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gregori [183]
2 years ago
6

Allen has a photo that is 6 in by 8 in. What will the dimensions of the photo be if he scales it down by a factor of ? (5 points

) Group of answer choices 3 in by 4 in 6 in by 8 in 12 in by 16 in 1.5 in by 2 in
Mathematics
1 answer:
Irina18 [472]2 years ago
7 0

Answer:

It will be 3 in by 4 in

Step-by-step explanation:

Original dimension is 6 in by 8 in

If he scale it down by a factor of 1/2

The new dimension will be

1/2(6) by 1/2(8)

= 3 by 4

= 3 in by 4 in

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Chau will run at most 28 miles this week. So far, he has run 18 miles. What are the possible numbers of additional miles he will
AysviL [449]

Answer:

Step-by-step explanation:

18 + x ≤ 28

x ≤ 28 - 18

x ≤ 10

1 ≤ x ≤ 10

3 0
3 years ago
Find the value of x.
Irina-Kira [14]
The 2 angles are complementary angles and need to equal 90 degrees.

Making an equation we have:

4x-10 + x = 90 degrees

Add like tems:
4x +x = 5x

Now we have:
5x-10 = 90

Add 10 to both sides:

5x = 100

Divide both sides by 5:

x = 100/5

X = 20


5 0
3 years ago
Read 2 more answers
Mrs. Fronkos class for evry 6 boys there are 8 girls. Write the ratio of boys to girls in simplest form
Rama09 [41]

Answer:

3:4

Step-by-step explanation:

there are 6 boys for every 8 girls, so the initial ratio is 6:8. however, this can be simplified by finding a number both 6 and 8 are evenly divisible by. In this case, you can divide the ratio by 2. 6/2 = 3 and 8/2 = 4, which is why the simplified form is 3:4

5 0
3 years ago
Solve equation <br> 9x^2+7x=3
Nonamiya [84]

I’m not for sure but if I had to guess I would say 115

3 0
3 years ago
In the isosceles △ABC m∠ACB=120° and AD is an altitude to leg BC . What is the distance from D to base AB , if CD=4cm?
7nadin3 [17]

Correct answer is: distance from D to AB is 6cm

Solution:-

Let us assume E is the altitude drawn from D to AB.

Given that m∠ACB=120° and ABC is isosceles which means

m∠ABC=m∠BAC = \frac{180-120}{2}=30

And AC= BC

Let AC=BC=x

Then from ΔACD , cos(∠ACD) = \frac{DC}{AC} =\frac{4}{x}

Since DCB is a straight line m∠ACD+m∠ACB =180

                                              m∠ACD = 180-m∠ACB = 60

Hence cos(60)=\frac{4}{x}

          x=\frac{4}{cos60}= 8

Now let us consider ΔBDE, sin(∠DBE) = \frac{DE}{DB} =\frac{DE}{DA+AB} = \frac{DE}{4+8}

DE = 12sin(30) = 6cm

7 0
3 years ago
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