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kati45 [8]
3 years ago
8

Can someone help me on number 4,5

Mathematics
1 answer:
dalvyx [7]3 years ago
4 0
For 4:
X=13, RST=155, RSU=102

For 5:
X=7, WXZ=38, ZXY=52
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Which is longer 2 meters or 200 cent a meters or 2.05 meters
Yuri [45]
Solving this requires use of conversions

we already know that
2.05 m > 2 m

all that's left is the question
200 cm vs. 2.05 meters

I we know our equivalencies, we know that
100 cm = 1 m

if this is so then
200 cm = 2 m

we already know that
2.05 m > 2 m

so it must also be that
2.05 m > 200 cm

2.05 \: meters
6 0
3 years ago
A pipe is 36 feet long. It needs to be cut into pieces that are each 3/4 feet long. How many pieces can be made from the pipe?
Ket [755]

108/4 = 27 pieces you multiply 36 x 3 = 108 and the denominator stays the same so 108/4 equals to 27 pieces

4 0
3 years ago
7x - 20 = 2x - 3(3x + 2)
kirill [66]

Answer:

7x-20=2x-3(3x+2)

We move all terms to the left:

7x-20-(2x-3(3x+2))=0

We calculate terms in parentheses: -(2x-3(3x+2)), so:

2x-3(3x+2)

We multiply parentheses

2x-9x-6

We add all the numbers together, and all the variables

-7x-6

Back to the equation:

-(-7x-6)

We get rid of parentheses

7x+7x+6-20=0

We add all the numbers together, and all the variables

14x-14=0

We move all terms containing x to the left, all other terms to the right

14x=14

x=14/14

x=1

   

   

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
84.87 is 115% of what
elena-s [515]
Divide 84.87 by 1.15, and you get 73.8

Then you can check your work by multiplying 73.8 by 1.15 and you get 84.87
6 0
3 years ago
równanie zmiany prędkość autokaru poruszającego się po prostym odcinku szosy i rozpoczynającego hamowanie od szybkości 20m/s ma
Rasek [7]

Answer:

s = 22.5 m

Step-by-step explanation:

the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.

v (t) = 25 - 5 t

at t = 0 , v = 20 m/s

Let the distance is s.

s =\int v(t) dt\\\\s =\int (25 - 5t)dt\\\\s= 25 t - 2.5 t^2 \\

Let at t = t, the v = 20

So,

20 = 25 - 5 t

t = 1 s

So, s = 25 x 1 - 2.5 x 1 = 22.5 m

3 0
2 years ago
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