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3 years ago

Name two sides that appear to be intersecting, but not perpendicular

Mathematics

2 answers:

Komok [63]3 years ago

8 0

Answer:

Two distinct lines intersecting each other at 90° or a right angle are called perpendicular lines. Here, AB is perpendicular to XY because AB and XY intersect each other at 90°. The two lines are parallel and do not intersect each other. They can never be perpendicular to each other.

Step-by-step explanation:

hope it helps, God bless!

GarryVolchara [31]3 years ago

6 0

Two distinct lines intersecting each
other at 90° or a right angle are called
perpendicular lines. Here, AB is
perpendicular to XY because AB and
XY intersect each other at 90° The
two lines are parallel and do not
intersect each other. They can never
be perpendicular to each other.

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Is p=2.5(600-500R)+-500 and p=-1250R+1000 a linear or nonlinear function? Explain.

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Answer:

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Step-by-step explanation:

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4 years ago

Find two consecutive odd integers whose sum is -88

frosja888 [35]

They are -35 and -53

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4 years ago

Verify that:

Lelu [443]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to verify that:

$\displaystyle \left(\cos(x)\right)\left(\cot(x)\right)=\csc(x)-\sin(x)$

Note that cot(x) = cos(x) / sin(x). Hence:

$\displaystyle \left(\cos(x)\right)\left(\frac{\cos(x)}{\sin(x)}\right)=\csc(x)-\sin(x)$

Multiply:

$\displaystyle \frac{\cos^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Recall that Pythagorean Identity: sin²(x) + cos²(x) = 1 or cos²(x) = 1 - sin²(x). Substitute:

$\displaystyle \frac{1-\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Split:

$\displaystyle \frac{1}{\sin(x)}-\frac{\sin^2(x)}{\sin(x)}=\csc(x)-\sin(x)$

Simplify:

$\csc(x)-\sin(x)=\csc(x)-\sin(x)$

Problem 2)

We want to verify that:

$\displaystyle (\csc(x)-\cot(x))^2=\frac{1-\cos(x)}{1+\cos(x)}$

Square:

$\displaystyle \csc^2(x)-2\csc(x)\cot(x)+\cot^2(x)=\frac{1-\cos(x)}{1+\cos(x)}$

Convert csc(x) to 1 / sin(x) and cot(x) to cos(x) / sin(x). Thus:

$\displaystyle \frac{1}{\sin^2(x)}-\frac{2\cos(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out the sin²(x) from the denominator:

$\displaystyle \frac{1}{\sin^2(x)}\left(1-2\cos(x)+\cos^2(x)\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (perfect square trinomial):

$\displaystyle \frac{1}{\sin^2(x)}\left((\cos(x)-1)^2\right)=\frac{1-\cos(x)}{1+\cos(x)}$

Using the Pythagorean Identity, we know that sin²(x) = 1 - cos²(x). Hence:

$\displaystyle \frac{(\cos(x)-1)^2}{1-\cos^2(x)}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor (difference of two squares):

$\displaystyle \frac{(\cos(x)-1)^2}{(1-\cos(x))(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Factor out a negative from the first factor in the denominator:

$\displaystyle \frac{(\cos(x)-1)^2}{-(\cos(x)-1)(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Cancel:

$\displaystyle \frac{\cos(x)-1}{-(1+\cos(x))}=\frac{1-\cos(x)}{1+\cos(x)}$

Distribute the negative into the numerator. Therefore:

$\displaystyle \frac{1-\cos(x)}{1+\cos(x)}=\displaystyle \frac{1-\cos(x)}{1+\cos(x)}$

3 0

3 years ago

What is 2^4/3 equal to

777dan777 [17]

16 squared with a three

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4 years ago

Read 2 more answers

What is the RANGE and the IQR for 17,18,22,17,16,21,19,and16

Delvig [45]

Answer:

so 1- 2 is negative 3 so minus tha by 3 than 4x2

Step-by-step explanation:

8 0

3 years ago