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AysviL
1 month ago
5

The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) us

ing a point and the slope. Which point did Harold use? When Harold wrote his equation, the point he used was (7, 3). When Harold wrote his equation, the point he used was (0, 7). When Harold wrote his equation, the point he used was (7, 0). When Harold wrote his equation, the point he used was (3, 7).
Mathematics
2 answers:
Zina [12.3K]1 month ago
8 0

We need to identify the point Harold utilized for this equation:

y = 3 (x-7)

Recall that the point-slope form of a line is expressed as:

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

Comparing Harold's equation to the standard form reveals the line's slope is m = 3..

Furthermore, it is clear that x_ {1} = 7 and y_ {1} = 0..

Therefore, Harold's equation was based on the point (7,0).

Answer:

The point (7,0) was used by Harold when writing his equation.

PIT_PIT [12.4K]1 month ago
8 0

Answer:

c

Step-by-step explanation:

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1 month ago
While in college, why did Euler work through advanced math books on his own? His father required it. The quality of education of
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8 0
29 days ago
Read 2 more answers
Given: KL ║ NM , LM = 45, m∠M = 50° KN ⊥ NM , NL ⊥ LM Find: KN and KL
Leona [12618]

Answer:

KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08\\ \\KN=45\sin 50^{\circ}\approx 34.47

Step-by-step explanation:

Given:

KL ║ NM,

LM = 45

m∠M = 50°

KN ⊥ NM  

NL ⊥ LM

To determine: KN and KL

1. Analyzing triangle NLM, we see it is a right triangle due to NL ⊥ LM. In this context,

LM = 45

m∠M = 50°

Consequently,

\tan \angle M=\dfrac{\text{opposite leg}}{\text{adjacent leg}}=\dfrac{NL}{LM}=\dfrac{NL}{45}\\ \\NL=45\tan 50^{\circ}

It is also true that

(angles LNM and M are complementary).m\angle LNM=90^{\circ}-50^{\circ}=40^{\circ}

2. Now considering triangle NKL, it also forms a right triangle as KN ⊥ NM. Within this triangle,

NL=45\tan 50^{\circ}

m\angle KLN=m\angle LNM=40^{\circ} (alternate interior angles)

m\angle KNL=90^{\circ}-40^{\circ}=50^{\circ} (angles KNL and KLN are complementary).

Thus,

\sin \angle KNL=\dfrac{\text{opposite leg}}{\text{hypotenuse}}=\dfrac{KL}{LN}=\dfrac{KL}{45\tan 50^{\circ}}\\ \\KL=45\tan 50^{\circ}\sin 50^{\circ}\approx 41.08

and

\cos \angle KNL=\dfrac{\text{adjacent leg}}{\text{hypotenuse}}=\dfrac{KN}{LN}=\dfrac{KN}{45\tan 50^{\circ}}\\ \\KN=45\tan 50^{\circ}\cos 50^{\circ}=45\sin 50^{\circ}\approx 34.47

3 0
29 days ago
Read 2 more answers
2. Santiago wants the lateral surfaces of the
Inessa [12570]

Answer:

Red paint will cover 262 square feet

of the ramp.

Step-by-step explanation:

The lateral surface area represents the surface area of the sides of the ramp without including the top and bottom surfaces

This ramp has three surfaces

The two sides are triangular with dimensions of length 20 and height 8.5

The back consists of a rectangular shape with length 12 and height 8.5

Step 1: Calculating the Area of Triangle

The area of the triangular side = \frac{1}{2} length \times height

The area of the triangle = \frac{1}{2} (20 \times 8.5)

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Area of both triangles = 85 \times 2  =  170

Step 2: Finding the Area of Rectangle

Area of rectangle = length \times height

therefore,

Area of the rectangular back = 12 \times 8.5 = 102  square feet

Step 3: Calculating the total lateral surface area

Overall Lateral Surface Area

= Area of triangles + Area of Rectangle

= 272 square feet

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