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ss7ja [257]
3 years ago
6

Write an expression that has three terms and simplifies to 4x-7.

Mathematics
1 answer:
nikklg [1K]3 years ago
4 0
You get 3x + 4x, +4- -3,and +3 + 4
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Please do attached question
kondaur [170]

Answer:

ummmmmmmmmmmmmmmmm

Step-by-step explanation:

hmmmmmmmmmmmm

6 0
3 years ago
the circumference of a circular painting is 40.82 ft. what is the radius of the painting? use 3.14 do not round
diamong [38]
Your answer would be about 20.41
7 0
3 years ago
How do you do this?<br> Please help
Assoli18 [71]

Answer:

36 pencils

Step-by-step explanation:

Let h and p represent the number of highlighters and the number of pencils, respectively.

Then h + p = 45, and h = 45 - p.

Tom paid a total of $30 for these supplies, with ($2/highligher)(h) + ($0.333/pencil) adding up to that amount.

substituting 45 - p for h in 2h + 0.333p = 30, we get:

2(45 - p) + 0.333p = 30, or

90 - 2p + 0.333p = 30

Combine the constants:  60 = 2p - 0.333p, or 60 = 1.667p

Then p = 60/1.667 = 35.9928, or 36.

Tom bought 36 pencils for $12, and 45-36, or 9, highlighters for $18, for a total purchase of $30.  This shows that these calculations are correct.

7 0
3 years ago
Read 2 more answers
A man travels 600km partly by train and partly by car. If he covers 400km by train and the rest by car, it takes him 6 hours and
11Alexandr11 [23.1K]

Answer:

A man travels 600km partly by train and partly by car. If he covers 400km by train and the rest by car, it takes him 6 hours and 30 minutes. But if he travels 200 km by train and rest by car, he takes half an hour longer.find the speed of the train and that of the car

Step-by-step explanation:

4 0
3 years ago
In tossing four fair dice, what is the probability of tossing, at most, one 3
11Alexandr11 [23.1K]

<u>Answer- </u>

In tossing four fair dice, the probability of getting at most one 3 is 0.86.

<u>Solution-</u>

The probability of getting at most one 3 is, either getting zero 3 or only one 3.

P(A) =The \ probability \ of \ getting \ zero \ 3 =(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6})=\frac{625}{1296} =0.48                    ( ∵ xxxx )

P(B) = The \ probability \ of \ getting \ only \ one \ 3 =(4)(\frac{1}{6})(\frac{5}{6})(\frac{5}{6})(\frac{5}{6}) = 0.38                    ( ∵ 3xxx, x3xx, xx3x, xxx3 )

P(Atmost one 3) = P(A) + P(B) = 0.48 + 0.38 = 0.86

7 0
3 years ago
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