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SSSSS [86.1K]
3 years ago
8

What is the answer to this problem? 5[8-(9x-7)]+6x=0

Mathematics
1 answer:
julia-pushkina [17]3 years ago
3 0
<span>5[8-(9x-7)]+6x=0
</span><span>5[8- 9x + 7]+6x=0
40 - 45x + 35 + 6x = 0
-39x = -75
     x = 1 36/39
     x = 1 12/13</span>
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Calculate the area of this trapezium.<br> 7.9 cm<br> 6.2 cm<br> 10.8 cm
cestrela7 [59]

Answer:

A = 57.97 cm²

Step-by-step explanation:

The area (A) of a trapezium is calculated as

A = \frac{1}{2} h (a + b)

where h is the perpendicular height and a, b the parallel bases

Here h = 6.2, a = 10.8 and b = 7.9, thus

A = \frac{1}{2} × 6.2 × (10.8 + 7.9) = 3.1 × 18.7 = 57.97 cm²

5 0
3 years ago
An urn contains 2 red marbles and 3 blue marbles. 1. One person takes two marbles at random from the urn and does not replace th
Ghella [55]

Answer:

A) The best way to picture this problem is with a probability tree, with two steps.

The first branch, the person can choose red or blue, being 2 out of five (2/5) the chances of picking a red marble and 3 out of 5 of picking a blue one.

The probabilities of the second pick depends on the first pick, because it only can choose of what it is left in the urn.

If the first pick was red marble, the probabilities of picking a red marble are 1 out of 4 (what is left of red marble out of the total marble left int the urn) and 3 out of 4 for the blue marble.

If the first pick was the blue marble, there is 2/4 of chances of picking red and 2/4 of picking blue.

B) So a person can have a red marble and a blue marble in two ways:

1) Picking the red first and the blue last

2) Picking the blue first and the red last

C) P(R&B) = 3/5 = 60%

Step-by-step explanation:

C) P(R&B) = P(RB) + P(BR) = (2/5)*(3/4) + (3/5)*(2/4) = 3/10 + 3/10 = 3/5

3 0
2 years ago
Parallelly lines/angle problem
rodikova [14]

Answer:

142 degrees

Step-by-step explanation:

4 0
2 years ago
What is 17/20 as a percentage?
Ksju [112]
The percentage is 85 percent
8 0
2 years ago
Read 2 more answers
how much material was used in the manufacture of 24000 celluloid dice,if each dice has an edge 1/4 inches?
Dovator [93]
<span>The volume of each celluloid die is (.25 x .25 x .25) = 0.015625 cubic inch.

To manufacture 24,000 of them, you need to start with <u>at least</u>

(24,000) x (0.015625) = <u>375 cubic inches</u>.

I don't know how celluloid is sold, so you should also keep in mind that
375 cubic inches = about 207.8 fluid ounces, or about 6.5 quarts.

I'm sure a bit more than that was used in the manufacture, since
there's always some wasted, spilled, or trimmed off of the edges. </span>
4 0
3 years ago
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