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3 years ago

Rewrite using a single positive exponent

Mathematics

1 answer:

fiasKO [112]3 years ago

5 0

Step-by-step explanation:

$\frac{ {7}^{4} }{7} = {7}^{4 - 1} = {7}^{3}$

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Each of these numbers represents the weight of a student in kilograms. What is the median of this data set?

aleksandrvk [35]

Answer:

Step-by-step explanation:

www.alcula.com/calculators/statistics/median/

4 0

3 years ago

Given the equation: = 4, which property should be applied first?

Ghella [55]

This is = 4 therefore no property is applied

3 0

3 years ago

What's the answer to the equation of 4x^2+25 and show yor work please and thank you

padilas [110]

4x^2+25=4x^2=-25=x^2=-25/4 has no solution. because, negative number do not live under radical except complex number

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3 years ago

The area of a mirror is 225 Square inches, it's width is

Zepler [3.9K]

NO. The mirror will not fit in a space that is 15 inches by 16 inches

Solution:

Given that area of mirror is 225 square inches

$width = 13\frac{3}{4} \text{ inches }$

Converting the above mixed fraction we get,

$width = 13\frac{3}{4} = \frac{13 \times 4 + 3}{4} = \frac{55}{4} \text{ inches }$

Let us find the length of mirror

The area of mirror is given as:

area of mirror = length x width

Substituting the given values,

$225 = length \times \frac{55}{4}\\\\length = 225 \times \frac{4}{55}\\\\length = 225 \times 0.0727\\\\length = 16.36$

Thus length of mirror is 16.36 inches

Will the mirror fit in a space that is 15 inches by 16 inches?

NO. The mirror will not fit in a space that is 15 inches by 16 inches

Because length of mirror is 16.36 inches whereas the given space is 15 inches long

16.36 > 15 so the length of mirror will not fit inside the space

Also width of mirror is $\frac{55}{4} = 13.75$ inches which is less than the given space whose width is 16 inches

13.75 < 16 so the width of mirror will not fit inside the space

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3 years ago

Show a number line 2/3 3/3

matrenka [14]

2|------|------|------|3/3 or 1 whole
1/3. 2/3

6 0

3 years ago