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baherus [9]
3 years ago
7

PLEASE HELP ASAP worth alot

Mathematics
1 answer:
vredina [299]3 years ago
4 0

Answer:

I believe it is C

Step-by-step explanation:

Hope this helps

and if it at all possible and you mark me brainlyest please

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A rectangular pyramid has a volume of 1,296 mc014-1.jpg, a height of 24 in., and a base length of 6 in. What is the width of its
stiks02 [169]
V=L X W X H

1,296=6 x w x 24
1,296=144 x w
1296/144=w
9=width
3 0
3 years ago
Read 2 more answers
Solve the equation for y. Identify the slope and y-intercept then graph the equation.
drek231 [11]

Answer:

y = 1        m = -3       b = 1

Step-by-step explanation:

y = mx + b                          y = -3x + 1

y = 1

m = -3

b = 1

5 0
3 years ago
How many different ways can you make 82 cents using current u.s. currency
andrew11 [14]
 <span>You can probably just work it out. 

You need non-negative integer solutions to p+5n+10d+25q = 82. 

If p = leftovers, then you simply need 5n + 10d + 25q ≤ 80. 


So this is the same as n + 2d + 5q ≤ 16 

So now you simply have to "crank out" the cases. 

Case q=0 [ n + 2d ≤ 16 ] 

Case (q=0,d=0) → n = 0 through 16 [17 possibilities] 
Case (q=0,d=1) → n = 0 through 14 [15 possibilities] 
... 
Case (q=0,d=7) → n = 0 through 2 [3 possibilities] 
Case (q=0,d=8) → n = 0 [1 possibility] 

Total from q=0 case: 1 + 3 + ... + 15 + 17 = 81 

Case q=1 [ n + 2d ≤ 11 ] 
Case (q=1,d=0) → n = 0 through 11 [12] 
Case (q=1,d=1) → n = 0 through 9 [10] 
... 
Case (q=1,d=5) → n = 0 through 1 [2] 

Total from q=1 case: 2 + 4 + ... + 10 + 12 = 42 

Case q=2 [ n + 2 ≤ 6 ] 
Case (q=2,d=0) → n = 0 through 6 [7] 
Case (q=2,d=1) → n = 0 through 4 [5] 
Case (q=2,d=2) → n = 0 through 2 [3] 
Case (q=2,d=3) → n = 0 [1] 

Total from case q=2: 1 + 3 + 5 + 7 = 16 

Case q=3 [ n + 2d ≤ 1 ] 
Here d must be 0, so there is only the case: 
Case (q=3,d=0) → n = 0 through 1 [2] 

So the case q=3 only has 2. 

Grand total: 2 + 16 + 42 + 81 = 141 </span>
3 0
3 years ago
How do you solve please help geometry
maxonik [38]
Your diagram doesn't clearly define what values correspond to what, can you explain.
7 0
3 years ago
Sebastian solved the radical equation y + 1 = but did not check his solution. (y + 1)2 = y2 + 2y + 1 = –2y – 3 y2 + 4y + 4 = 0 (
Softa [21]

Answer:

There are no true solutions to the equation.

Step-by-step explanation:

<u><em>The correct equation is</em></u>

y+1=\sqrt{-2y-3}

Solve for y

squared both sides

(y+1)^2=(-2y-3)

(y^2+2y+1)=(-2y-3)

y^2+2y+1+2y+3=0

y^2+4y+4=0

(y+2)(y+2)=0

y=-2

<em>Verify</em>

substitute the value of y in the original expression

-2+1=\sqrt{-2(-2)-3}

-1=1 ----> is not true

therefore

There are no true solutions to the equation.

4 0
3 years ago
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