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Leona [35]
3 years ago
7

Pls help with this question. WILL MARK BRAINLIEST + 10 POINTS!!!!! :D#7

Mathematics
1 answer:
Mandarinka [93]3 years ago
4 0
The answer is the second choice π/2+360 ; π/2-360
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In isosceles pqr with base , pq = 2x + 3, and pr = 9x - 11. what is the value of x?
Anon25 [30]
First, draw and label a rough sketch of an isosceles triangle. Now put in the values of pq=2x+3 and pr=9x-11 

You know that the triangle is isosceles, therefore, the sides pq and pr are identical.  

Now, make the equations equal to each other to find the value of x. 

2x+3=9x-11 
3+11=9x-2x 
14=7x 
14/7=x 
2=x 

Therefore the value of x=2. 

Hope I helped :) 
8 0
3 years ago
Find the value of n. The diagram is not to scale.
lesya692 [45]

Answer:

use symolab

Step-by-step explanation:

7 0
3 years ago
Tell whether the relation is a function. {(-5, -3), (-1, 3), (4, 3), (8, 8)}
VladimirAG [237]

Given:

The relation is {(-5, -3), (-1, 3), (4,3), (8, 8)}.

To find:

Whether the given relation is a function or not.

Solution:

A relation is called a function, if there exist a unique output for each input.

We have, a relation given as

{(-5, -3), (-1, 3), (4,3), (8, 8)}

Here, all x-values are different. So, each x-value has a unique y-value.

Thus, there exist a unique output for each input.

Therefore, the given relation is a function.

8 0
3 years ago
What is the value of p+q
Korvikt [17]

ax^2+bx+c=0\\\\x_1,\ x_2-solutions\\\\x_1+x_2=\dfrac{-b}{a}\\--------------------------\\\\4x^2-7x-2=0\\\\a=4,\ b=-7,\ c=-2\\\\p,\ q-solutions\\\\p+q=\dfrac{-(-7)}{4}=\dfrac{7}{4}

6 0
3 years ago
Solve the following quadratic by<br> completing the square.<br> y = x^2 - 6x+7
kozerog [31]

Answer:

x= 3+\sqrt{2}, \quad x= 3 -\sqrt{2}

Step-by-step explanation:

<u>Given quadratic equation</u>:

y = x^2 - 6x+7

To complete the square, begin by adding and subtracting the square of half the coefficient of the term in x:

\implies y = x^2 - 6x+\left(\dfrac{-6}{2}\right)^2-\left(\dfrac{-6}{2}\right)^2+7

\implies y = x^2 - 6x+\left(-3\right)^2-\left(-3\right)^2+7

\implies y = x^2 - 6x+9-9+7

Factor the perfect square trinomial:

\implies y=(x-3)^2-9+7

\implies y=(x-3)^2-2

To solve the quadratic, set it to zero and solve for x:

\implies (x-3)^2-2=0

\implies (x-3)^2-2+2=0+2

\implies (x-3)^2=2

\implies \sqrt{(x-3)^2}=\sqrt{2}

\implies x-3= \pm\sqrt{2}

\implies x-3+3= 3\pm\sqrt{2}

\implies x= 3\pm\sqrt{2}

Therefore, the solution to the given quadratic equation is:

x= 3+\sqrt{2}, \quad x= 3 -\sqrt{2}

6 0
1 year ago
Read 2 more answers
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