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Alexxx [7]
3 years ago
10

One kind of hard candy sells for $.89 per kilogram another sells for $1.10 per kilogram how many kilograms of each kind a need t

o be used for 30 kg of a mixture to sell for $.96 per kilogram
Mathematics
1 answer:
Furkat [3]3 years ago
4 0

Answer:

20kg of $0.89 candy

10kg of $1.10 candy

Step-by-step explanation:

Candy 1 = 0.89 per kg

Candy 2 = 1.10 per kg

Total kilogram, kg = 30

Let candy 1 = x ; candy 2 = (30 - x) ;

0.89x + 1.10(30 - x) = 0.96(30)

0.89x + 33 - 1.10x = 28.8

0.89x - 1.10x = 28.8 - 33

-0.21x = - 4.2

x = 4.2 / 0.21

x = 20

20kg of $0.89 candy

(30 - x) = (30 - 20) = 10kg

10kg of $1.10 candy

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A person is standing 50 yards from the base of a building. The angle of elevation from the person to the top of the building is
Whitepunk [10]

Answer:

The height of the building is H = 11.65 yards

Step-by-step explanation:

Distance from the base of the building to the person = 50 yards

Angle of elevation = 25°

From Δ ABC

⇒ \tan 25 = \frac{H}{25}

⇒ H = 25 \tan 25

⇒ H = 11.65 yards

Therefore  the height of the building is H = 11.65 yards

6 0
3 years ago
Car X weighs 136 pounds more than car Z. Car Y weighs 117 pounds more than car Z. The total weight of all three cars is 9439 pou
Aleksandr [31]

Let x, y and z denote the weighs of car X, car Y and car Z, respectively.

We know that car X weighs 136 more than car Z, this can be express by the equation:

x=z+136

We also know that Y weighs 117 pounds more than car Z, this can be express as:

y=z+117

Finally, we know that the total weight of all the cars is 9439, then we have:

x+y+z=9439

Hence, we have the system of the equations:

\begin{gathered} x=z+136 \\ y=z+117 \\ z+y+z=9439 \end{gathered}

To solve the system we can plug the values of x and y, given in the first two equations, in the last equation; then we have:

\begin{gathered} z+136+z+117+z=9439 \\ 3z=9439-136-117 \\ 3z=9186 \\ z=\frac{9186}{3} \\ z=3062 \end{gathered}

Now that we have the value of z we plug it in the first two equations to find x and y:

\begin{gathered} x=3062+136=3198 \\ y=3062+117=3179 \end{gathered}

Therefore, car X weighs 3198 pound, car Y weighs 3179 pounds and car Z weighs 3062 pounds.

4 0
1 year ago
How much higher is checkpoint 4 than checkpoint 1<br> checkpoint 1= -86<br> checkpoint 4= 2,177
Rina8888 [55]
Answer is 2263.
Checkpoint 1 is -86 
Checkpoint 4 is 2177
Subtract -86 from 2177....2177-(-86)
   =2263
2263 is how much higher checkpoint 1 (-86) is from checkpoint 4 (2177)

Hope this helps! :)
6 0
3 years ago
Read 2 more answers
Using fraction busting solve for X <br><br>1/x² + 1/x = 6/x² ​
777dan777 [17]
The solution for this question would be. x += 5 hope it helps, and im truly am sorry if its incorrect!
8 0
3 years ago
Which could be the first step in solving this system of equations by substitution?
Ainat [17]

Answer:

The first step is to replace the y in y-x=15 with 7x (we can do this because the first equation tells us that they're equal)

solve and get

(2.5,17.5)

or x= 2.5 y=17.5

Step-by-step explanation:

The first step is to recognize that y=7x which means that we can just replace the y in the second equation with 7x

7x-x=15

6x=15

x=2.5

Then we can solve for y

y=7(2.5)

y=17.5

3 0
3 years ago
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