Olga [200]

Answer:

They would recieve cents as change.

Step-by-step explanation:

They bought 2 sandwiches, 1 salad, 1 piece of fruit and 2 drinks

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Which means all items are not more 20.03

If they would receive change, it would be in cents.

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3 years ago
Help!!!! please!!!! ill give brainliest!!!!
Anna [14]

Answer: The 2nd option is correct.

Step-by-step explanation:

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3 0
3 years ago
From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of th
wel

Answer:

\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}

Step-by-step explanation:

The <u>width</u> of a square is its <u>side length</u>.

The <u>width</u> of a circle is its <u>diameter</u>.

Therefore, the largest possible circle that can be cut out from a square is a circle whose <u>diameter</u> is <u>equal in length</u> to the <u>side length</u> of the square.

<u>Formulas</u>

\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}

\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}

\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}

If the diameter is equal to the side length of the square, then:
\implies \sf r=\dfrac{1}{2}s

Therefore:

\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}

So the ratio of the area of the circle to the original square is:

\begin{aligned}\textsf{area of circle} & :\textsf{area of square}\\\sf \dfrac{1}{4}\pi s^2 & : \sf s^2\\\sf \dfrac{1}{4}\pi & : 1\end{aligned}

Given:

  • side length (s) = 6 in
  • radius (r) = 6 ÷ 2 = 3 in

\implies \sf \textsf{Area of square}=6^2=36\:in^2

\implies \sf \textsf{Area of circle}=\pi \cdot 3^2=28\:in^2\:\:(nearest\:whole\:number)

Ratio of circle to square:

\implies \dfrac{28}{36}=\dfrac{7}{9}

5 0
2 years ago
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