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photoshop1234 [79]
2 years ago
10

A piece of string is 122 inches long bill wants to cut it into 2 pieces so that one piece is 3 times as long as the other how lo

ng is each piece
Mathematics
1 answer:
allochka39001 [22]2 years ago
3 0
122/4=30.5
30x3=91.5

Piece 1= 30.5in.
Piece 2= 91.5in
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Please answer me<br />bis two fifths of c.<br />4a = 30<br />Work out the ratio a:b:c<br />Give your ans
irina [24]

b is two fifths of c, so we make this a ratio with c = 1.

The ratio of b:c is 2/5 : 1

We also have 4a = 3c, rewrite this ratio so c is 1 by dividing both sides by 4,

so we get the ratio of a to c as 3/4 : 1

Now we get the a:b:c ratio of 3/4 : 2/5 : 1 now we can change the fractions to whole numbers, first by multiplying the 3 numbers by 4 to get get rid of the denominator of 4 in a:

3 : 8/5 : 4

Now multiply the 3 numbers by 5 to remove the denominator  of 5 in b:

15 : 8 : 20

Now we can check using the equations:

b is 2/5 of c:  20 x 2/5 = 40/5 = 8 This is true.

4a = 3c: 4(15) = 3(20) = 60 = 60, this is also true.

The ratios is 15 : 8: 20

7 0
2 years ago
Please Help Me With This Question
irina1246 [14]

Answer:

4/4 quarters

Step-by-step explanation:

hope this will help you

7 0
2 years ago
Read 2 more answers
A bag contains 10 black marbles and 6 red marbles. If a red
bonufazy [111]

Answer:

66.67% probability of selecting a black marble from the bag.

Step-by-step explanation:

Initially there are 16 marbles, 10 of which are black and 6 of which are red.

A red marble is drawn and not put back. So now there are 15 marbles, of which 10 are black.

So there is a 10/15 = 2/3 = 0.666% 7 = 66.67% probability of selecting a black marble from the bag.

7 0
3 years ago
There are 60 chocolates in a box
nika2105 [10]
There could be 35 toffees in the box.
8 0
3 years ago
One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p
scZoUnD [109]

Answer:

c^2 = 9dp

Step-by-step explanation:

Given

dx^2 + cx + p = 0

Let the roots be \alpha and \beta

So:

\alpha = 2\beta

Required

Determine the relationship between d, c and p

dx^2 + cx + p = 0

Divide through by d

\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0

x^2 + \frac{c}{d}x + \frac{p}{d} = 0

A quadratic equation has the form:

x^2 - (\alpha + \beta)x + \alpha \beta = 0

So:

x^2 - (2\beta+ \beta)x + \beta*\beta = 0

x^2 - (3\beta)x + \beta^2 = 0

So, we have:

\frac{c}{d} = -3\beta -- (1)

and

\frac{p}{d} = \beta^2 -- (2)

Make \beta the subject in (1)

\frac{c}{d} = -3\beta

\beta = -\frac{c}{3d}

Substitute \beta = -\frac{c}{3d} in (2)

\frac{p}{d} = (-\frac{c}{3d})^2

\frac{p}{d} = \frac{c^2}{9d^2}

Multiply both sides by d

d * \frac{p}{d} = \frac{c^2}{9d^2}*d

p = \frac{c^2}{9d}

Cross Multiply

9dp = c^2

or

c^2 = 9dp

Hence, the relationship between d, c and p is: c^2 = 9dp

8 0
3 years ago
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