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Murrr4er [49]
2 years ago
14

Can someone help me with my algebra 2 assignment.

Mathematics
1 answer:
gulaghasi [49]2 years ago
5 0
What’s the assignment
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Please help! It’s fairly easy!
icang [17]

Answer:

see below

Step-by-step explanation:

This would be true if the lines are parallel.  We have no information about the lines, so we cannot assume the lines are parallel, so we cannot use the corresponding angles theorem

8 0
3 years ago
Read 2 more answers
A total of 708 tickets were sold for the school play they were either adult tickets or student tickets there were 58 more studen
zhenek [66]
X = adult tickets and x+58 = student tickets

Create the equation from the problem.
x + x + 58 = 708

Combine like terms.
2x +58 = 708

Subtract 58.
2x = 650

Divide by 2.
x = 325

Therefore, your answer is:
there were 325 adult tickets sold.

(Adding 58 to that number will get you the student tickets sold which is 383 if they ask.)
3 0
3 years ago
How do I simplify the expression 7^0+7^2 over (7^3)^2
AlekseyPX

\frac{7^0+7^2}{(7^3)^2}=\frac{1+49}{7^6} = \frac{50}{117649}

Ok done. Thank to me :>

7 0
2 years ago
What are the solutions of the equation x4+3x2+2=0? use u substitution to solve
sweet [91]

Answer: x = ± i , x = ±√2i are solution .

Step-by-step explanation:

Given : x^{4} +3x^{2} +2=0, use u substitution to solve.

To find : what are the solutions of the equation .

Solution : We have given that x^{4} +3x^{2} +2=0.

Let consider x^{2} = u .

Substitute u =  x^{2} in given equation .

u^{2} + 3u +2=0.

On factoring

u^{2} + 2u+1u +2=0.

Taking common

u( u+2) +1 (u+2) = 0.

On grouping

(u+1) (u+2) =0

Now, u+1 = 0 ⇒ u = -1.

         u+2 = 0 ⇒ u = -2.

In term of x, plugging x^{2} = u.

x^{2} = -1   ;  x^{2} = -2.

taking square root both side

x = ± i

x = ±√2i.

Therefore, x = ± i , x = ±√2i are solution .

8 0
3 years ago
Find the value of x.
Gala2k [10]

Answer:

\large\boxed{2\sqrt7}

Step-by-step explanation:

Look at the picture.

ΔADC and ΔACB are similar. Therefore the corresponding sides are in proportion:

\dfrac{AC}{AD}=\dfrac{AB}{AC}

We have

AC=x,\ AD=2,\ AB=2+12=14

Substitute:

\dfrac{x}{2}=\dfrac{14}{x}           <em>cross multiply</em>

x^2=(2)(14)\\\\x^2=28\to x=\sqrt{28}\\\\x=\sqrt{4\cdot7}\\\\x=\sqrt4\cdot\sqrt7\\\\x=2\sqrt7

5 0
3 years ago
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