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3 years ago

12

Tim buys a torch and a battery. The torch costs 11 times as much as the battery. Tim pays with a £10 note and gets £4.84 change.

How much did the battery cost?

1 answer:

algol [13]3 years ago

6 0

The cost of the battery is £0.43

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A cylinder has a volume of 157 cubic centimeters.

ziro4ka [17]

Answer:

4.47100222

Step-by-step explanation:

157/2.5=62.8

62.8/pi= 19.98986085

square root the answer above and you get 4.47100222 for the radius

5 0

3 years ago

A zookeeper predicted the weight of a new baby elephant to be 250 pounds when it was born. The elephant actually weighed 282 pou

worty [1.4K]

About 11% - I first subtracted 250 from 282, to get 32. I then did 32/282 to get .11.... You round that out to about 11%

3 0

3 years ago

Solve the system useing elimination 4x+8y=16 6x-8y=4

nydimaria [60]

Answer:

x=2 and y=1

Step-by-step explanation:

Add the two equations to remove "y" and solve for x:

4x+8y=16

6x-8y=4

________

10x=20

x=2

Plug x=2 into one of the original equations to get the value of "y":

4x+8y=16

4(2)+8y=16

8+8y=16

8y=8

y=1

Therefore, x=2 and y=1

6 0

3 years ago

I tell you these facts about a mystery number, $c$: $\bullet$ $1.5 < c < 2$ $\bullet$ $c$ can be written as a fraction wit

makkiz [27]

Answer:

Possible answer: $\displaystyle c = \frac{16}{10} = \frac{8}{5} = 1.6$ .

Step-by-step explanation:

Rewrite the bounds of $c$ as fractions:

The simplest fraction for $1.5$ is $\displaystyle \frac{3}{2}$ . Write the upper bound $2$ as a fraction with the same denominator:

$\displaystyle 2 = 2 \times 1 = 2 \times \frac{2}{2} = \frac{4}{2}$ .

Hence the range for $c$ would be:

$\displaystyle \frac{3}{2} < c < \frac{4}{2}$ .

If the denominator of $c$ is also $2$ , then the range for its numerator (call it $p$ ) would be $3 < p < 4$ . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than $2$ .

To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)

$\displaystyle \frac{3}{2} = \frac{2 \times 3}{2 \times 2} = \frac{6}{4}$ .

$\displaystyle \frac{4}{2} = \frac{2\times 4}{2 \times 2} = \frac{8}{4}$ .

At this point, the difference between the numerators is now $2$ . That allows a number ( $7$ in this case) to fit between the bounds. However, $\displaystyle \frac{1}{c} = \frac{4}{7}$ can't be written as finite decimals.

Try multiplying the numerator and the denominator by a different number.

$\displaystyle \frac{3}{2} = \frac{3 \times 3}{3 \times 2} = \frac{9}{6}$ .

$\displaystyle \frac{4}{2} = \frac{3\times 4}{3 \times 2} = \frac{12}{6}$ .

$\displaystyle \frac{3}{2} = \frac{4 \times 3}{4 \times 2} = \frac{12}{8}$ .

$\displaystyle \frac{4}{2} = \frac{4\times 4}{4 \times 2} = \frac{16}{8}$ .

$\displaystyle \frac{3}{2} = \frac{5 \times 3}{5 \times 2} = \frac{15}{10}$ .

$\displaystyle \frac{4}{2} = \frac{5\times 4}{5 \times 2} = \frac{20}{10}$ .

It is important to note that some expressions for $c$ can be simplified. For example, $\displaystyle \frac{16}{10} = \frac{2 \times 8}{2 \times 5} = \frac{8}{5}$ because of the common factor $2$ .

Apparently $\displaystyle c = \frac{16}{10} = \frac{8}{5}$ works. $c = 1.6$ while $\displaystyle \frac{1}{c} = \frac{5}{8} = 0.625$ .

8 0

3 years ago

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The difference of a number and 12 is 4

lora16 [44]

16, since the inverse operation of subtraction is addition add 12 and 4.

5 0

3 years ago

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