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2 years ago

Name two pairs of congruent angles using the figure shown.

Mathematics

1 answer:

Troyanec [42]2 years ago

4 0

Option 1, they are congruent angles because they are base angles of a triangle. Therefore, they are congruent

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We want to find the zeros of this polynomial: p(x)= 3x^3 – 3x^2– 18x

bulgar [2K]

Answer:

x=0 x=3 x=-2

Step-by-step explanation:

p(x)= 3x^3 – 3x^2– 18x

Factor out the greatest common factor, 3x

p(x)= 3x (x^2 – x– 6)

Factor inside the parentheses

What 2 numbers multiplies to -6 and adds to -1

-3*2 = -6

-3+2 = -1

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

Setting the function equal to zero to find the zeros

0 = 3x (x-3)(x+2)

Using the zero product property

3x = 0 x-3 =0 x+2 =0

x=0 x=3 x=-2

3 0

3 years ago

Read 2 more answers

It takes a turtle 3 1/4 hours to walk 1 1/2 miles. How many hours does it take to walk one mile?

fomenos

195min/1.5 miles so 130 min / 1 mile or 2h 10min (2 1/6h)

3 0

3 years ago

If 3x - 5 = 22, find the value of 2x. a. 6 b. 24 c. 12 d. 18

Softa [21]

Answer:

d. 18

Step-by-step explanation:

3x - 5 = 22

3x = 27

x = 9

Then, 2x is equal to:

= 2(9)

= 18

3 0

3 years ago

Harold uses the binomial theorem to expand the binomial (3x^5 - 1/9y^3)^4

riadik2000 [5.3K]

<h3>The complete question:</h3>

Harold uses the binomial theorem to expand the binomial $(3x^5 -\dfrac{1}{9}y^3)^4$

(a) What is the sum in summation notation that he uses to express the expansion?

(b) Write the simplified terms of the expansion.

Answer:

(a). $(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}( -\dfrac{1}{9}y^3)^k $$$

(b). $(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561}$

Step-by-step explanation:

(a).

The binomial theorem says

$(x+y)^n=$$\sum_{k=0}^{n} \binom{n}{k}x^{n-k}y^k $$$

For our binomial this gives

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=$$\sum_{k=0}^{n} \binom{4}{k}x^{4-k}y^k $$}$

(b).

We simplify the terms of the expansion and get:

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}y^k $$= \binom{4}{0}(3x^5)^{4-0}(-\dfrac{1}{9}y^3 )^0+\binom{4}{1}(3x^5)^{4-1}(-\dfrac{1}{9}y^3 )^1+\\\\\binom{4}{2}(3x^5)^{4-2}(-\dfrac{1}{9}y^3 )^2+\binom{4}{3}(3x^5)^{4-3}(-\dfrac{1}{9}y^3 )^3+\binom{4}{4}(3x^5)^{4-4}(-\dfrac{1}{9}y^3 )^4$

$$$\sum_{k=0}^{n} \binom{4}{k}(3x^5)^{4-k}(-\frac{1}{9}y^3 )^k $$= (3x^5)^{4}+4(3x^5)^{3}(-\frac{1}{9}y^3 )+6(3x^5)^{2}(-\frac{1}{9}y^3 )^2+\\\\4(3x^5)(-\frac{1}{9}y^3 )^3+(-\frac{1}{9}y^3 )^4$

$\boxed{(3x^5 -\dfrac{1}{9}y^3)^4=81x^{20}-12x^{15}y^3+\dfrac{2x^{10}y^6}{3}-\dfrac{4x^5y^9}{243}+\frac{y^{12}}{6561} }$

3 0

3 years ago

What is the product of 8/15 times 6/5 and 1/3

nordsb [41]

The answer would be 16/75

3 0

3 years ago

Read 2 more answers