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marishachu [46]
3 years ago
13

What is the length of the hypotenuse

Mathematics
2 answers:
lisov135 [29]3 years ago
8 0

Answer:

C=17

Step-by-step explanation:

15^2+8^2=c^2

C^2=289

C=17

igomit [66]3 years ago
6 0
The hypotenuse is 17cm you find it by using Pythagorean’s Theorem
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L.C.M of x2 + 3x + 2 and x2 + 4x + 3
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Answer:

Step-by-step explanation:

x^2 + 3x + 2 = (x+1)(x+2)

x^2 + 4x + 3 = (x+1)(x+3)

LCM = (x+1)(x+2)(x+3) = (x^2 + 3x + 2) (x+3) = x^3 + 6x^2 + 11x + 6

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Explain in detail with at least two sentences how to represent the product of A and B geometrically given that A=15+56i and B =
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Basically each part of the first complex number gets multiplied by each part of the second complex number. You are multiplying 2 binomials.

(15 + 56i(7 + 4i)

= 105 + 60i + 392i + 224i^2

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= - 119 + 452i


7 0
3 years ago
How do I find the volume of a prism?
frez [133]

Answer:

470 cm³

Step-by-step explanation:

Decompose the prism into two rectangular prism.

Prism 1 will have the following dimensions and volume:

Length = 10 cm

Height = 8 cm

Width = 9 - 5 = 4 cm

Volume = L × W × H

Volume = 10 × 8 × 4 = 320 cm³

Prism 2:

Length = 10 cm

Height = 3 cm

Width = 5 cm

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6 0
3 years ago
A box contains 15 resistors. twelve of them are labelled 50ω and the other three are labeled 100ω. what is the probability that
Igoryamba

Answer : P(second resistor is 100ω , given that the first resistor is 50ω) is given by

\frac{1}{5}

Explanation :

Since we have given that

Total number of resistors =15

Number of resistors labelled with 50ω = 12

Number of resistors labelled with 100ω =3

Let A: Event getting resistor with 50ω

B: Event getting resistor with 100ω

Since A and B are independent events .

So,

P(A\cap B)=P(A).P(B)

Now, According to question , we can get that

P(A)= \frac{12}{15}=\frac{4}{5}\\\\P(B)=\frac{3}{15}=\frac{1}{5}

So,

P(A\cap B)=P(A).P(B)\\\\P(A\cap B)=\frac{4}{5}\times \frac{1}{5}\\\\P(A\cap B)=\frac{4}{25}

So, by using the conditional probability , which state that

P(B\mid A)=\frac{P(A\cap B)}{P(A)}

P(B\mid A)=\frac{\frac{4}{25}}{\frac{4}{5}}\\\\P(B\mid A)=\frac{5}{25}\\\\P(B\mid A)=\frac{1}{5}

So, P(second resistor is 100ω , given that the first resistor is 50ω) is given by

\frac{1}{5}


4 0
3 years ago
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