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3 years ago

What is the 6th factor in 42

Mathematics

1 answer:

torisob [31]3 years ago

6 0

1, 2, 3, 6, 7, 14, 21, 43

6th one is 14

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If three pears and four oranges cost $4.85 and three pears and ten oranges cost $8.75 what is the cost of a pear and an orange

marishachu [46]

Answer:

1 pear = $0.75; 1 orange = $0.65

Step-by-step explanation:

(1) 3P + 4O = 4.85

(2) 3P + 10O = 8.75 Eqn (2) - Eqn (1)

3P + 10O – 3P – 4O = 8.75 – 4.85 Combine like terms

6O = 3.90 Divide each side by 6

O = $0.65 Substitute into Eqn (1)

3P + 4×0.65 = 4.85

3P + 2.60 = 4.85 Subtract 2.60 from each side

3P = 2.25 Divide each side by 3

P = $0.75

Oranges cost $0.65 each and pears are $0.75 each

4 0

3 years ago

Diego cut 7 smaller boards of equal length from a board that is 9 1/3 feet long. How long is each of the 7 smaller boards? Write

Temka [501]

It would be 9 1/3 divided by 7/1 or 28/3 * 1/7

It would give you 28/21 which would simplify to 4/3 and still leave you with an improper fraction

4 0

3 years ago

In a circle of radius 10 cm, a sector has an area of 40 (pi) sq. cm. What is the degree measure of the arc of the sector?

Rashid [163]

Answer:

144°

Step-by-step explanation:

First, find the area of the circle, with the formula A = $\pi$ r²

Plug in 10 as the radius, and solve

A = $\pi$ r²

A = $\pi$ (10²)

A = 100 $\pi$

Using this, create a proportion that relates the area of the sector to the degree measure of the arc.

Let x represent the degree measure of the arc of the sector:

$\frac{40\pi }{100\pi }$ = $\frac{x}{360}$

Cross multiply and solve for x:

100 $\pi$ x = 14400 $\pi$

x = 144

So, the degree measure of the sector arc is 144°

6 0

3 years ago

Read 2 more answers

How to do this problem??

lorasvet [3.4K]

We know that the sum of the inner angles of any triangle is 180º

72º + (7x + 3)º + (3x + 5)º = 180º
72º + 7xº + 3º + 3xº + 5º = 180
7xº + 3xº = 180º - 72º - 3º - 5º
10xº = 100º
$x^0 = \frac{100^0}{10^0}$
$x^0 = 10\to\:\boxed{\boxed{x = 10^0}}\end{array}}\qquad\quad\checkmark$

The sum of the external angle (9y + 1)º with inner angle (3x + 5) = 180 °, <span>Replace the measure of "x" found:
</span>
(9y + 1)º + (3x + 5)º = 180º
9yº + 1º + 3xº + 5º = 180º
9yº + 1º + 3.(10)º + 5º = 180º
9yº + 1º + 30º + 5º = 180º
9yº = 180º - 1º - 30º - 5º
9yº = 144º
$y^0 = \frac{144^0}{9^0}$
$y^0 = 16\to\:\boxed{\boxed{y = 16^0}}\end{array}}\qquad\quad\checkmark$

Answer:
<span>The measures of "x" and "y" are respectively: 10º and 16º</span>

6 0

3 years ago

Last year during November there was a heatwave. The average daily temperature went from 57°F to 76°F. What was the percent incre

lesya692 [45]

Answer:

hhhh

Step-by-step explanation:

hhhhhh

8 0

3 years ago