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Svet_ta [14]
3 years ago
10

Determine whether the graphs of the given equations are parallel, perpendicular, or neither. Explain.

Mathematics
1 answer:
mr_godi [17]3 years ago
3 0

Answer:

They are neither parallel nor perpendicular.

it isn't parallel because they cross

it's not perpendicular because they don't cross at a 90 degree angle

Step-by-step explanation:

Hope this helps you and btw u should download desmos it's an amazing calculator that graphs these things for you!!! hope u have a wonderful day and stay safe and healthy:)))

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Given triangle ABC with vertices A(16,0), B(9,2), and C(2,0).
Margaret [11]
Given:

Triangle Point coordinates or vertices:
A(16,0)
B(9,2)
C(2,0)

Required:

Side with slope -2/7

Solution:

To know which side has the slope required, we need to calculate the slope of each paired point by the two-point slope formula:

m = (y2-y1)/(x2-x1)

We first try points A and B.

m1 = (2-0)/(9-16)

m1 = -2/7

Thus, the side with the slope -2/7 is side AB.
4 0
3 years ago
Simplify the left side of equation so it looks like the right side. cos(x) + sin(x) tan(x) = sec (x)
uranmaximum [27]

Step-by-step explanation:

Consider LHS

\cos(x)  +  \sin(x)  \tan(x)  =  \sec(x)

Apply quotient identies

\cos(x)  +   \sin(x) \times  \frac{ \sin(x) }{ \cos(x) }  =  \sec(x)

Multiply the fraction and sine.

\cos(x)  +  \frac{ \sin {}^{2} (x) }{ \cos(x) }  =  \sec(x)

Make cos x a fraction with cos x as it denominator.

\cos(x)  \times  \cos(x)  =  \cos {}^{2} (x)

so

\frac{ \cos {}^{2} (x) }{ \cos(x) }  +  \frac{ \sin {}^{2} (x) }{ \cos(x) }  =  \sec(x)

Pythagorean Identity tells us sin squared and cos squared equals 1 so

\frac{1}{ \cos(x) }  =  \sec(x)

Apply reciprocal identity.

\sec(x)  =  \sec(x)

7 0
3 years ago
(-4,-1), (4, 5)<br> What’s the slope
kati45 [8]

Answer: m = 3/4

Step-by-step explanation:

7 0
3 years ago
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beks73 [17]

Answer:

5/101/180

Step-by-step explanation:

6 0
3 years ago
Prove that the triangle EDF is isosceles. Give reasons for your answer.
Gekata [30.6K]

Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

8 0
3 years ago
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