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olga nikolaevna [1]
3 years ago
13

The value of x is? Help I cant find the answer to this

Mathematics
1 answer:
Sindrei [870]3 years ago
5 0
X=3

45x=25x+(57+x)
45x=26x+57

Subtract 26x from both sides

19x=57

Divide both sides by 19

X=3
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2. Find the perimeter of a rectangle of<br> Length 13m with an area of 65m
I am Lyosha [343]

Answer:

Perimeter = 36m

Step-by-step explanation:

Area = l x w

65m = 13m x w

5m = w

Perimeter = 2(l + w)

Perimeter = 2(13m + 5m)

Perimeter = 36m

3 0
3 years ago
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How much fencing will John need? Use 22/7 for x.
Serjik [45]

Well 22/7 is 3.14 and so on but since there is a little more than 3 you need to make sure he covers up all the fencing places... SO John will need 4 fences

4 0
3 years ago
Solve the system using combination method to get a solution (x,y)<br> 4x+7y=10<br> -4x-6y=28
PIT_PIT [208]
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3 0
3 years ago
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9 – 3 (2+6)÷6 -2 × 5? step by step please
RoseWind [281]
9-3(2+6)/6-2x5

9-3x8/6-2x5

9-3x8/6-10

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5 0
3 years ago
What is the probability that the number of “HEADs” on these four coins is equal to 3? (4 points)
slavikrds [6]

Answer:

a. \frac{1}{4}  

Step-by-step explanation:  

We are asked to find the probability of getting 3 heads on 4 flips.

\text{Probability}=\frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}

Since we know that flipping a fair coin has 2 equally likely possible outcomes, so flipping four coins will have 2*2*2*2=16 possible outcomes.

Sample space of possible outcomes.

HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT,

THHH, THHT, THTH,THTT, TTHH, TTHT, TTTH, TTTT.

We can see that there are 4 favorable outcomes of getting heads.

\text{Probability of getting 3 heads}=\frac{4}{16}

\text{Probability of getting 3 heads}=\frac{1}{4}

Therefore, the probability of getting 3 heads on 4 coins will be \frac{1}{4} and option a is the correct choice.

5 0
3 years ago
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