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rjkz [21]
3 years ago
7

Need help with this geometry

Mathematics
1 answer:
andrezito [222]3 years ago
7 0
Could you send a picture so I can help?
You might be interested in
Which of the algebraic expressions are not linear expressions? Select all that apply. a. 3x - 2 b. -9y c. -7x^2^ + 4x d. -1/2x^2
Sedaia [141]

Answer:

c, d and e

Step-by-step explanation:

A linear expression is an expression in which sum of exponents of variables is at most 1 in each of its terms.

Option a: 3x - 2

Here, the exponent of the variable is 1.

So, it is a linear expression.

Option b: -9y

Here, the exponent of the variable is 1.

So, it is a linear expression.

Option c: -7x^2 + 4x

Here, the highest exponent of the variable is 2.

So, it is NOT a linear expression.

Option d: -\frac{1}{2} x^2

Here, the exponent of the variable is 2.

So, it is NOT a linear expression.

Option e: 8xy

Here, sum of exponents of variables is 2.

So, it is NOT a linear expression.

Hence, c, d and e are NOT algebraic expressions.

6 0
3 years ago
Give the first 3 multiples of 12
algol [13]
Answer: 1,12 2,6 3,4
7 0
3 years ago
What is 2 1/6 divided by 3 +5 1/2
irakobra [83]
<h2>the correct answer is...6.22222222222(:</h2>
3 0
3 years ago
What solution do 2x2 - 13x + 21 = 0 and 2x2 + 9x - 56 = 0 have in common? Round your answer to the nearest tenth if necessary.
lys-0071 [83]
So if we want to know the common solution(s) to a system of 2 equations, So we can just set both equations equal to each other and solve for the x value(s). That’s where I start below;

2x^2-13x+21 = 2x^2+9x-56
2x^2 cancels out and moving everything to one side and anything with an x variable to the other side we have then;
-22x=-77
22x=77 by cancelling the negative signs
x=77/22 therefore x=7/2 or 3.5

Hope this helps you. Any questions please ask.
4 0
3 years ago
Write an equation for the line that passes through (4,-6) and (-7,-6). Is the line parallel or perpendicular to the y-axis?
alisha [4.7K]

Answer:

y = -6, perpendicular

Step-by-step explanation:

You can use the points to find the slope of the line:

m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} =\frac{-6-(-6)}{-7-4} =\frac{-6+6}{-7-4}=\frac{0}{-11}=0

Then, using point-slope form, choose one set of coordinates to use for y1 and x1:

y-y_{1}=m(x-x_{1})

y-(-6)=0(x-4)

y+6=0

y=-6

Since the slope of the line is zero, it's a horizontal line. The y-axis is a vertical line, which means that the line is perpendicular to the y-axis.

7 0
3 years ago
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