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Savatey [412]
2 years ago
10

Clay puts $50 on his bus pass. Every time he rides the bus, $3.00 is deducted from his pass. How much money is left on his pass

after he takes 9 buss rides?
Mathematics
1 answer:
daser333 [38]2 years ago
6 0

Answer:

23$ is left on his pass

Step-by-step explanation:

9x3=27 50-27=23

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The length of a rectangle is 4 centimeters less than twice it’s width. The perimeter of the rectangle is 34vm. What are the dime
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l = 2w - 4

Because we're solving for 2l + 2w, that can be simplified to

2(2w - 4) + 2w = 34

4w - 8 + 2w = 34

6w - 8 = 34

6w = 42

w = 7

Knowing this, we can input w:

2(7) + 2l = 34

14 + 2L = 34

2l = 20

l = 10

L = 10, W = 7, Option C

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The correct answer is...?
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<em>Answer:</em>

<em>The answer to both problems is </em><em><u>GRAPH</u></em>

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LOTS OF POINTS GIVING BRAINLIEST I NEED HELP PLEASEE
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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.

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The angles of two pentagons are all equal to 108°. The length of one side of each pentagon measures 11 cm.
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There is not enough information to determine whether the pentagons are congruent.

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