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3 years ago

What is the highest common factor (HCF) of 9 and 15?

Mathematics

1 answer:

Iteru [2.4K]3 years ago

5 0

Answer:

Step-by-step explanation:

9's factor is 1, 3, 9.

and 15's factor is 1, 3, 5, 15.

so, the answer is 3.

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Solve the system of linear equations using substitution 2y=2x+12 y=-2-3

velikii [3]

Answer:

x=-11, y=-5. (-11, -5).

Step-by-step explanation:

2y=2x+12

y=-2-3

-----------------

simplify 2y=2x+12 into y=x+6

y=-5

------------------

-5=x+6

x=-5-6=-11

6 0

3 years ago

Use the lists of multiples to answer the problem. Multiples of 8: 8, 16, 24, 32, 40, . . . Multiples of 12: 12, 24, 36, 48, 60,

patriot [66]

Answer:

24 minutes

Step-by-step explanation:

Look at the lists of multiples and find the first instance of the same number appearing in both lists.

Multiples of 8: 8, 16, 24, 32, 40, ...

Multiples of 12: 12, 24, 36, 48, 60, ...

The first number that appears in both lists is 24, so the answer is 24.

Answer: 24 minutes

7 0

3 years ago

Read 2 more answers

Please help ! -9v-6+7v=6 what is v?

DENIUS [597]

-9v-6+7v=6
-2v-6=6
-2v=6+6
-2v=12
v=-6

8 0

3 years ago

Help OvO with this math question-

harkovskaia [24]

Answer:

Step-by-step explanation:

40 x 80 = 3200

120 x 100 = 12000

3200 + 12000 = 15200

6 0

2 years ago

Read 2 more answers

Mark's school is selling tickets to the annual dance competition. On the first day of ticket sales the school sold 4 senior tick

erica [24]

The cost of each senior ticket is $ 5 and cost of each child ticket is $ 12

Solution:

Let "a" be the price of each senior ticket

Let "b" be the price of each child ticket

On the first day of ticket sales the school sold 4 senior tickets and 4 child tickets for a to total of 68

Thus a equation is framed as:

4 senior tickets x price of each senior ticket + 4 child tickets x price of each child ticket = 68

$4 \times a + 4 \times b = 68$

4a + 4b = 68 ---------- eqn 1

The school took in 120 on the second day by selling 12 senior tickets and 5 child tickets

Similarly, we frame a equation as:

$12 \times a + 5 \times b = 120$

12a + 5b = 120 ---------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 1 by 3

12a + 12b = 204 -------- eqn 3

Subtract eqn 2 from eqn 3

12a + 12b = 204

12a + 5b = 120

( - ) --------------

7b = 84

b = 12

Substitute b = 12 in eqn 1

4a + 4(12) = 68

4a + 48 = 68

4a = 20

a = 5

Thus cost of each senior ticket is $ 5 and cost of each child ticket is $ 12

4 0

3 years ago