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shutvik [7]
3 years ago
15

Round to the nearest thousand. 88,651 6th grade mafs

Mathematics
2 answers:
adelina 88 [10]3 years ago
7 0

Answer:

90,000

Step-by-step explanation:

88 = 90

7nadin3 [17]3 years ago
6 0

Answer: 89,000

Step-by-step explanation:

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I think it's C. sjisebd ddhisuaja. usually get ebe.

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Which graph shows a dilation?
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Where are the graphs??
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The length of a rectangle is 3 centimeters less than four times its width. Its area is 10 square centimeters. Find the dimension
Kay [80]

Answer:

W = 2 cm

L = 5 cm

Step-by-step explanation:

A rectangle is a four sided shape with 4 perpendicular angles. It has two pairs of parallel sides which are equal in distance: width and length. Its area, the amount of space inside it, can be found using the formula A = l*w. If the area is 10 cm² and the length is "3 cm less than 4 times the width" or 4w - 3, you can substitute and solve for w.

A = l*w

10 = (4w - 3)(w)

10 = 4w² - 3w

Subtract 10 from both sides to make the equation equal to 0. Then solve the quadratic by quadratic formula.

4w² - 3w - 10 = 0

Substitute a = 4, b = -3 and c = -10.

w = \frac{3 +/- \sqrt{(-3)^2 - 4(4)(-10)} }{2(4)} = \frac{3 +/- \sqrt{9 +160)} }{8} =  \frac{3 +/- \sqrt{169} }{8} = \frac{3+/-13}{8}

There are two possible solutions which can be found.

3 + 13 / 8 = 16/ 8 = 2

3 - 13 / 8 = -10/8 = -5/4

Since w is a side length or distance, it must be positive so w = 2 cm.

If the width is 2 cm then the length is 4(2) - 3 = 8 - 3 = 5 cm.

8 0
3 years ago
A box contains 3 coins. One coin has 2 heads and the other two are fair. A coin is chosen at random from the box and flipped. If
Blababa [14]

Answer: Our required probability is \dfrac{1}{2}

Step-by-step explanation:

Since we have given that

Number of coins = 3

Number of coin has 2 heads = 1

Number of fair coins = 2

Probability of getting one of the coin among 3 = \dfrac{1}{3}

So, Probability of getting head from fair coin = \dfrac{1}{2}

Probability of getting head from baised coin = 1

Using "Bayes theorem" we will find the probability that it is the two headed coin is given by

\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times \dfrac{1}{2}+\dfrac{1}{3}\times 1}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{1}{6}+\dfrac{1}{6}+\dfrac{1}{3}}\\\\=\dfrac{\dfrac{1}{3}}{\dfrac{2}{3}}\\\\=\dfrac{1}{2}

Hence, our required probability is \dfrac{1}{2}

No, the answer is not \dfrac{1}{3}

5 0
3 years ago
Is <br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B3%7D%7B6%7D%20" id="TexFormula1" title=" \frac{3}{6} " alt=" \frac{3}{6} "
STALIN [3.7K]

Hey!

-----------------------------------------

Answer: Yes 3/6 is rational!

-----------------------------------------

Why? Well, because 3 and 6 are integers that don't repeat. 3/6 simplifies to 1/2 or 0.5 which doesn't repeat so it's rational.

-----------------------------------------

Hope This Helped! Good Luck!

8 0
4 years ago
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