1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nata [24]
3 years ago
8

Mikaela placed a frame around a print that measures 10 inches by 10 inches. The area of just the frame itself is 69 square inche

s. What is the width of the frame?

Mathematics
1 answer:
olchik [2.2K]3 years ago
3 0

Answer:

1.5 in

Step-by-step explanation:

Let x be the width of the frame.

Side of print=10 in

Area of frame=69 square in

We have to find the width of the frame.

Side of frame=10+x+x=10+2x

Area of square=(side)^2

By using the formula

Area of print=10\times 10=100in^2

Area of frame with print=(10+2x)^2

Area of frame=Area of frame with print-Area of print

69=(10+2x)^2-100

(10+2x)^2=69+100=169

(10+2x)=\sqrt{169}=\pm 13

10+2x=13

Because Side is always positive.

2x=13-10=3

x=\frac{3}{2}=1.5

Hence, width of frame=1.5 in

You might be interested in
BRAINIST & 20 POINTS!!
valentina_108 [34]

Answer:

y=2/3x-5

Step-by-step explanation:

First, let's put this equation into slope-intercept form.

2x-3y=24

-3y=-2x+24

y=2/3x-8

For two lines to be parallel, they must have the same slope. Therefore our slope is 2/3.

Now, we must find b. To do that we can plug the y and the x in.

-7=2/3(-3)+b

-7=-2+b

-5=b

y=2/3x-5

8 0
3 years ago
Read 2 more answers
What is 3.4 rounded to the nearest whole number?
ivolga24 [154]

Answer:3

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Two similar triangles are shown below:
lana [24]
The answer is b. angles p and s; angles q and r

3 0
3 years ago
Read 2 more answers
What is the maximum number of songs she can downloads using the voucher
lozanna [386]

You need to be more specific and show me the problem.

6 0
3 years ago
The table show the age in years of employees in a company
adelina 88 [10]

Answer:

A. 24 ≤ a < 26.

B. 22.5

Step-by-step explanation:

A. Determination of the modal class interval.

Mode is the class with the highest frequency.

From the table given above, the highest frequency is 8, therefore the class will the highest frequency is:

24 ≤ a < 26.

B. To obtain the mean, we must determine the class mark. This is illustrated below:

Class >>>>> class mark >>> frequency

18 – 19 >>>> 18.5 >>>>>>>>> 3

20 – 21 >>> 20.5 >>>>>>>> 2

22 – 23 >>> 22.5 >>>>>>>> 7

24 – 25 >>> 24.5 >>>>>>>> 8

26 >>>>>>>> 26 >>>>>>>>> 0

The mean is given by the summation of the product of the class mark and frequency divided by the total frequency. This is illustrated below:

Mean = [(18.5x3) + (20.5x2) + (22.5x7) + (24.5x8) + (26x0)] / (3+2+7+8+0)

Mean = (55.5 + 41 + 157.5 + 196 + 0)/20

Mean = 450/20

Mean = 22.5

Therefore, the mean age is 22.5

4 0
3 years ago
Read 2 more answers
Other questions:
  • HELP WILL MARK BRAINLIEST
    5·1 answer
  • What is 5+5 I’m having lots of trouble
    7·2 answers
  • 3t(t+2)-3t^2 for t=19
    11·1 answer
  • Solve the compound inequality and graph the solutions.<br><br> −3 &lt; 3x ≤ 12
    9·1 answer
  • A small rural town has experienced a population decrease at a rate of 25 people every 5 years since 1970. If the population of t
    13·1 answer
  • Which equation is equivalent to 4.5 = 2a?<br> a = 2.5<br> a = 2.25<br> 4.5a = 2<br> a + 2 = 4.5
    13·2 answers
  • Kendra was asked to find the number of students who planned to attend the evening symphony performance. She was told that 40 per
    15·2 answers
  • It was estimated that 825 people would attend the soccer game , but 817 people actually attended what is the percent error , to
    10·2 answers
  • What are the intercepts for 7x - 5 = 4y -6<br>​
    7·2 answers
  • I Usually don't ask for help but I need help I give 35 points if you help me and give me the answer also with the explanation Th
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!