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fiasKO [112]
4 years ago
11

The school store buys granola bars for $0.40 each and sells them for 0.65 .What is the percent markup?

Mathematics
1 answer:
adell [148]4 years ago
4 0

The percent markup is 62.5%

The work is provided in the image attached.

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Given f(x) = 3x – 2 what is f (2x)?
Naily [24]

Answer:

6x - 2

Step-by-step explanation:

f(2x) = 3(2x) - 2

f(2x) = 6x - 2

3 0
3 years ago
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Find an equation of the line that has the given slope and passes through the given point.
masha68 [24]

Answer:

 y = 3x +21

Step-by-step explanation:

m= 3

The y-intercept of the line that passes through the given point  (-7,0) is

 0 = -7(3) + b

b= 21

The equation of the line that passes through the given point  (-7,0) is

 y =mx +b

 y = 3x +21

6 0
3 years ago
The measures of one of the angles of a triangle is 35.The sum of the measures of the other two angles is 145 and the difference
jok3333 [9.3K]

Answer:

The measures of the angles are 65° and 80°

Step-by-step explanation:

One of the angles is 35°

We know that the other two angles have a difference of 15°

With this knowledge we can write an equation.

2x symbolizes the two angles

+15 symbolizes the difference between the two angles

145 = 2x + 15

145 - 15 = 130

2x = 130

x = 65

65 is the size of one of the angles.

Next, we add 15 to 65 and we get 80

80 is our second angle

8 0
4 years ago
How are rates and ratios related
Dovator [93]
Rates are the percentage and ratios are like this 3:1. so, in that 3:1 ratio, there is 3 to 1 cookies to cups of milk. So there is 75% cookies and 25% milk.

I hope i helped you, comment and ill answer

7 0
4 years ago
Read 2 more answers
A slope field produces...
balandron [24]

It's A. B doesn't really make sense ("derivative of the differential equation" is somewhat nonsensical, "derivative of an equation" is not meaningful).

More to the point: Slope fields are used to visualize solutions to differential equations of the form

<em>y'</em> = <em>f(x</em>, <em>y)</em>

You take some point (<em>x</em>, <em>y</em>) and evaluate <em>y'</em> at the point. This gives the slope of the line tangent to the particular solution to the DE that passes through the point (<em>x</em>, <em>y </em>).

Sample several points and evaluate <em>y'</em> at those points and you get several different slopes.

Simple example:

<em>y'</em> = <em>x</em> ² - <em>y</em> ² = (<em>x</em> + <em>y </em>) (<em>x</em> - <em>y</em> )

Let's take the points (1, 1), (-1, 0), and (2, -2), at which we get slopes

<em>y'</em> = <em>f</em> (1, 1) = 0

<em>y'</em> = <em>f</em> (-1, 0) = 1

<em>y'</em> = <em>f</em> (2, -2) = 0

From here, you can get particular solutions that pass through a certain point by interpolating the slopes of the tangents to the solution. I've attached a slope field for the example here at the points listed above. (In the order red, green, black). Each light gray arrow in the background shows the slope of the tangent line.

If you're still unsure how the slope field is generated, I suggest looking up videos on the subject. The process is a bit difficult to describe without a dynamic visual aid.

3 0
3 years ago
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