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docker41 [41]
3 years ago
6

2 Points

Mathematics
1 answer:
Westkost [7]3 years ago
5 0

Answer:

B

Step-by-step explanation:

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What are the solutions of the system? Solve by graphing. y=x^2-2x-1, y = -2
pychu [463]
I used desmos online.

(1,-2)
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Two Parallel lines are intersected by a transversal /1 and /2 are alternate exterior angles. The measure of /1 is (5x - 16) the
Alecsey [184]

Answer:

C. m/1=m/2=54

Step-by-step explanation:

these two angles are equal hence:

5x-16=4x-2=-16+2 =-14

4x-5x=-x

-x=-14=x=14

then you substitute the value of 'x' with 14

therefore:(5×14-16)=54

(4×14-2)=54

m/1=m/2=54

6 0
3 years ago
1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2
vlabodo [156]
Use math induction proof method will be usefully 

for n=1 will get 

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suppose true for n=k and prove for k = k+1 

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7 0
3 years ago
1. In a karate class there are 22 students. 15 of the students are male. What 6 is the ratio of females to the total number of s
Over [174]

Answer:

15:22

the ratio of men to women in a karate class is 3.3:1. If there are 100 women, how many men are there?Step-by-step explanation:

5 0
3 years ago
What are the zeros for the function
Greeley [361]

The zeroes of the function are x = -3, \ \ x =2\sqrt{ 3}i,  \ \ x =-2\sqrt{ 3}i.

Solution:

Given function:

f(x)=x^{3}+3 x^{2}+12 x+36

<u>To find the zeros of the function:</u>

⇒ f(x) = 0

\Rightarrow x^{3}+3 x^{2}+12 x+36=0

\Rightarrow (x^{3}+3 x^{2})+(12 x+36)=0

Take x² as common in first bracket and take 12 as common in 2nd bracket.

\Rightarrow x^2(x+3 )+12( x+3)=0

Now, take common term (x + 3) outside.

\Rightarrow (x+3 ) (x^2+12)=0

<em>Using zero factor principle, If ab = 0 then a = 0 or b = 0.</em>

x+3=0,  \ \ x^2+12=0

x = –3

x^2+12=0

x² = –12

x² = 2² × 3 × –1

Taking square root on both sides, we get

\sqrt{x^2} =\pm\sqrt{2^2\times 3\times -1}

x =\pm2\sqrt{ 3\times -1}

we know that \sqrt{-1} =i.

x =\pm2\sqrt{ 3}i

x =2\sqrt{ 3}i,  \ \ x =-2\sqrt{ 3}i

Hence the zeroes of the function are x = -3, \ \ x =2\sqrt{ 3}i,  \ \ x =-2\sqrt{ 3}i.

7 0
3 years ago
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