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DerKrebs [107]
3 years ago
12

What is the value of (Negative three-fourths) Superscript negative 4?

Mathematics
2 answers:
Goryan [66]3 years ago
8 0

Answer:

-3.16049382716

Step-by-step explanation:

-3/4 = -0.75

-0.75 ^ -4 = -3.16049382716

ZombieBoy
2 years ago
Ah yes, Pi is the answer
svetoff [14.1K]3 years ago
5 0

D) 256/81

I just took the quiz on edge

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What is the solution to -2(5y-5)-3y
Alenkasestr [34]

10-13y

hope this helps

7 0
3 years ago
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The value of t can be determined by using the expression 2.75s. Which table represents the relationship between the values of t
ad-work [718]
This isn't a table, but you can easily make it into one.

s          t
-2     -5.5
-1      -2.75
0         0
1        2.75
2        5.5
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3 years ago
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Find the value of x in each Figure. Please help, thank you!!
Archy [21]

Answer:

#5  39 = x

Sides:

x - 15

2x

2x

Total of sides = 180

180 = (x-15) + 2x + 2x

180 = x - 15 + 4x  

180 = -15 + 5x     (add 15 to each side to remove -15)

195 = 5x     (divide by 5 on each side to get x)

39 = x

#7  24 = x

Sides:

2x - 4

3x + 5  

2x + 11

Total of sides = 180

180 = (2x - 4) + (3x + 5) + (2x + 11)

180 = 2x - 4 + 3x + 5 + 2x + 11

180 = 7x + 12      (subtract 12 to each side to remove 12)

168 = 7x      (divide by 7 on each side to get x)

24 = x

#8  37 = x

Sides:

x

x + 4

3x - 9

Total of sides = 180

180 = (x) + (x + 4) + (3x - 9)

180 = x + x + 4 + 3x - 9

180 = 5x - 5      (add 5 to each side to remove -5)  

185 = 5x      (divide by 5 on each side to get x)

37 = x

Step-by-step explanation:

5 0
3 years ago
A circle has centre (3,0) and radius 5. The line y = 2x + k intersects the circle in two points. Find the set
lara [203]

A circle with center (3,0) and radius 5 has equation

(x-3)^2+y^2=25 \iff x^2 + y^2- 6 x  = 16

If we substitute y=2x+k in this equation, we have

x^2+(2x+k)^2-6x=16 \iff 5x^2+(4k-6)x+k^2-16=0

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

\Delta=b^2-4ac=(4k-6)^2-4\cdot 5\cdot (k^2-16)>0

The expression simplifies to

-4 (k^2 + 12 k - 89)>0 \iff k^2 + 12 k - 89

The solutions to the associated equation are

k^2 + 12 k - 89=0 \iff k=-6\pm 5\sqrt{5}

So, the parabola is negative between the two solutions:

-6-5\sqrt{5}

8 0
3 years ago
What is the solution to the following equation?
cupoosta [38]

The answer is 8. you are basically adding 13 to -5. so it would be 13-5=8.

3 0
3 years ago
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