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katovenus [111]
4 years ago
15

A garden is rectangular with an area of 6/7 square miles. If the the length of the garden is 3/5 miles, with its width? Answer w

ith a fraction in simplest form.
Mathematics
2 answers:
Alik [6]4 years ago
6 0
Find the common denominator and sub tract3/5 
Olegator [25]4 years ago
3 0
Area= length x width
area=6/7, length=3/5, width=?

6/7=3/5 x width
6/7 divided by 3/5= 6/7 x 5/3 (you have to multiply by the recipricol, or flip the fraction.)

6/7 x 5/3=30/21= 1 3/7=width
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Let f(x)=16x^5-48x^4-8x^3 and g(x)= 8x^2. find f(x)/g(x)
Gekata [30.6K]

Answer:

2x³ - 6x² - x

Step-by-step explanation:

Divide each term in f(x) by g(x), that is

\frac{16x^5}{8x^2} - \frac{48x^4}{8x^2} - \frac{8x^3}{8x^2}

= 2x^{(5-2)} - 6x^{(4-2)} - x^{(3-2)}

= 2x³ - 6x² - x

3 0
4 years ago
Read 2 more answers
Find the values of x y and z
andrey2020 [161]

9514 1404 393

Answer:

  • x = 18√2
  • y = 18√3
  • z = 36

Step-by-step explanation:

These "special" right triangles have side length ratios that it is useful to remember.

<u>45°-45°-90° triangle</u>

Sides have the ratios 1 : 1 : √2. That is, x is √2 times as long as the side length shown as 18.

  x = 18√2

<u>30°-60°-90° triangle</u>

Shortest to longest, sides have the ratios ...

  1 : √3 : 2

That is, y is √3 times the length of the side marked 18, and z is 2 times the length of the side marked 18.

  y = 18√3

  z = 2·18 = 36

4 0
3 years ago
the PE clasd has 12 boys and 8 girls what is the maximum number of teams that the teacher can divide the class into so that each
marysya [2.9K]

<u>Answer:</u>

4

<u>Step-by-step explanation:</u>

We are given that a PE class has 12 boys and 8 girls. The teacher needs to divide in maximum number of teams so that each team has an equal number of boys and girls.

For this, we will factors of 12 and 8 and their greatest common factor.

12 = 4 × 3 = 2 × 2 × 3

8 = 4 × 2 = 2 × 2 × 2

The greatest common factor is 4, therefore the maximum number of teams with equal number of boys and girls is 4.

7 0
3 years ago
Para embalar una caja se emplea 4,2 m de cinta adhesiva. ?Cuántas cajas se podrán embalar con tres rollos que tienen 3 hm, 7 dam
kompoz [17]

For this case, we perform the conversions:

First roll:

1 \ Hectometer --------> 100 \ meters\\3 \ Hectometer --------> x

x = \frac {3 * 100} {1}\\x = 300 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 300 meters of adhesive tape.

1 -----------> 4.2

c -----------> 300

c = \frac {300 * 1} {4.2}\\c = 71.42857143\\c = 71

You can pack 71 boxes.

Second roll:

1 \ Decametro --------> 10 \ meters\\7 \ Decameter --------> x\\x = \frac {7 * 10} {1}\\x = 70 \ meters.

We make a rule of three to determine the number of "c" boxes that can be packed with 70 meters of adhesive tape.

1 -----------> 4.2

c -----------> 70

c = \frac {70 * 1} {4.2}\\c = 16.66666667\\c = 16

You can pack 16 boxes.

Third roll:

1 -----------> 4.2

c -----------> 50

c = \frac {50 * 1} {4.2}\\c = 11.9047619\\c = 11

You can pack 11 boxes.

Thus, in total you can pack11 + 16 + 71 = 98 \ boxes

Answer:

98 boxes

4 0
4 years ago
Can someone please help with this ?
creativ13 [48]

Answer:     -9n+20

This is the same as 20-9n

================================================

Explanation:

The jump from 11 to 2 is "minus 9"

The jump from 2 to -7 is also "minus 9".

Assuming this pattern continues on, we have an arithmetic sequence with

  • a = 11 = first term
  • d = -9 = common difference

The nth term can be found like so

a_n = a + d(n-1)\\\\a_n = 11 + (-9)(n-1)\\\\a_n = 11 -9n + 9\\\\a_n = -9n+20\\\\

Let's check the answer by trying n = 3

a_n = -9n+20\\\\a_3 = -9*3+20\\\\a_3 = -27+20\\\\a_3 = -7\\\\

This shows the third term is -7, which matches what the original sequence shows. The answer is partially confirmed. I'll let you check the other values of n. You should get 11 when trying n = 1, and you should get 2 when trying n = 2.

4 0
2 years ago
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