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3 years ago

SOMEONE PLEASE HELP ME WITH THIS ALL YOU HAVE TO DO IS ANSWER THE QUESTION PLEASE

Mathematics

1 answer:

ratelena [41]3 years ago

5 0

When I clicked the image it said not found...

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During a field trip, 60 students are put into equal-sized groups. Use 60 ÷ 5 = 12 to answer this question. If 12 represents the

wel

Answer:

the groups

Step-by-step explanation:

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3 years ago

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What is 692,004 in expanded form

rjkz [21]

600000+90000+2000+4=692004

I think that's how you write it.

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4 years ago

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Please help on number 15

borishaifa [10]

....................................................................

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3 years ago

Two right triangles are shown. First, find the missing leg of the triangle on the right. Then you have two lengths to use to fin

jasenka [17]

Answer:

x = 30m

Step-by-step explanation:

GIVEN :-

There are 2 right-angled triangles.

In the triangle on the right side ,

Hypotenuse = 26m
Base = 10m

In the triangle on the left side ,

Base = 18m

TO FIND :-

Length of the common perpendicular to both triangles.
Length of the hypotenuse of the triangle on the left side.

FACTS TO KNOW BEFORE SOLVING :-

$Hypotenuse = \sqrt{Base^2 + Perpendicular^2}$ .
$Perpendicular = \sqrt{Hypotenuse^2 - Base^2}$

SOLUTION :-

In the triangle on the right side ,

$Perpendicular = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24m$

⇒ Perpendicular of the triangle on right side = Perpendicular of the triangle on left side.

In the triangle on the left side ,

$Hypotenuse = \sqrt{24^2 + 18^2} = \sqrt{576 + 324} = \sqrt{900} = 30m$

∴ x = 30m

3 0

3 years ago

A few seconds ago i posted a really hard question, but this one is WAY more hard, whats 2 + 4? (i doubt that anyones brain is bi

Gelneren [198K]

Answer:

Dude why did you do this to us it's too hard

Step-by-step explanation:

P.S. It' 6

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3 years ago