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Elena-2011 [213]
3 years ago
12

A store is having a sale on jelly beans and trail mix. For 8 pounds of jelly beans and 3 pounds of trail mix, the total is $18.

For 2 pounds of jelly beans and 5 pounds of trail mix, the total cost is $13. Find the cost for each pound of jelly beans and each pound of trail mix.
Mathematics
1 answer:
topjm [15]3 years ago
6 0

Answer: each pound of jelly beans costs 1.5 dollars, or $1.50, and each pound of trail mix costs 2 dollars, or $2

Step-by-step explanation:

j = jellybeans

t = trail mix

2j + 5t = 13

8j + 3t = 18

To solve this, we want to isolate a variable, and solve for the other. Let's do that with each equation.

2j + 5t = 13

2j = 13 - 5t

j = (13 - 5t) / 2

8j + 3t = 18

8j = 18 - 3t

j = (18 - 3t) / 8

Now that we have isolated the variable j in each equation, we can set the two right-sides equal to each other, and solve for t.

(13 - 5t) / 2 = (18 - 3t) / 8

By multiplying both sides by 8, this becomes simpler to look at.

8 * (13 - 5t) / 2 = 18 - 3t

4 * (13 - 5t) = 18 - 3t

4 * 13 - 4 * 5t = 18 - 3t

52 - 20t = 18 - 3t

52 = 18 + 17t

34 = 17t

34 / 17 = t

t = 2

Now that we have solved for t, we can trivially plug it into one of the other equations to solve for j. Let's choose the first equation, as it doesn't matter which we choose.

t = 2

2j + 5t = 13

2j + 5 * 2 = 13

2j + 10 = 13

2j = 13 - 10

2j = 3

j = 1.5

So, each pound of jelly beans costs 1.5 dollars, or $1.50, and each pound of trail mix costs 2 dollars, or $2.

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Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st
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Answer:

\boxed{2^{\frac{802}{27}} \cdot 3^9}

Step-by-step explanation:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

\left(\dfrac{1}{3^2}  \cdot \dfrac{1}{4^5}  \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3

Also,

\boxed{a^{0} = 1, a\neq 0 }

\left(\dfrac{1}{9}  \cdot \dfrac{1}{1024}  \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

Note

\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq  0}

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

Note

\boxed{\dfrac{1}{\dfrac{1}{a} }  = a}

9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)

Once

9216=2^{10}\cdot 3^2 \implies  9216^3=2^{30}\cdot 3^6

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And

$4^{\frac{4}{27}} = 2^{\frac{8}{27} $

We have

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Also, once

\boxed{\dfrac{c^a}{c^b}=c^{a-b}}

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As

30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27}  =\dfrac{802}{27}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3

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The general solution of tan(b·x) = 2, given that the smallest positive solution is x = 0.3, is presented as follows;

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The given smallest positive solution of tan (b·x) = 2 is x = 0.3

The general solution of tan (b·x + c) = m, is given as follows;

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α = arctan(m) = x₀

The minimum positive value of the general solution is therefore presented as follows;

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The general solution of tan(b·x) = 2, is therefore;

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Learn more about the general solution of a sine function here:

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Possible rational roots will be,

x = \frac{\pm \text{factor of 35}}{\pm \text{Facors of 66}}

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