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kvasek [131]
3 years ago
15

Evaluate using the given values b=1, c=2, d=3, p=-3, q=-2, r=-1 a) 2b+c-d

Mathematics
1 answer:
IRISSAK [1]3 years ago
5 0
The answer should be one
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Samantha earns $22 per hour as a plumbing apprentice. How much does she earn per minute in cents?
m_a_m_a [10]
A. She would use hours to minutes, which you can do by dividing 22 by 60.
b. 22 / 60 = .366 repeating, or about .37 per minute
3 0
3 years ago
Use point-slope form − 1 = ( − 1) to find the equation of the line that passes through the point (−3,5) and has slope = −2 . Wri
Tom [10]

Answer:

y - 5 = -2(x + 3)

Step-by-step explanation:

When you write an equation in point-slope form, you only need two things: a point and a slope.

Given:

point: (-3, 5)

slope (m): -2

The standard point-slope equation is

y - y₁ = m(x - x₁)

Plug in what you know.

y - (5) = -2(x - (-3))

Simplify.

y - 5 = -2(x + 3)

This is your equation.

Learn with another example:

brainly.com/question/24436844

8 0
2 years ago
Simplify this expression 3p+2p-p​
madam [21]

Answer:

Step-by-step explanation:

3p + 2p - p

= 3p + p

= 4p

(add the co-efficients)

3 0
3 years ago
Read 2 more answers
If perpendiculars from any point within an angle on its arms are equal, prove that it lies on the bisector of that angle
Oxana [17]
Your question can be quite confusing, but I think the gist of the question when paraphrased is: P<span>rove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?

Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector

4 0
4 years ago
What is the equation of a line that passes through the point (8, 1) and is perpendicular to the line whose equation is y=−2/3x+5
swat32

Let k:y=m_1x+b_1 and l:y=m_2x+b_2

l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}

We have y=-\dfrac{2}{3}x+5\to _1=-\dfrac{2}{3}

Therefore

m_2=-\dfrac{1}{-\frac{2}{3}}=\dfrac{3}{2}

We have the equation of a line: y=\dfrac{3}{2}x+b.

Put the coordinates of the point (8, 1) to the equation of a line:

1=\dfrac{3}{2}(8)+b

1=(3)(4)+b

1=12+b       <em>subtract 11 from both sides</em>

-11=b\to b=-11

Answer: \boxed{y=\dfrac{3}{2}x-11}

3 0
3 years ago
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