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8_murik_8 [283]
3 years ago
7

Krishaun has $5, $10, and $20 coins worth a total of $560. She has the same number of $5 coins as $10 coins, and the same number

of $10 coins as $20 coins. How many $20 coins does Krishaun have?
a.16
b.15
c.20
d.28
e.17
Mathematics
1 answer:
valkas [14]3 years ago
4 0

f = t = n five ten and twenty

5n + 10n + 20n = 560

35n = 560

n = twenty dollar coins = 16

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Negative 5 is fewer than 6 less than a number k
lara [203]

Answer:

k > 1 is the solution.

Step-by-step explanation:

We are given that 'Negative 5 is fewer than 6 less than a number k', so algebraically it means that - 5 < k - 6.

Then this is a linear inequality of single variable k and we have to solve it.

Now, we have

- 5 < k - 6

⇒ - 5 + 6 < k - 6 + 6 {Adding 6 to both sides}

⇒ 1 < k

⇒ k > 1

This is the solution of the given inequality. (Answer)

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Write and solve an equation.
monitta

Answer:

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Step-by-step explanation:

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3 years ago
Which of the numbers, 3/12 or 2/9, is greater? How do you know?
nata0808 [166]

Answer:

3/12 > 2/9

Step-by-step explanation:

Well, 3/12 in decimal form is 0.25, and 2/9 in decimal form is around 0.23 so that would make 3/12 the greater fraction.

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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
3 years ago
when constricting parallel and perpendicular lines, how are the steps similar intersecting arcs are created and connected
Slav-nsk [51]

Answer:

Intersecting arcs are created and connected.

Copy an angle by creating arcs with a compass

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