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zepelin [54]
2 years ago
12

4\sqrt(3)-3\sqrt(12)+2\sqrt(75)

Mathematics
1 answer:
postnew [5]2 years ago
5 0

Answer:

8√3 Decimal Form: 13.85640646

Step-by-step explanation:

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Select the property of equality used to arrive at the conclusion. If x = 4, then 5x = 20 the addition property of equality the m
shtirl [24]

Answer:

The Division Property of Equality

Step-by-step explanation:

<u>The Addition Property of Equality:</u> When you add something to one side of the equation, you must add the same thing to the other side.

<u>The Subtraction Property of Equality:</u> When you subtract something from one side of the equation, you must subtract the same thing from the other side.

<u>The Multiplication Property of Equality:</u> When you multiply something to one side of the equation, you must multiply the same thing to the other side.

<u>The Division Property of Equality:</u> When you divide something from one side of the equation, you must divide the same thing from the other side.

In this case, you have to divide both sides of the equation by 5 to get x = 4. That means that the division property of equality was used.

I hope this helps! Have a great day!

8 0
2 years ago
Read 2 more answers
1. (01.02)
Anton [14]

Answer:4

Step-by-step explanation:

f(1)=6(1)+2

=8

g(f(1))=2(8)+4/5

=20/5

=4

6 0
2 years ago
I need help on these as well
kvv77 [185]

Answer:

5. 2

6. -1

7. undefined

Step-by-step explanation:

6 0
3 years ago
If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w
MatroZZZ [7]

Let x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11} be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.

2. If the mean weight of the 7 other players is 202 lb, then

\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.

3. From the previous statements you have that

x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.

Add these two equalities and then divide by 11:

x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.

Answer: the mean weight of the 11-person team is 208\dfrac{10}{11}\ lb.

4 0
2 years ago
Log base 8 of 1 over 32
Vanyuwa [196]
Upload a pic like take a pic
8 0
3 years ago
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