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3 years ago

Can a right triangle be formed using these squares

Mathematics

1 answer:

Ilya [14]3 years ago

7 0

Yes. The two smaller squares have a sum of 169 which is the value of the larger square.

a^2 + b^2 = c^2

25 + 144 = 169

It can be done. Notice the figure below shows you how to arrange the squares to give the answer of a^2 + b^2 = c^2

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victus00 [196]

Answer:

40 marbles

Step-by-step explanation:

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2 years ago

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2 years ago

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2 years ago

Y=2x^2+12x+14 in vertex form

balandron [24]

Answer:

y = 2(x + 3)² - 4

Step-by-step explanation:

The equation of a parabola in vertex form is

y = a(x - h)² + k

where (h, k) are the coordinates of the vertex and a is a multiplier

Using the method of completing the square

y = 2x² + 12x + 14 ← factor out 2 from the first 2 terms

= 2(x² + 6x) + 14

To complete the square

add/subtract ( half the coefficient of the x- term)² to x² + 6x

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3 years ago

Suppose p(a) = 0.40 and p(a 

sveticcg [70]

Between the probability of union and intersection, it's not clear what you're supposed to compute. (I would guess it's the probability of union.) But we do know that

$P(A\cup B)+P(A\cap B)=P(A)+P(B)$

For parts (a) and (b), you're given everything you need to determine $P(B)$ .

For part (c), if $A$ and $B$ are mutually exclusive, then $P(A\cap B)=0$ , so $P(A\cup B)=P(A)+P(B)$ . If the given probability is $P(A\cup B)=0.55$ , then you can find $P(B)=0.15$ . But if this given probability is for the intersection, finding $P(B)$ is impossible.

For part (d), if $A$ and $B$ are independent, then $P(A\cap B)=P(A)\cdot P(B)$ .

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3 years ago