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andreev551 [17]
3 years ago
12

Which set of side lengths form a right triangle?

Mathematics
1 answer:
astraxan [27]3 years ago
4 0
Since we don't want to use the Pythagorean Theorem on all of these, first we look for some Pythagorean Triples. Its very helpful to know some of the common ones, so here's a list:

3, 4, 5
5, 12, 13
7, 24, 25
8, 15, 17

So we either look for these or a multiple of these numbers.

The last one is the only one that fits because 15, 20, 25 is a multiple of the triple 3, 4, 5 by multiplying all of them by 5. 

Thus, 15, 20, 25 forms a right triangle. 
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Simplify the expression (8n^4-6n+8n^2)+(6n^4-3n+7)
Marysya12 [62]

Answer:

=14n⁴+8n²-9n+7

Step-by-step explanation:

(8n⁴-6n+8n²)+(6n⁴-3n+7)

=8n⁴-6n+8n²+6n⁴-3n+7

=8n⁴+6n⁴+8n²-6n-3n+7

=14n⁴+8n²-9n+7

8 0
2 years ago
How can you use similar triangles to find the missing parts of a triangle?
REY [17]
I think the blanks would be, "sides" and "angles."
6 0
3 years ago
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Work out the volume of a prism with an area of 10m squared and a length of 8m
Jobisdone [24]
Is it 80 m I think that’s the answer
4 0
3 years ago
A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it
nata0808 [166]

Answer:

P(A) = 3/20

Step-by-step explanation:

P(A)=P(blue)P(head)=(3/10)(1/2)=3/20

as there are 10 cards in total, out of which 3 are blue so the probability to get the blue card is, P(blue) = 3/10. and the probability of getting a head when a coin is tossed is P(head) = 1/2.

So in total

P(A) = P(blue)*p(head) = (3/10)*(1/2) = 3/20 = 0.15

8 0
3 years ago
Brandon graphed one line in a system of equations. The system has only one solution, (0,3). Which equations could also be part o
AlladinOne [14]

Answer:

y = 3 and y = x + 3

Step-by-step explanation:

The equation y = 3 whose graph is a line parallel to x-axis can be a part of the system.

Consider the equation y = x + 3.

Substitute (0, 3).

3 = 0 + 3

So, (0, 3) lies on y = x + 3 and this can also be a part of the system.

6 0
3 years ago
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