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schepotkina [342]
3 years ago
12

I spent 3/4 of my money and then I spent 1/5 of what was left. 

Mathematics
1 answer:
Alborosie3 years ago
6 0
1/20 is the fraction that you'll have left
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What is the slope of the line whose equation is 12=4x−6y?<br><br> Enter your answer in the box.
swat32
Slope= 2/3
Explanation- adding 6y on both sides and subtracting 12 on both sides of eq, 12 + 6y -12=4x+6y-6y-12
6y = 4x-12
Divide by 6 on both sides of eq, we get y=4/6 x -12/6
y=2/3 x-2
5 0
2 years ago
Convert the following equation into slope intercept form. 3x + 21y = 21 Stars
RideAnS [48]

Answer:

y = -(1/7)x + 1

Step-by-step explanation:

3x + 21y = 21 \\ 21y = 21 - 3x \\ y =  \frac{21 - 3x}{21}  \\ y = 1 -  \frac{1}{7} x \\ y =  -  \frac{1}{7} x + 1

6 0
2 years ago
Please help me the pic might not load
ladessa [460]
<h3>Answer: B) 25 = 10 + w</h3>

5 0
3 years ago
The perimeter of a rectangle is 50 centimeters. If the width of the rectangle is 9 centimeters
nordsb [41]

Answer:

Length = 17 cm

Width = 8 cm

Step-by-step explanation:

Let the length be x cm

Therefore, Width = (x - 9) cm

Perimeter of rectangle = 2(l +w)

50 = 2[x + (x - 9)]

50/2 = 2x - 9

25 = 2x - 9

25+ 9 = 2x

2x = 34

x = 34/2

x = 17 cm

x - 9 = 17 - 9 = 8 cm

Therefore,

Length = 17 cm

Width = 8 cm

Thus, the dimensions of the rectangle are 17 cm and 8 cm.

6 0
2 years ago
Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located
arsen [322]
Check the picture below.

\bf \textit{using the pythagorean theorem}&#10;\\\\&#10;c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\&#10;-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}&#10;\\\\\\&#10;csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

3 0
3 years ago
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