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4 years ago

Jayla ate 75% of the jelly beans in the bag. If she ate 24 jelly beans how many were in the bag?

Mathematics

1 answer:

maksim [4K]4 years ago

7 0

Answer:

32 jelly beans were in the bag

Step-by-step explanation:

We know 75% is 3/4 and that she ate 24, so divide 24 by 3 then multiply it by 4:

24/3= 8

8 x 4 = 32

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3 years ago

Aiden owns a food truck that sells tacos and burritos. He sells each taco for $4.75 and each burrito for $7. Yesterday Aiden mad

nevsk [136]

Number of tacos sold is 65 and number of burritos sold is 40

<h3><u>Solution:</u></h3>

Given that Aiden sells each taco for $4.75 and each burrito for $7

Let the number of tacos sold be "t" and number of burritos sold be "b"

Given that Aiden sold 25 more tacos than burritos

t = b + 25 ---- eqn 1

Also given that yesterday Aiden made a total of $588.75 in revenue

number of tacos sold x cost of each tacos + number of burritos sold x cost of each burritos = 588.75

$t \times 4.75 + b \times 7 = 588.75$

4.75t + 7b = 588.75 ----- eqn 2

Substitute eqn 1 in eqn 2

4.75(b + 25) + 7b = 588.75

4.75b + 118.75 + 7b = 588.75

11.75b = 470

b = 40

Substitute b = 40 in eqn 1

t = 40 + 25

t = 65

Thus the number of tacos sold is 65 and number of burritos sold is 40

5 0

3 years ago

Lin runs 5 miles in 40 minutes.

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3 0

3 years ago

Use Cramer Rule to solve the following system: 8x−5y=70 and 9x+7y=3

nlexa [21]

Answer:

$(x,y) = (5,-6)$

Step-by-step explanation:

$\underline{\textbf{Determinant of a matrix.}}\\\\\text{For a}~ 2 \times 2 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2\\b_1&b_2 \end{vmatrix} = a_1b_2 - a_2b_1\\\\\\\text{For a}~ 3 \times 3 ~ \text{matrix,}\\\\\begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2&b_3\\c_2&c_3 \end{vmatrix} - a_2 \begin{vmatrix} b_1&b_3\\c_1&c_3 \end{vmatrix}+ a_3 \begin{vmatrix} b_1&b_2\\c_1&c_2 \end{vmatrix}\\\\\\$

$~~~~~~~~~~~~~~~~~~=a_1(b_2c_3-b_3c_2) -a_2(b_1c_3-b_3c_1) +a_3(b_1c_2-b_2c_1)$

$\underline{\textbf{Cramer's Rule to solve a system of two equations.}}\\\\\text{Consider the system of two equations:}\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_1x + b_1 y= c_1\\\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~a_2x +b_2 y = c_2\\\\\text{Here,}\\\\x = \dfrac{D_x}{D}= \dfrac{\begin{vmatrix} c_1&b_1\\c_2&b_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\\\ y= \dfrac{D_y}{D}= \dfrac{\begin{vmatrix} a_1&c_1\\a_2&c_2 \end{vmatrix}}{\begin{vmatrix} a_1&b_1\\a_2&b_2 \end{vmatrix}}\\\\$

$\underline{\textbf{Solution:}}\\\\~~~~~~~~~~~~~~~~~~~~~~~8x-5y = 70~~~~~~...(i)\\\\~~~~~~~~~~~~~~~~~~~~~~~9x +7y = 3~~~~~~~...(ii)\\\\\text{Applying Cramer's rule:}\\\\x = \dfrac{D_x}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 70& -5 \\3&7 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{70(7) -(-5)(3)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{490+15}{56+45}\\\\\\~~=\dfrac{505}{101}\\\\\\~~=5$

$y = \dfrac{D_y}{D}\\\\\\~~=\dfrac{\begin{vmatrix} 8& 70 \\9&3 \end{vmatrix}}{\begin{vmatrix} 8& -5\\ 9& 7\end{vmatrix}}\\\\\\~~=\dfrac{(8)(3) -(70)(9)}{(8)(7)-(-5)(9)}\\\\\\~~=\dfrac{24-630}{56+45}\\\\\\~~=-\dfrac{606}{101}\\\\\\~~=-6$

$\textbf{Hence, the solution to the system of equation is}~ (x,y) = (5,-6)$

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2 years ago

Whats an equation slop 0.5, passes through(6,4)

4vir4ik [10]

$\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{4})~\hspace{10em} slope = m\implies 0.5 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=0.5(x-6) \\\\\\ y-4=0.5x-3\implies y=0.5x+1$

5 0

3 years ago