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Marizza181 [45]
3 years ago
9

If a tree in a 1 in. : 2 ft. scale drawing is 10 inches tall, how tall is the actual tree?

Mathematics
1 answer:
Citrus2011 [14]3 years ago
4 0

Answer:

20 feet tall

Step-by-step explanation:

Since it is 1 in: 2ft and the drawing is 10 in, you multiply 10 by 2 and get 20 feet.

(hope this helps!)

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Which of the following lists of fractions are in order from least to greatest? Select TWO that apply.
lora16 [44]

Answer:

i think its a 1/4 then 2/3 and 3/10

4 0
3 years ago
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How can you use distributive property to find 3/4 times 2 and 1/3
zubka84 [21]

Answer:

1 3/4

Step-by-step explanation:

You can set it up like the distrubtive property first

3/4(2+ 1/3)    Next you use the distirbutive property instead of adding inside the parenthesis and you multiply the 3/4 by 2 and then the 3/4 by 1/3

Use 2 as 2/1 in fraction problems

3/4 * 2/1        Now multiply the numerators together and the denomenators together

3 * 2 = 6          4 * 1 =4

6/4

3/4 *1/3

3 * 1 =3

4 * 3 =12

3/12

Now we have to make them both like terms (same denominator) so we can add them, we will multiply the 6/4 by 3 (Both top and bottom)

6 * 3 = 18               4 *3 =12

18/12

Now we add the fractions

3/12 +18/12 = 21/12

Reduce that and you get

21/12 = 1 9/12  or 1 3/4

5 0
3 years ago
Find the slope between (-4,1) and (-2,-2)
alukav5142 [94]

Answer:

-1.5

Step-by-step explanation:

y=-1.5x-5

7 0
3 years ago
Expansion Numerically Impractical. Show that the computation of an nth-order determinant by expansion involves multiplications,
posledela

Answer:

  • number of multiplies is n!
  • n=10, 3.6 ms
  • n=15, 21.8 min
  • n=20, 77.09 yr
  • n=25, 4.9×10^8 yr

Step-by-step explanation:

Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...

  mpy[n] = n·mp[n-1]

  mpy[2] = 2

So, ...

  mpy[n] = n! . . . n ≥ 2

__

If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...

  10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10

Then the larger matrices take ...

  n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min

  n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years

  n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years

_____

For the shorter time periods (less than 100 years), we use 365.25 days per year.

For the longer time periods (more than 400 years), we use 365.2425 days per year.

8 0
3 years ago
A student is making cookies for a bake sale. He sells 8 snickerdoodles and 12 chocolate chip
lions [1.4K]

Answer:

Snickerdoodles are 0.22$ each and chocolate chip are 0.43$ each

Step-by-step explanation:

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You can now take that equation and plug it into equation number 2:

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solve that equation by multiplying 5 through the brackets: 5x-10/3x+34.6/12=3.25

add common terms: 5/3x=4.4/12

find the value of x, you will get x=0.22

plug the value of x (0.22) into one of your equations and solve for the y-value

8 0
3 years ago
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