All categories

2 years ago

I really don’t understand what the questions asking? How do I solve this

Mathematics

1 answer:

Eddi Din [679]2 years ago

5 0

Answer:

$17.85

Step-by-step explanation:

Let the dollar amount per shirt be represented by s.

Let the dollar amount per pants be represented by p.

You have the following system:

3s+2p=104.81

2s+1p=61.33

If I multiply the bottom equation by -2, I can set the system up for elimination since the equations are already and the same form and I will have 2p and -2p in the same column there. 2p+(-2p)=0

Multiplying equation 2 by -2:

3s+2p=104.81

-4s-2p=-122.66

-----------------------Add the equations:

-1s =-17.85

Multiply both sides by -1:

s=17.85

So one shirt cost $17.85

You might be interested in

Evaluate |-a^2|, given a = 5, b = -3, and c = -2.

geniusboy [140]

The sunset is 25. how this helps!

7 0

2 years ago

Read 2 more answers

Find the slope of the line passing through the points (2,-9) and )1,-3)

nignag [31]

Hi there!

$\large\boxed{\text{slope =} 6}$

We can calculate slope using the following formula:

$slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Plug in the corresponding points:

$slope = \frac{-9-(-3)}{2-1}$

Simplify:

$slope = \frac{-6}{1} = -6$

3 0

2 years ago

Read 2 more answers

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

6 0

3 years ago

Evaluate: sin (225degree)

PtichkaEL [24]

Sin(225)

225 degrees is in the 3rd quadrant. And in the third quadrant only tangent is positive, sin is negative.

For third quadrant, basic angle = 225 - 180 = 45

sin225 = - sin45

sin225 = - sin45 = -0.7071

sin 225° ≈ -0.7071

7 0

2 years ago

What is the difference between 3.9 and -1.6

mars1129 [50]

If you subtract the two you will get 5.5

8 0

3 years ago

Read 2 more answers