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azamat
4 years ago
9

Is each line parallel, perpendicular, or neither parallel or perpendicular to a line whose slops is -3/4?

Mathematics
1 answer:
Ymorist [56]4 years ago
7 0

Answer:

Line M is neither.

Line N is perpendicular.

Line P is neither.

Line Q is parallel.

Step-by-step explanation:

Parallel lines are lines whose slope are the same. Line q is the exact same slope. It is parallel.

Perpendicular lines are lines whose slopes are the negative reciprocal of each other \frac{a}{b}, -\frac{b}{a}. Line n is 4/3 and is perpendicular.

Line m and p are neither.


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In the figure, angle B measures 44°, angle A measures 62°, and angle E measures 50°.
Gre4nikov [31]
We don't need the figure
angle b = 44 degrees
angle a = 62 degrees
angle e = 50 degrees
angle f = unknown
we know that 
angle a + b + e + f = 180 degrees
50 + 44 + 62 + f =180 degrees
f= 180-50-44-62
but here there is only one blank so we have to add 44 and 62 to make one number that is 106
therefore, f = 180-50-106
if you further want to solve it angle f is 24
3 0
3 years ago
Which equation represents a quadratic relation?
Finger [1]

Answer:

a) y = 4x^2

Step-by-step explanation:

Because in a) y = 4x^2 the degree of equation is 2.

7 0
3 years ago
Find the area of the circle. Round your answer to the nearest hundredth.
Hoochie [10]
Area is 706.858, rounded is 707
6 0
3 years ago
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What are the zeros of the quadratic function f(x) = 2x2 – 10x – 3?
Natasha2012 [34]

Answer:

x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}  are zeroes of given quadratic equation.

Step-by-step explanation:

We have been a quadratic equation:

2x^2-10x-3

We need to find the zeroes of quadratic equation

We have a formula to find zeroes of a quadratic equation:

x=\frac{b^2\pm\sqrt{D}}{2a}\text{where}D=\sqrt{b^2-4ac}

General form of quadratic equation is ax^2+bx+c

On comparing general equation with b given equation we get

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On substituting the values in formula we get

D=\sqrt{(-10)^2-4(2)(-3)}

\Rightarrow D=\sqrt{100+24}=\sqrt{124}

Now substituting D in  x=\frac{b^2\pm\sqrt{D}}{2a} we get

x=\frac{(-10)^2\pm\sqrt{124}}{2\cdot 2}

x=\frac{100\pm\sqrt{124}}{4}

x=\frac{100\pm2\sqrt{31}}{4}

x=\frac{50\pm\sqrt{31}}{2}

Therefore, x=\frac{50+\sqrt{31}}{2},\frac{50-\sqrt{31}}{2}



5 0
4 years ago
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Mama L [17]
The answer is A.
I used the formula for finding the vertex of a parabola and I got the first answer choice. So A is correct. :)
6 0
4 years ago
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