All categories

3 years ago

Ignore my errors please. how would i solve these problems... please explain step by step

Mathematics

1 answer:

chubhunter [2.5K]3 years ago

5 0

1.

Make the fractions improper.

72/12 x 19/6 = 19.

2.

Area = L x W

L = 3 1/6 x
W = 6 7/12

= 19.

#1 = 19 feet
#2 = 9 3/4 feet

You might be interested in

What is x?: 8+9.8x+9=3.8x - 16+3.8x

Alika [10]

Answer:

-15

Step-by-step explanation:

8 0

3 years ago

The expression below shows how to convert 12 gallons to cups with one part missing

Basile [38]

Your answer is c!!!
Hdjdjndjdndnndndndnndn

7 0

3 years ago

Read 2 more answers

3y=-2x-19 -5x-3y=-2

nevsk [136]

3y=-2x-19 -5x-3y=-2-5x - (-2x - 19) = -2
-5x + 2x + 19 = -2
-3x = -21
x = 7

3y = -2(7) - 19
3y = -14 - 19
3y = 33
y = 11

4 0

3 years ago

Read 2 more answers

1. In ABC, A=45 degrees, B=75 degrees, and BC=12. What is the length of AB?

Aleksandr-060686 [28]

We know that sum of the interior angles of any triangle equal 180° :

So :

A + B + C = 180°

45° + 75° + C = 180°

120° + C = 180°

Both sides minus ( 120°) :

C = 180° - 120°

C = 60°

_________________________________

According to the theorem of sinuses :

$\frac{AB}{ \sin(C) } = \frac{BC}{ \sin(A) } \\$

$\frac{AB}{ \sin(60) } = \frac{12}{ \sin(45) } \\$

$\frac{AB}{ \frac{ \sqrt{3} }{2} } = \frac{12}{ \frac{ \sqrt{2} }{2} } \\$

$Multiply \: both \: sides \: by \: \frac{ \sqrt{3} }{2}$

$AB = \frac{24}{ \sqrt{2} } \times \frac{ \sqrt{3} }{2} \\$

$AB = \frac{12 \sqrt{3} }{ \sqrt{2} } \\$

$AB = \frac{2 \times 6 \sqrt{3} }{ \sqrt{2} } \\$

$AB = \frac{ ({ \sqrt{2} })^{2} \times 6 \sqrt{3} }{ \sqrt{2} } \\$

$AB = \frac{ \sqrt{2} \times \sqrt{2} \times 6 \sqrt{3} }{ \sqrt{2} } \\$

$AB = \frac{ \sqrt{2} \times 6 \sqrt{3} }{1} \\$

$AB = \sqrt{2} \times 6 \sqrt{3}$

$AB = 6 \sqrt{2 \times 3}$

$AB = 6 \sqrt{6}$

_________________________________

I think this is the correct answer.

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

4 0

4 years ago

Heyo random person scrolling.Please help me?

Ierofanga [76]

Let's do

$\\ \sf\longmapsto 2sin^260cos60tan^245$

$\\ \sf\longmapsto 2\left(\dfrac{\sqrt{3}}{2}\right)^2\times \dfrac{1}{2}\times (1)^2$

$\\ \sf\longmapsto 2\times \dfrac{(\sqrt{3})^2}{2^2}\times \dfrac{1}{2}\times 1$

$\\ \sf\longmapsto \dfrac{3}{2}\times \dfrac{1}{4}$

$\\ \sf\longmapsto \dfrac{3}{4}$

Trigonometric values

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} $ \sin $ & 0 & $\dfrac{1}{2 }$ & $\dfrac{1}{ \sqrt{2} }$ & $\dfrac{ \sqrt{3}}{2}$ & 1 \\ \cline{1-6} $ \cos $ & 1 & $ \dfrac{ \sqrt{ 3 }}{2} } $ & $ \dfrac{1}{ \sqrt{2} } $ & $ \dfrac{ 1 }{ 2 } $ & 0 \\ \cline{1-6} $ \tan $ & 0 & $ \dfrac{1}{ \sqrt{3} } $ & 1 & $ \sqrt{3} $ & $ \infty $ \\ \cline{1-6} \cot & $ \infty $ &$ \sqrt{3} $ & 1 & $ \dfrac{1}{ \sqrt{3} } $ &0 \\ \cline{1 - 6} \sec & 1 & $ \dfrac{2}{ \sqrt{3}} $ & $ \sqrt{2} $ & 2 & $ \infty $ \\ \cline{1-6} \csc & $ \infty $ & 2 & $ \sqrt{2 } $ & $ \dfrac{ 2 }{ \sqrt{ 3 } } $ & 1 \\ \cline{1 - 6}\end{tabular}}$

6 0

3 years ago