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3 years ago

6 sweets cost 42p altogether how much do 8 sweets cost

2 answers:

8 0

The best thing to do is to find the cost of one sweet, and to do this, divide 42 by 7.
42/7= 6
Therefore, one sweet costs 6p.
To find the cost of 8 sweets, you've got to multiply 6 by 8, and this gives you 54p.
Therefore, 8 sweets cost 54p
This is basically just finding unit rate and multiplying.
<span>Hope this helps :)</span>

Viefleur [7K]3 years ago

5 0

42 by 7.
42/7= 6
1 sweet = 6p*8=48p

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You run 4 3/4 miles through a park each day if you ran that distance five times last week how many miles did you run

Svetlanka [38]

4.75 miles x 5 = 23.75 / 23 3/4 miles

5 0

3 years ago

Find the co-ordinates of the point of trisection of the line joining the points A(1,8) and B(4,2).

Licemer1 [7]

<u>coordinates of the point are (2,6) and (3,4)</u>

Answer:

Solution given:

let the given point be A(1,8) and B(4,2).

P and Q are the two points on AB such that

AP=PQ=QB=k

now

comparing AP and PB

AP=k

PB=2k

ratio of AP and PB = $\frac{1k}{2k}$ = ratio 1:2

now

finding p

for this

$(m_1,m_2)=(1,2)$

For AB

$(x_1,y_1)=(1,8)$

$(x_2,y_2)=(4,2)$

now by using division formula

$(x,y)=(\frac{m_1x_2+m_2x_1}{m_1+m2},\frac{m_1y_2+m_2y_1}{m_1+m2})$

$=(\frac{1*4+2*1}{1+2},\frac{1*2+2*8}{1+2})=(2,6)$

similarly

Q divides AB

Ratio of AQ and QB = $\frac{2k}{1k}$ = ratio 2:1

$(x_1,y_1)=(1,8)$

$(x_2,y_2)=(4,2)$

$(m_1,m_2)=(2,1)$

by using division formula

$(a,b)=(\frac{m_1x_2+m_2x_1}{m_1+m2},\frac{m_1y_2+m_2y_1}{m_1+m2})$

$=(\frac{2*4+1*1}{2+1},\frac{2*2+1*8}{2+1})=(3,4)$

4 0

3 years ago

Points B, D, and F are midpoints of the sides of ΔACE. EC = 38 and DF = 16. Find AC. The diagram is not to scale.

Lunna [17]

Answer: The measure of AC is 32.

Explanation:

It is given that the Points B, D, and F are midpoints of the sides of ΔACE. EC = 38 and DF = 16.

The midpoint theorem states that the if a line segments connecting two midpoints then the line is parallel to the third side and it's length is half of the third side.

Since F and D are midpoints of AE and EC respectively.

So by midpoint theorem length of AC is twice of DF.

$AC=2\times DF$