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sweet [91]
3 years ago
14

Mrs. Ramiriz is making pies to sell at the local farmer’s market. It costs her $5 to make each pie, plus a one-time cost of $30

for baking supplies. She plans to sell the pies for $12 each. Which equation can be used to find the number of pies she needs to sell to break even?
Mathematics
1 answer:
gregori [183]3 years ago
5 0
The answer is 5x + 30 = 12x

x - the number of pies

It costs her $5 to make each pie<span>, plus a one-time cost of $30 for baking supplies:
</span>COST: f(c) = 5x + 30

<span>She plans to sell the pies for $12 each:
PROFIT: f(p) = 12x

</span><span>To find the number of pies she needs to sell to break even:
f(c) = f(p)
5x + 30 = 12x</span>
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The ratio of Ed's toy cars to Pete's toy cars was initially 5:2. After Ed gave 30 toy cars to Pete, they each had an equal numbe
Slav-nsk [51]

Answer:

140 toy cars

Step-by-step explanation:

The ratio of Ed's toy car to Pete's toy car is initially given as 5:2

Ed gave Pete a total number of 30 cars

Let x represent the greatest common factor that exists between both number

Number of Ed's car is represented as 5x

Number of Pete car is represented as 2x

Since they each have an equal number of cars which is 30 then we can solve for x as follows

5x-30=2x+30

Collect the like terms

5x-2x= 30+30

3x= 60

Divide both sides by the coefficient of x which is 3

3x/3=60/3

x=20

Ed's car is 5x, we substitute 20 for x

5(20)

= 100 cars

Pete car is 2x,we substitute 20 for x

2(20)

= 40 cars

Therefore, the total number of cars can be calculated as follows

= 100+40

= 140 toy cars

Hence they have 140 toy cars altogether

5 0
3 years ago
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Find the midpoint of the segment connecting (4,7) to (-2,3).
Vadim26 [7]

Answer:

<h2>Midpoint (1 , 5)</h2>

Step-by-step explanation:

Let (x , y) be the coordinates of the midpoint then:

x=\frac{4+(-2)}{2} = 1\\\\y=\frac{7+3}{2} =5

6 0
3 years ago
Help because I DONT know how to even start
Ierofanga [76]

Slope is rise/run or y2-y1/x2-x1. So to find the slope, find two points where the point is exact, then do the equation. That's the best I can explain, sorry if this doesn't help at all.

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3 years ago
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This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, pl
Lostsunrise [7]

First,

tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)

and given that 90° < <em>θ </em>< 180°, meaning <em>θ</em> lies in the second quadrant, we know that cos(<em>θ</em>) < 0. (We also then know the sign of sin(<em>θ</em>), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < <em>θ</em>/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(<em>θ</em>/2) > 0 and sin(<em>θ</em>/2) > 0.

Now recall the half-angle identities,

cos²(<em>θ</em>/2) = (1 + cos(<em>θ</em>)) / 2

sin²(<em>θ</em>/2) = (1 - cos(<em>θ</em>)) / 2

and taking the positive square roots, we have

cos(<em>θ</em>/2) = √[(1 + cos(<em>θ</em>)) / 2]

sin(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / 2]

Then

tan(<em>θ</em>/2) = sin(<em>θ</em>/2) / cos(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / (1 + cos(<em>θ</em>))]

Notice how we don't need sin(<em>θ</em>) ?

Now, recall the Pythagorean identity:

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

Dividing both sides by cos²(<em>θ</em>) gives

1 + tan²(<em>θ</em>) = 1/cos²(<em>θ</em>)

We know cos(<em>θ</em>) is negative, so solve for cos²(<em>θ</em>) and take the negative square root.

cos²(<em>θ</em>) = 1/(1 + tan²(<em>θ</em>))

cos(<em>θ</em>) = - 1/√[1 + tan²(<em>θ</em>)]

Plug in tan(<em>θ</em>) = - 12/5 and solve for cos(<em>θ</em>) :

cos(<em>θ</em>) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(<em>θ</em>/2) and tan(<em>θ</em>/2) :

sin(<em>θ</em>/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(<em>θ</em>/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0
2 years ago
p varies jointly as q and the square of r, and p = 200 when q = 2 and r = 3. Find p when q = 5 and r= 2.
Margarita [4]

Step-by-step explanation:

If a variables varies jointly, we can just divide it by the other variables in relation to it.

For example, since p variables jointly as q and square of r, then

\frac{p}{q {r}^{2} }  = k

where k is a constant

First, let find k. Substitute p= 200

q= 2, and r=3.

\frac{200}{2(3) {}^{2} }  = k

\frac{200}{18}  = k

\frac{100}{9}  = k

Now, since we know our constant, let find p.

\frac{p}{q {r}^{2} }  =  \frac{100}{9}

Q is 5, and r is 2.

\frac{p}{5( {2}^{2}) }  =  \frac{100}{9}

\frac{p}{20}  =  \frac{100}{9}

p =  \frac{2000}{9}

3 0
1 year ago
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