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fredd [130]
3 years ago
11

Which graph represents a line with a slope of and a y-intercept equal to that of the line y = x – 2? HELP!!!!!!!!!!!!!!!!!!!!!!!

!!
30 POINTS!!!

AND BRAINLEIST!!!!

Mathematics
2 answers:
Arlecino [84]3 years ago
7 0
None of the graphs have a slope and y-intercept equal to y = x - 2.

y = x - 2 has a slope of 1 and y-intercept of -2, none of the graphs you shown have a slope of 1.
____ [38]3 years ago
5 0

The answer is the last graph. (Ik this is super late but for anyone else who needs it here you go)

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Hurry ASAP!!!!! I only need help with number 6 please explain how you get the answer to! The instructions for his equation is: s
SVETLANKA909090 [29]

Answer:

-ax + 6by + 10

Step-by-step explanation:

(2ax + 3by + 5) - (3ax - 3by - 5)

Subtrahend is the value AFTER the subtraction sign.  So in this example the subtrahends enclosed in the 2nd bracket are 3by and 5.

So subtracting by changing the signs of the subtrahends:

= 2ax + 3by + 5 - 3ax + 3by + 5

Collecting like terms:

= 2ax - 3ax + 3by + 3by + 5 + 5

Combining like terms:

= -ax + 6by + 10

5 0
3 years ago
Which is closest to the circumference of the CD?
seropon [69]

Answer:

idk but heres an equation

Step-by-step explanation:

Since  

2

π

r

=

37.68

,

r

=

18.84

π

Area=  

π

⋅

(

18.84

)

2

π

2

=  

354.9456

3.14

= 113.04

5 0
3 years ago
Read 2 more answers
Consider a particle moving around a circle with a radius of 38cm. It rotates from 10 degrees to 100 degrees in 11 seconds. Calcu
kifflom [539]

Step-by-step explanation:

Given that,

Radius of circle, r = 38 cm = 0.38 m

It rotates form 10 degrees to 100 degrees in 11 seconds i.e.

\theta_i=10^{\circ}=0.174\ rad

\theta_f=100^{\circ}=1.74\ rad

Let \omega is the angular velocity of the particle such that, \omega=\dfrac{\omega_f-\omega_i}{t}

\omega=\dfrac{1.74-0.174}{11}

\omega=0.142\ rad/s

We need to find the instantaneous velocity of the particle. The relation between the angular velocity and the linear velocity is given by :

v=r\times \omega

v=0.38\times 0.142

v = 0.053 m/s

So, the instantaneous velocity of the particle is 0.053 m/s. Hence, this is the required solution.  

6 0
3 years ago
LOTS OF POINTS GIVING BRAINLIEST I NEED HELP PLEASEE
Sidana [21]

Answer:

Segment EF: y = -x + 8

Segment BC: y = -x + 2

Step-by-step explanation:

Given the two similar right triangles, ΔABC and ΔDEF, for which we must determine the slope-intercept form of the side of ΔDEF that is parallel to segment BC.

Upon observing the given diagram, we can infer the following corresponding sides:

\displaystyle\mathsf{\overline{BC}\:\: and\:\:\overline{EF}}

\displaystyle\mathsf{\overline{BA}\:\: and\:\:\overline{ED}}

\displaystyle\mathsf{\overline{AC}\:\: and\:\:\overline{DF}}

We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.

<h2>Slope of Segment BC:</h2>

In order to solve for the slope of segment BC, we can use the following slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}  }

Use the following coordinates from the given diagram:

Point B:  (x₁, y₁) =  (-2, 4)

Point C:  (x₂, y₂) = ( 1,  1 )

Substitute these values into the slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{1\:-\:4}{1\:-\:(-2)}\:=\:\frac{-3}{1\:+\:2}\:=\:\frac{-3}{3}\:=\:-1}

<h2>Slope of Segment EF:</h2>

Similar to how we determined the slope of segment BC, we will use the coordinates of points E and F from ΔDEF to find its slope:

Point E:  (x₁, y₁) =  (4, 4)

Point F:  (x₂, y₂) = (6, 2)

Substitute these values into the slope formula:

\displaystyle\mathsf{Slope\:(m)\:=\:\frac{y_2 \:-\:y_1}{x_2 \:-\:x_1}}\:=\:\frac{2\:-\:4}{6\:-\:4}\:=\:\frac{-2}{2}\:=\:-1}

Our calculations show that segment BC and EF have the same slope of -1.  In geometry, we know that two nonvertical lines are <u>parallel</u> if and only if they have the same slope.  

Since segments BC and EF have the same slope, then it means that  \displaystyle\mathsf{\overline{BC}\:\: | |\:\:\overline{EF}}.

<h2>Slope-intercept form:</h2><h3><u>Segment BC:</u></h3>

The <u>y-intercept</u> is the point on the graph where it crosses the y-axis. Thus, it is the value of "y" when x = 0.

Using the slope of segment BC, m = -1, and the coordinates of point C, (1,  1), substitute these values into the <u>slope-intercept form</u> (y = mx + b) to solve for the y-intercept, <em>b. </em>

y = mx + b

1 = -1( 1 ) + b

1 = -1 + b

Add 1 to both sides to isolate b:

1 + 1 = -1 + 1 + b

2 = b

Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 2.

Therefore, the linear equation in <u>slope-intercept form of segment BC</u> is:

⇒  y = -x + 2.

<h3><u /></h3><h3><u>Segment EF:</u></h3>

Using the slope of segment EF, <em>m</em> = -1, and the coordinates of point E, (4, 4), substitute these values into the <u>slope-intercept form</u> to solve for the y-intercept, <em>b. </em>

y = mx + b

4 = -1( 4 ) + b

4 = -4 + b

Add 4 to both sides to isolate b:

4 + 4 = -4 + 4 + b

8 = b

Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 8.

Therefore, the linear equation in <u>slope-intercept form of segment EF</u> is:

⇒  y = -x + 8.

8 0
2 years ago
Describe how the graph of the equation relates to the graph of y = x2.<br> f(x) = -x + 5)2
Kobotan [32]

Answer:

Step-by-step explanation:

It's y = x^2, not y = x2.  The graph of y = x^2 is a parabola with vertex (0, 0) and which opens up.

The graph of y = (x + 5)^2 has the same shape as that of y = x^2, but has been translated 5 units to the left.

6 0
3 years ago
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