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erik [133]
3 years ago
12

2r+8-r=-7 Need help ASAP

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
8 0

Answer:

r= (-15) is the answer to the question ok

Mumz [18]3 years ago
6 0

Answer:

2r + 8 - r = 7

r + 8 = 7

r = -1

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10.) state whether the given side lengths can form a triangle
Verizon [17]

Answer:

A. 5,8,12, side lengths can form a triangle.

B. 20, 18, 2, side lengths can not form a triangle.

C. 15, 26 , 9, side lengths can not form a triangle.

D. 4.75, 12.25, 16.25, side lengths can form a triangle.

Step-by-step explanation:

We are given three sides in options A, B , C and D.

We need to check if the given sides would form a triangle or not.

Note: Sum of two sides is always greater than third sides in a triangle.

So, we need to check the sum of two sides of given sides length to check if sum is greater than third side length.

A. 5,8,12.

5+8 is greater than 12.

5+12 is greater than 8.

12+8 is greater than 5.

Therefore A. 5,8,12, side lengths can form a triangle.

B. 20, 18, 2

18+2 is not greater than 20.

Therefore B. 20, 18, 2, side lengths can not form a triangle.

C. 15, 26 , 9

15+9 is less than 26.

Therefore C. 15, 26 , 9, side lengths can not form a triangle.

D. 4.75, 12.25, 16.25.

4.75+12.25 is greater than 16.25.

4.75+16.25 is greater than 12.25.

12.25+16.25 is greater than 4.75.

Therefore D. 4.75, 12.25, 16.25, side lengths can form a triangle.


3 0
3 years ago
The volume of a cylinder is 81 cubic feet. What is the volume of a cone that fits exactly inside the cylinder? Provide an explan
expeople1 [14]
V ( cylinder ) = r² π h
V ( cone ) = 1/3 r² π h
A cone that fits exactly inside the cylinder has the volume:
V ( cone ) = 1/3 V ( cylinder = 1/3 · 81 = 27 ft³
8 0
3 years ago
What is 9 3/10- 5 7/10 equal
Scorpion4ik [409]
T\The answer would be 3 6/10, or 3.6.
3 0
4 years ago
Using these complex zeros (1,1,-1/2,2+i,2-i) factor f(x)=-2x^5 +11x^4 -22x^3 +14x^2 +4x -5
Marianna [84]
\bf \begin{cases}
x=1\implies &x-1=0\\
x=1\implies &x-1=0\\
x=-\frac{1}{2}\implies 2x=-1\implies &2x+1=0\\
x=2+i\implies &x-2-i=0\\
x=2-i\implies &x-2+i=0
\end{cases}
\\\\\\
(x-1)(x-1)(2x+1)(x-2-i)(x-2+i)=\stackrel{original~polynomial}{0}
\\\\\\
(x-1)^2(2x+1)~\stackrel{\textit{difference of squares}}{[(x-2)-(i)][(x-2)+(i)]}

\bf (x^2-2x+1)(2x+1)~[(x-2)^2-(i)^2]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)-(-1)]
\\\\\\
(x^2-2x+1)(2x+1)~[(x^2-4x+4)+1]
\\\\\\
(x^2-2x+1)(2x+1)~[x^2-4x+5]
\\\\\\
(x^2-2x+1)(2x+1)(x^2-4x+5)

of course, you can always use  (x-1)(x-1)(2x+1)(x²-4x+5)  as well.
7 0
3 years ago
I WILL GIVE BRAINLIEST<br> Who will get this question right
Ksju [112]
If 101 use 3/5 for his Artwork, then the space remaining for 102 is:

5/5 - 3/5 = 2/5 (since the whole wall represents 100% or 5/5 or 1)

6 0
3 years ago
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