Ask question

Ask question

All categories

3 years ago

8

A candy store sells caramels and milk chocolate by the pound the table below shows the total cost in dollars for a pound of each

type of candy the store sells
Caramels: $9.28
Milk Chocolate: $12.80

How much more is the cost for 1 3/4 pounds of milk chocolate than the cost for 1 3/4 pounds of caramels?

2 answers:

kari74 [83]3 years ago

4 0

Answer:

$ 6.16

Step-by-step explanation:

Given,

The cost of each pounds of Caramels = $ 9.28,

So, the cost of $1\frac{3}{4}$ pounds of caramels = $1\frac{3}{4}\times 9.28$

$=\frac{7}{4}\times 9.28$

$=7\times 2.32$

= $ 16.24,

Now, the cost of each pounds of milk chocolate = $ 12.80,

So, the cost of $1\frac{3}{4}$ milk chocolate = $1\frac{3}{4}\times 12.80$

$=\frac{7}{4}\times 12.80$

$=7\times 3.2$

= $ 22.4,

∵ 22.4 > 16.24,

Thus, the cost of $1\frac{3}{4}$ pounds of milk chocolate > The cost of $1\frac{3}{4}$ pounds of caramels,

Also, 22.4 - 16.24 = 6.16,

Hence, the cost for 1 3/4 pounds of milk chocolate is $ 6.16 more than the cost for 1 3/4 pounds of caramels.

Rasek [7]3 years ago

3 0

Answer:

$6.16

Step-by-step explanation:

1 3/4 Caramels cost is $16.24

1 3/4 Chocolate cost is $22.40.

22.40 - 16.24 = $6.16

<em>$6.16 more</em>

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>

You might be interested in

Which property justifies the following steps:

ad-work [718]

Step 3:
14/7x-6/3=5
2x-2=5
2x=5+2
2x=7
x=7/2

6 0

3 years ago

Read 2 more answers

∠xangle, x and \angle y∠yangle, y are supplementary angles. \angle y∠yangle, y measures 88^\circ88

Yuliya22 [10]

Answer:

92∘

Step-by-step explanation:

Supplementary angles combine to create a straight angle. So, they have a sum of 180.

∠x + ∠y

∠x + 88 --------------> 180

=92∘

4 0

3 years ago

A particle moving in a planar force field has a position vector x that satisfies x'=Ax. The 2×2 matrix A has eigenvalues 4 and 2

andrey2020 [161]

Answer:

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

Step-by-step explanation:

Consider the provided matrix.

$v_1=\begin{bmatrix}-3\\1 \end{bmatrix}$

$v_2=\begin{bmatrix}-1\\1 \end{bmatrix}$

$\lambda_1=4, \lambda_2=2$

The general solution of the equation $x'=Ax$

$x(t)=c_1v_1e^{\lambda_1t}+c_2v_2e^{\lambda_2t}$

Substitute the respective values we get:

$x(t)=c_1\begin{bmatrix}-3\\1 \end{bmatrix}e^{4t}+c_2\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-3c_1e^{4t}-c_2e^{2t}\\c_1e^{4t}+c_2e^{2t} \end{bmatrix}$

Substitute initial condition $x(0)=\begin{bmatrix}-6\\1 \end{bmatrix}$

$\begin{bmatrix}-3c_1-c_2\\c_1+c_2 \end{bmatrix}=\begin{bmatrix}-6\\1 \end{bmatrix}$

Reduce matrix to reduced row echelon form.

$\begin{bmatrix} 1& 0 & \frac{5}{2}\\ 0& 1 & \frac{-3}{2}\end{bmatrix}$

Therefore, $c_1=2.5,c_2=1.5$

Thus, the general solution of the equation $x'=Ax$

$x(t)=2.5\begin{bmatrix}-3\\1\end{bmatrix}e^{4t}-1.5\begin{bmatrix}-1\\1 \end{bmatrix}e^{2t}$

$x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

The required position of the particle at time t is: $x(t)=\begin{bmatrix}-7.5e^{4t}+1.5e^{2t}\\2.5e^{4t}-1.5e^{2t}\end{bmatrix}$

6 0

3 years ago

Which functions have a y-intercept that is greater than the y-intercept of the function g(x) = |x + 3| + 4? Check three

Marina CMI [18]

Answer:

A

Step-by-step explanatio

If i were to give my judge on this <9M+13=5^9

7 0

2 years ago

N + 3m; use m = 3, and n = 1

musickatia [10]

Answer:

10

Step-by-step explanation:

plug 3 into m and plug 1 into n
order of operations say you multiple first so you multiple 3 times 3 which equals 9
then you add 1 and 9 which equals 10

4 0

3 years ago