If 13 candy bars cost $21.97, how much do 20 candy bars cost?

1 answer:

5 0

Answer: $33.80

Step 1: Find out the cost for one candy bar

To find out the cost for one candy bar, you will need to divide $21.97 by 13. Another way to say this would be by rewriting the equation as 13x=21.97. After dividing, you see that one candy bar is $1.69.

Step 2: Find out the cost for 20 candy bars

To figure out the cost of 20 candy bars, simply multiply the price for one by 20. Another way to say this would be by rewriting this as 1.69x20=33.8.

Step 3: Your answer!

By completing the above steps, you have figured out the answer of $33.80.

Hope this helps!

You might be interested in

I upped the steaks plz answer this, I will give brainiest

Bad White [126]

Answer:

$\displaystyle x = \frac{3}{4}$

Step-by-step explanation:

we are given that

$\displaystyle P = \left(x, - \frac{ \sqrt{7} }{4} \right)$

we want to figure out x

remember that,

In unit circle x coordinate is considered cos and y coordinate is considered sin

so in order to figure out x we should figure out the value of cos to do we can consider Pythagoras trig indentity

given by

$\displaystyle \sin ^{2} ( \theta) + { \cos}^{2} ( \theta) = 1$

we are given sin=-√7/4 thus

substitute:

$\displaystyle \left( - \frac{ \sqrt{7} }{4} \right) ^{2} + { \cos}^{2} ( \theta) = 1$

simplify square:

$\displaystyle \frac{7}{16} + \cos ^{2} ( \theta) = 1$

move left hand side expression to right hand side and change its sign:

$\displaystyle \cos ^{2} ( \theta) = 1 - \frac{7}{16}$

simplify Substraction:

$\displaystyle \cos ^{2} ( \theta) = \frac{9}{16}$

square root both sides:

$\displaystyle \cos ^{} ( \theta) = \pm\frac{3}{4}$

since in unit circle cos is x and P belongs to Q:IV negative cos isn't applicable

hence,

$\displaystyle x = \frac{3}{4}$

5 0

3 years ago

Which expression results from using the distributive property 5(4x)

nignag [31]

20x would be ur answer.

6 0

1 year ago

Read 2 more answers

Answer this please .

Monica [59]

Answer:

$\frac{3}{5}x+2=\frac{2}{3}x\\\\\frac{3}{5}x+2-2=\frac{2}{3}x-2\\\\\frac{3}{5}x=\frac{2}{3}x-2\\\\\frac{3}{5}x-\frac{2}{3}x=\frac{2}{3}x-2-\frac{2}{3}x\\\\-\frac{1}{15}x=-2\\\\\left(-\frac{1}{15}x\right)\left(-15\right)=\left(-2\right)\left(-15\right)\\\\x=30$