All categories

3 years ago

Helllllpppppp!

Mathematics

2 answers:

Akimi4 [234]3 years ago

8 0

Answer:

Step-by-step explanation:

The system of equations is:

● 3x + 2y = 7

● 2x + 4y = 8

Where x is the price of a single coffee and y is the price of a single coffee

We will solve this simultaneous equations using the graphing method:

The solutions is the intersection point of to the lines.

(Picture below)

The intersection point is (1.5,1.25)

andreyandreev [35.5K]3 years ago

5 0

Answer:

x = 1.5

y =1.25

Step-by-step explanation:

3x+2y=7

2x+4y=8

Multiply the first equation by -2

-2(3x+2y=7 )

-6x-4y = -14

Add this to the second equation

-6x-4y = -14

2x+4y=8

----------------

-4x = -6

Divide by -4

-4x/-4 = -6/-4

x = 1.5

Now find y

3x+2y=7

3(1.5) + 2y = 7

4.5 +2y = 7

Subtract 4.5 from each side

2y = 2.5

Divide by 2

y = 1.25

You might be interested in

If shoes are $20 each and the markup is 150%, how much will that cost?

MArishka [77]

150% to decimal = 1.50 since 100% = 1
1.50 x 20 = 30
30 + 20 = 50

3 0

3 years ago

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.