Can somebody tell me the answer and I will pay u

One bar of candy A and two bars of candy B have 774 calories. Two bars of candy A and one bar of candy B contains 786 calories.

vampirchik [111]

◆ Define the variables:

Let the calorie content of Candy A = a
and the calorie content of Candy B = b

◆ Form the equations:

One bar of candy A and two bars of candy B have 774 calories. Thus:

a + 2b = 774

Two bars of candy A and one bar of candy B contains 786 calories

2a + b = 786

◆ Solve the equations:

From first equation,
a + 2b = 774
=> a = 774 - 2b

Put a in second equation
2×(774-2b) + b = 786
=> 2×774 - 2×2b + b = 786
=> 1548 - 4b + b = 786
=> -3b = 786 - 1548
=> -3b = -762
=> b = -762/(-3) = 254 calorie

◆ Find caloric content:

Caloric content of candy B = 254 calorie

Caloric content of candy A = a = 774 - 2b = 774 - 2×254 = 774 - 508 = 266 calorie

5 0

3 years ago

35 points! plz help ASAP<br />I have worked the problem<br />a-(2+3) =-2<br />

Firlakuza [10]

a-(2+3)=-2

a-5=-2

a=3

You add 5 to -2 to get your answer.

6 0

3 years ago

Read 2 more answers

One factor of f(x) = x3 − 14x2 + 61x − 84 is (x − 7). What are the zeros of the function?

Slav-nsk [51]

Answer:

-7

Step-by-step explanation:

Because zero is the x, so x=-7 then it's -7

8 0

3 years ago

A researcher weighed pumpkins grown across a certain region. The data are rounded to the nearest pound and displayed in this his

icang [17]

Answer:

I think it would be c

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter and area in square units

ad-work [718]

Check the picture below.

the triangle has that base and that height, recall that A = 1/2 bh.

now as for the perimeter, you can pretty much count the units off the grid for the segment CB, so let's just find the lengths of AC and AB,

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&C&(~ 6 &,& 2~)
\end{array}~~~ 
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{(6-2)^2+(2-8)^2}\implies AC=\sqrt{4^2+(-6)^2}
\\\\\\
AC=\sqrt{16+36}\implies AC=\sqrt{52}\implies AC=\sqrt{4\cdot 13}
\\\\\\
AC=\sqrt{2^2\cdot 13}\implies AC=2\sqrt{13}$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&A&(~ 2 &,& 8~) 
% (c,d)
&B&(~ 16 &,& 2~)
\end{array}\\\\\\
AB=\sqrt{(16-2)^2+(2-8)^2}\implies AB=\sqrt{14^2+(-6)^2}
\\\\\\
AB=\sqrt{196+36}\implies AB=\sqrt{232}\implies AB=\sqrt{4\cdot 58}
\\\\\\
AB=\sqrt{2^2\cdot 58}\implies AB=2\sqrt{58}$

so, add AC + AB + CB, and that's the perimeter of the triangle.

8 0

4 years ago