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Pepsi [2]
3 years ago
5

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

Mathematics
1 answer:
Iteru [2.4K]3 years ago
5 0

Answer:

3

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 <br>2y=2x+12<br>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 ! <br><br> -9v-6+7v=6 <br> 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:

C

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

<em><u>Solution:</u></em>

Let "a" be the price of each senior ticket

Let "b" be the price of each child ticket

<em><u>On the first day of ticket sales the school sold 4 senior tickets and 4 child tickets for a to total of 68</u></em>

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

<em><u>The school took in 120 on the second day by selling 12 senior tickets and 5 child tickets</u></em>

Similarly, we frame a equation as:

12 \times a + 5 \times b = 120

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

<em><u>Let us solve eqn 1 and eqn 2</u></em>

<em><u>Multiply eqn 1 by 3</u></em>

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

<em><u>Subtract eqn 2 from eqn 3</u></em>

12a + 12b = 204

12a + 5b = 120

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

7b = 84

b = 12

<em><u>Substitute b = 12 in eqn 1</u></em>

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
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