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Klio2033 [76]
3 years ago
13

Solve for x. Round to the nearest tenth of a degree, if necessary.

Mathematics
1 answer:
kotegsom [21]3 years ago
5 0

Answer:

52.6°

Step-by-step explanation:

cos(theta)= adj/hyp

hyp=28

adj=17

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If line ab is parallel to line cd and the slope of line AB = -3/5, what is the slope of CD?
Greeley [361]

Answer:

-3/5

Step-by-step explanation:

Parallel lines have the same slope.  If ab has a slope of -3/5 and is parallel to cd, the cd has a slope of -3/5

8 0
3 years ago
Read 2 more answers
Item 5 Use the axis of symmetry to find the reflection of each point. The reflection of (−3,−3) ( − 3 , − 3 ) is (, ). The refle
GenaCL600 [577]

Answer:

dddddddddddddddddddddddd

Step-by-step explanation:

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4 0
2 years ago
This question is not that hard but I’m getting rushed
Vsevolod [243]
14:21 one is the right answer
4 0
2 years ago
Type the correct answer in each box. If necessary, use / for the fraction bar(s).
Eduardwww [97]

Answer:

x=5/6 and y=8/5

(5/6 , 8/5)

Step-by-step explanation:

You can use the elimination method! The point of this method is to add or subtract one equation from the other, multiplying them by constants if necessary, to cancel out one variable so you can solve for the other. Then, you can plug the value you got for your solution into either equation to solve for the other. I'll demonstrate.

Look at the two equations and the coefficients of both variables. You have 12x and -6x -- 6*2=12, so this is perfect. (You could also eliminate the y because 5*3=15, but I'll show you by eliminating the x instead.)

Here's what that looks like:

(-6x+1y=3)2

-12x+10y=6

So we just multiplied the second equation by 2 on both sides. Let's see how that helps us.

12x+15y=34

-12x+10y=6

If we add the two equations now, x will be canceled out and we can solve for y.

   12x+15y=34

+(-12x+10y=6)

___________

     0x+25y=40

           25y=40

0x=0, so we can get rid of the x. Now, we need to solve for y.

25y=40

y=40/25

You probably know how to simplify fractions, so divide both the numerator and the denominator by 5 to get y=8/5.

Now you can use this value in either equation and solve for x. I'll use the first. (This is called substitution.)

12x+15(8/5)=34

12x+3(8)=34 <-- What I did here is cancel out the 5 in the denominator with 15 to leave 3, because 5*3=15.

12x+24=34

12x=10

x=10/12

x=5/6 (Divide the numerator and denominator by 2.)

You can write your answer as a point, too. (5/6 , 8/5)

8 0
2 years ago
The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to pro
oksian1 [2.3K]

Here is the full question

A snack food company wishes to have a cylinder package for it's almond and cashew mix. The cylinder must contain 120 cm³ worth of product. The base and sides will cost $.01 per cm2 to produce but the top, which is plastic and resealable, will cost $.02 per cm2 to produce. What should the dimensions be to minimize cost?

Answer:

The radius and height are both dimension in the cylinder; in order to minimize the cost

radius = 2.515 cm

height = 18.93 cm

Step-by-step explanation:

We denote the radius of the cylinder to be = r

and the height of the cylinder = h

The volume of a cylinder is known to be = πr²h

Also, from the question; we are also told that the cylinder contains 120 πcm³

i.e πr²h = 120π

Dividing both sides with π; we have:

r²h = 120

h = \frac{120}{r^2}

The base and sides will cost $.01 per cm² to produce

Total cost of the base and side c_1 = 0.01 ( πr² + 2πrh)

but the top, which is plastic and resealable, will cost $.02 per cm² to produce.

i.e

cost of the top cylinder c_2 = 0.02 ( πr²)

Overall Total cost = c_1 + c_2

= 0.01 ( πr² + 2πrh) + 0.02 ( πr²)

= 0.01 πr² + 0.02 πrh + 0.02 πr²

= 0.03 \pi r^2 + 0.02 \pi r (\frac{120}{r^2} )

= 0.03 \pi r^2 + 2.4 \frac{\pi}{r}

Taking the differentiation to find the radius dimension to minimize cost; we have:

\frac{dc}{dr} =0 ⇒ 0.06 \pi r^2 - \frac{2.4 \pi}{r^2} =0

0.06 \pi r^2 = \frac{2.4 \pi}{r^2}

r^4 = \frac{2.4 \pi}{0.06 \pi}

r^4 = 40

r = \sqrt[4]{40}

r= 2.515 cm

However, \frac{d^2c}{dr^2}=0.12 \pi r + \frac{4.8 \pi}{r^3}

\frac{d^2c}{dr^2}|__{r= 2.515}} =0.12 \pi (2.515) + \frac{4.8 \pi}{(2.515)^3} >0

Therefore; we can say that the cost is minimum at r = 2.515 since it is positive.

To determine the height ; we have:

h = \frac{120}{r^2} \\h = \frac{120}{(2.515)^2}

h = \frac{120}{6.34}

h = 18.93 cm

7 0
3 years ago
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