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ollegr [7]
3 years ago
10

17% of what is 156? I got 918. just checking my answer..

Mathematics
1 answer:
balu736 [363]3 years ago
4 0
17% of 156 = 0.17 x 156
17% of 156  = 26.52

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Answer: 26.52
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i dont know if this is correct

AT (X)= 6X^2 another part to this equation is: AT/6= X^2 and the answer is X= AT/6^1/2b.

8 0
4 years ago
A girl gave her cousin $2.15 in nicels and dimes . Altogether, she gave her cousin 28 coins . How many of them were dimes? How m
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Answer: 15 dimes and 13 nickels

Step-by-step explanation:

7 0
4 years ago
Create a word problem that involves speed
lana66690 [7]

A hockey puck is sliding on frictionless ice on an infinite hockey rink.
Its speed is 36 km/hour.  How far does the puck slide in 10 seconds ? 

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        (36 km/hr) x (0.6214 mi/km)  =  <em>22.37 mi/hr</em>


6 0
4 years ago
The endpoints of are A(2, 2) and B(3, 8). is dilated by a scale factor of 3.5 with the origin as the center of dilation to give
goldfiish [28.3K]

Answer:

The given line segment whose end points are A(2,2) and B(3,8).

Distance AB is given by distance formula , which is

if we have to find distance between two points (a,b) and (p,q) is

=  \sqrt{(p-a)^2+(q-b)^2}

AB= \sqrt{(3-2)^2+(8-2)^2}=\sqrt{1+36}=\sqrt{37} = 6.08 (approx)

Line segment AB is dilated by a factor of 3.5 to get New line segment CD.

Coordinate of C = (3.5 ×2, 3.5×2)= (7,7)

Coordinate of D = (3.5×3, 3.5×8)=(10.5,28)

CD = AB × 3.5

CD = √37× 3.5

     = 6.08 × 3.5

= 21.28 unit(approx)

2. Slope of line joining two points (p,q) and (a,b) is given by

m=\frac{q-b}{p-a}

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As the two lines are coincident , so their slopes are equal.

Slope of line AB=Slope of line CD = 6




6 0
4 years ago
Read 2 more answers
FUNDRAISING A school is raising money by selling calendars for $20 each. Mrs. Hawkins promised a party to whichever of her Engli
____ [38]

Rj, this is the solution to question 19:

• 1st period class sold 60 calendars in total

,

• 2nd period class sold 123 calendars in total

We calculate the average per week, this way:

• 1st period class sold 60/4 = 15 calendars per week

,

• 2nd period class sold 123/4 = 30.75 calendars per week

In consequence the multiplication equation that represents this situation is:

15w + 30.75w = 183, where w is the number of weeks Mrs. Hawkins' classes sold calendars.

This is the solution to question 20:

We already know the average number of calendars that 1st and 2nd period classes sold per week. Now, let's calculate the average for 3nd 4th peiod classes, this way:

• 3rd period class sold 89 calendars in total

,

• 4th period class sold 126 calendars in total

Therefore, the average is:

• 3rd period class sold 89/4 = 22.25 calendars per week

,

• 4th period class sold 126/4 = 31.5 calendars per week

Summarizing, we have:

15 + 30.75 + 22.25 + 31.5 = 99.5 calendars was the average number sold by all her classes in a week.

•

7 0
1 year ago
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