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3 years ago

An athlete jumps these 4 distances in a long jump competition. Which jump does not round to 4.7 meters?

Mathematics

1 answer:

frozen [14]3 years ago

3 0

His third jump

4.647 does not round to 4.7
The 4 is under 5 so it rounds down to 4.6

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-2x+4=10 <br> 3/4x =2<br> Plz help ill give brainly

Archy [21]

Answer:

$a) \: - 2x + 4 = 10 \\ 4 = 10 + 2x \\ 4 - 10 = 2x \\ - 6 = 2x \\ \frac{ - 6}{2} = x \\ - 3 = x \: or \: x = - 3 \\ \\ b) \: \frac{3}{4x} = 2 \\ 3 = 2 \times 4x \\ 3 = 8x \\ x = \frac{3}{8} \\ 0.375 = x \: \: or \: x = 0.375$

Step-by-step explanation:

please mark me brainliest

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3 years ago

Factor this expression completely<br><br>-21x + 28

pickupchik [31]

-21x+28 is factored into:

-7(3x-4)

3 0

3 years ago

Read 2 more answers

Find the missing side of this right

I am Lyosha [343]

Answer:

Step-by-step explanation:

square of 15 + square of x = square of 21

225 + square of x = 441

441 - 225 = 216

x = 216

6 0

3 years ago

Calvin had 30 minutes in time-out. For the first 23 1/3 minutes, Calvin counted spots on the ceiling. For the rest of the time,

scoundrel [369]

Answer: $\frac{20}{3}\ minutes$ or $6 \frac{2}{3}\ minutes$

Step-by-step explanation:

For this exercise you can convert the mixed number to an improper fraction:

1. Multiply the whole number part by the denominator of the fraction.

2. Add the product obtained and the numerator of the fraction (This will be the new numerator).

3. The denominator does not change.

Then:

$23\frac{1}{3}= \frac{(23*3)+1}{3}= \frac{70}{3}\ minutes$

You know that he had 30 minutes in time-out, he counted spots on the ceiling for $\frac{70}{3}$ minutes and the rest of the time he made faces at his stuffed tiger.

Then, in order to calculate the time Calvin spent making faces at his stuffed tiger, you need to subract 30 minutes and $\frac{70}{3}$ minutes:

$30\ min-\frac{70}{3}=(\frac{3(30)-70}{3})=\frac{20}{3}\ minutes$ or $6 \frac{2}{3}\ minutes$

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3 years ago

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In a group of 60 students, 18 students take algebra i, 20 students take algebra ii, and 10 students take both subjects. how many

nadya68 [22]

Answer:

Step-by-step explanation:

The number that take algebra i includes those who take both subjects, as does the number taking algebra ii. Then the number taking algebra i alone is ...

18 -10 = 8 . . . . . take only algebra i

So the number taking any kind of algebra is ...

20 + 8 = 28

and the number not taking either subject is ...

60 -28 = 32 . . . not taking either subject

8 0

3 years ago