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8_murik_8 [283]
3 years ago
7

Factorise the following: a) x + 3x - 4 b) x - 2x - 3 c) x + 2x - 8

Mathematics
1 answer:
KatRina [158]3 years ago
6 0

Answer:

a.x+3x-4

=4x-4

1(x-1).

b.x-2x-3

=x-3

c. x+2x-8

=3x-8

x(3-8)

x(-5)

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in a right triangle, the length of a hypotenuse is 20 inches and the length of one leg is 15 inches. what is the length of the o
Nata [24]
You'll have to use the Pythagorean theorem:

c^{2}  = a^{2} + b^{2}

In your case:

c^{2} = 20^{2} + 15^{2}
c^{2} = 400 + 225

c^{2} = 625
c =  \sqrt{625}
c = 25 inches


Answer:

\boxed{\bf~The~answer~is~25~in.}


Hope it helped,

Happy homework/ study/ exam!


5 0
3 years ago
Find the volume of a cube with area of 1532m2
skelet666 [1.2K]

Answer:

4080.01

Step-by-step explanation:

V = 6 A2/3/36 square rooted = 6 × 1532 2/3/36 square rooted = 4080.01042

5 0
2 years ago
3.16 x 10x10x10 please tell me the answer please
kkurt [141]

Your answer would be 3160.

7 0
3 years ago
Paul and Richard are dividing 36.25 divided by 5 Paul says that the first digit of the quotient is in the ones Place Richardson
shepuryov [24]
36.25/5 = 7.25

The first digit of the quotient is in the ones. Paul is correct.
7 0
3 years ago
Read 2 more answers
Each of the 25 balls in a certain box is either red, blue, or white and has a number from 1 to 10 painted on it. If one ball is
dem82 [27]

Answer:

Probability ball: white - P(W)P(W);

Probability ball: even - P(E)P(E);

Probability ball: white and even - P(W&E).

Probability ball picked being white or even: P(WorE)=P(W)+P(E)-P(W&E).

(1) The probability that the ball will both be white and have an even number painted on it is 0 --> P(W&E)=0 (no white ball with even number) --> P(WorE)=P(W)+P(E)−0P(WorE)=P(W)+P(E)−0. Not sufficient

(2) The probability that the ball will be white minus the probability that the ball will have an even number painted on it is 0.2 --> P(W)−P(E)=0.2P(W)−P(E)=0.2, multiple values are possible for P(W)P(W) and P(E)P(E) (0.6 and 0.4 OR 0.4 and 0.2). Can not determine P(WorE)P(WorE).

(1)+(2) P(W&E)=0 and P(W)−P(E)=0.2P(W)−P(E)=0.2 --> P(WorE)=2P(E)+0.2P(WorE)=2P(E)+0.2 --> multiple answers are possible, for instance: if P(E)=0.4P(E)=0.4 (10 even balls) then P(WorE)=1P(WorE)=1 BUT if P(E)=0.2P(E)=0.2 (5 even balls) then P(WorE)=0.6P(WorE)=0.6. Not sufficient.

Answer: E.

6 0
3 years ago
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