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3 years ago

How to round 295,245 to the nearest integer?

Mathematics

1 answer:

marissa [1.9K]3 years ago

6 0

That's already rounded to the nearest integer.
To the nearest ten, that would be 295250.

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The perimeter of a parallelogram is 72 meters.The width of the parallelogram is four meters less than its length. Find the lengt

Naya [18.7K]

Parallelogram have 4 sides.

Use the distributive property. 2 multiply the last and negative 4 in the parentheses.
2l + 2(l - 4) = 72 m

Combine Like terms
2l + 2l - 8 = 72 m
4l - 8 = 72 m
Get l all by itself by adding 8 to both sides
4l - 8 = 72 m
+8. +8

Divide both sides by 4
4l = 80
4. 4

l= 20

The width of the parallelogram is four meters less than its length.

4 sides
2 length = 20
2 width = 16

I got 16 by subtracting 4 from 20

Comment if you need any other help :)

4 0

3 years ago

Which mathematical sentence most Accurately expresses the information in the problem below?

d1i1m1o1n [39]

Answer:

Don't quote me on this but it's probably C. 12c ≥ 88

Step-by-step explanation:

This is because he packages 88 eggs <em>into </em>cartons of 12.

88/12 is 7.33333... so it makes sense to have a greater or equal amount of eggs. If you multiply 7.333333... by 12, each additional 3 gets you closer to 88 so again, makes sense to have more.

5 0

3 years ago

The U.S. Department of Housing and Urban Development (HUD) uses the median to report the average price of a home in the United

MatroZZZ [7]

Answer:

Step-by-step explanation:

given that the U.S. Department of Housing and Urban Development (HUD) uses the median to report the average price of a home in the United States.

We know that mean, median and mode are measures of central tendency.

Mean is the average of all the prices while median is the middle entry when arranged in ascending order.

Mean has the disadvantage of showing undue figure if extreme entries are there. i.e. outlier affect mean.

Suppose a price goes extremely high, then mean will fluctuate more than median.

So median using gives a reliable estimate since median gives the middle price and equally spread to other sides.

3 0

3 years ago

Help me? idk the answer :P

Alexus [3.1K]

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
\\\\
-------------------------------\\\\
$

$\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
-------------------------------\\\\
\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4$

$\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
-------------------------------\\\\
\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}
\\\\\\
\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}$

$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$

7 0

3 years ago

Which statement is true? A. 2x – 1 = 13 and x = 6 B. 2x – 1 = 13 and x = 7 C. 2x + 1 = 13 and x = 7 D. 2x – 1 = 13 and x = 11

alexandr1967 [171]

Answer:

B. 2x – 1 = 13 and x = 7

Step-by-step explanation:

We are given 4 equations and a solution for each. We have to tell which of the given solution satisfies the given equation.

Option A.

2x -1 = 13 and x = 6

Using this value in the equation, we get:

2(6) -1 = 13

12 - 1 = 13

11 = 13, which is not true. Hence this option is not valid

Option B.

2x - 1 = 13 and x = 7

Using the value in the equation, we get:

2(7) - 1 =13

14 - 1 =13

13 = 13, which is true. Hence this option is valid.

Option C.

2x + 1 =13 and x = 7

Using the value in the equation, we get:

2(7) + 1 = 13

15 = 13, which is not true. So this option is not valid

Option D.

2x - 1 = 13 and x = 11

Using this value in the equation, we get:

2(11) - 1 = 13

21 = 13, which is not true. Hence this option is not valid.

6 0

3 years ago