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3 years ago

WILL GIVE BRAINlest!!

Mathematics

2 answers:

miss Akunina [59]3 years ago

3 0

Answer:

C. i think :)

Step-by-step explanation:

sorry if I'm wrong :(

hope it helps

ArbitrLikvidat [17]3 years ago

3 0

I think it’s c hope this helped

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HURRY PLEASEEEE

Natasha_Volkova [10]

Answer:

Last one.

Step-by-step explanation:

8 0

3 years ago

Simplify six-hundred thirty over 10

spin [16.1K]

630/10

Lets divide!

630/10 = 63
10/10 = 1

63/1 = 63

Answer: 63

3 0

4 years ago

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Decide if this statement is true or false. x = 2 if and only if | x1 = 2 A.true B. False

diamong [38]

Answer:

true, do u need an explanation?

6 0

2 years ago

Please answer this!!!!

lora16 [44]

Answer:

He could have asked the girls as well, and could have asked the entire class instead of just 10 students.

Step-by-step explanation:

By only asking 10 boys, he is not getting a complete representation of the whole class, as their are 30 students in the class.

6 0

3 years ago

Where will her cut be located? Round to the nearest tenth. x = (StartFraction m Over m + n EndFraction) (x 2 minus x 1) + x 1 A

Arte-miy333 [17]

Answer:

The correct answer is 25.2 in.

Step-by-step explanation:

It is given that number line goes from 0 to 60 which can be used to represent a ribbon of length = 60 inches.

2 inches of the ribbon are frayed so actual length = 58 inches

Please refer to the attached image for the ribbon.

A is at 0

C is at 60

B is at 2

P is the point to divide the remaining ribbon in the ratio 2:3.

Part AB of the ribbon is frayed.

BP: PC = 2:3

Let BP = 2 $x$ and PC = 3 $x$

Now, BP + PC = BC = 58 = 2 $x$ + 3 $x$ = 5 $x$

So,

$5x =58\\\Rightarrow x =11.6$

BP = $2\times x = 2 \times 11.6 = 23.2\ inches$

Location of the Cut = 2 + 23.2 = 25.2 inches

Alternatively, we can use the formula directly:

$x = \dfrac{m} {m + n } (x_2 - x_1) + x_1$

$x_1 = 2\\x_2 = 60$

m: n is the ratio 2:3

$x = \dfrac{2} {2 +3 } (60- 2) + 2\\\Rightarrow 0.4 (58 )+2\\\Rightarrow 23.2+2 \\\Rightarrow \bold{25.2\ inches}$

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4 years ago

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