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3 years ago

Express 3080 as a product of prime factors

1 answer:

n200080 [17]3 years ago

4 0

Prime Factors of 3080 =2, 2, 2, 5, 7, 11

Which is the same as = 23 x 5 x 7 x 11

Prime Factors Tree of 3080

3080
/ \
2 1540
/ \
2 770
/ \
2 385
/ \
5 77
/ \
7 11
/ \
11 1

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Which equation represents the hyperbola in general form?

valina [46]

ANSWER

$9 {y}^{2} - 4 {x}^{2}- 36y + 16x - 16 = 0$

EXPLANATION

The standard equation of the hyperbola is

$\frac{ {(y - 2)}^{2} }{4} - \frac{ {(x - 2)}^{2} }{9} = 1$

We multiply through by 36 to obtain:

$9 {(y - 2)}^{2} - 4( {x - 2)}^{2} = 36$

We now expand to get,

$9( {y}^{2} - 4y + 4) - 4( {x}^{2} - 4x + 4) = 36$

Expand :

$9 {y}^{2} - 36y + 36 - 4 {x}^{2} + 16x - 16 = 36$

To get the general form, we equate everything to zero to get,

$9 {y}^{2} - 4 {x}^{2}- 36y + 16x - 16 = 0$

The correct choice is D.

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3 years ago

What is the solution of the equation?

elena-s [515]

Answer:

Step-by-step explanation:

its c=5

6y+18+3=51

6y+21=51

6y=51-21

6y=30

y=5

4 0

3 years ago

Read 2 more answers

We want to find the zeros of this polynomial:

fredd [130]

Answer:

x = -3 and x = -3/2

Step-by-step explanation:

After writing down the polynomial, split it; put a line between 3x^2 and -18x. Look and 2x^3 + 3x^2 and -18x - 27 separately and factor them both:

p(x) = 2x^3 + 3x^2 - 18x -27

p(x) = x^2(2x+3) -9(2x+3)

Now notice how x^2 and -9 have the same factor (2x+3). That means x^2 and -9 can go together:

p(x) = (x^2 - 9)(2x+3)

Factor it once more because there's a difference of squares:

p(x) = (x+3)(x-3)(2x+3)

Now just plug in whatever makes the each bracket equal 0:

x = -3, x = 3, and x = -3/2

Those are your zeros.

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3 years ago

(8a^3÷27x^-3)1÷3 simplify

Damm [24]

Answer: (8a^3x^3)/81

Step-by-step explanation: Simplifying the expression :)

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3 years ago

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A restaurant menu has five kinds of soups, seven kinds of main courses, six kinds of desserts, and five kinds of drinks. If a cu

ryzh [129]

Answer: 1050

Step-by-step explanation:

Number of combinations of selecting r things out of n = $^nC_r=\dfrac{n!}{r!(n-r)!}$

such that $^nC_1=n$

Given: A restaurant menu has 5 kinds of soups, 7kinds of main courses, 6 kinds of desserts, and 5 kinds of drinks.

If a customer randomly selects one item from each of these four categories, then by fundamental counting principle , the number of different outcomes are possible = $^5C_1\times \ ^7C_1\times\ ^6C_1\times\ ^5C_1 =5\times7\times6\times5=1050$

hence, total number of outcomes = 1050

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3 years ago

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