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garik1379 [7]
3 years ago
15

The length of each side of a cube is x cm. If x is an interference,why can’t the volume of the cube equal 15 cm

Mathematics
1 answer:
slamgirl [31]3 years ago
5 0

we don't know what the value of "X" is, so therefore, we cannot conclude that it is 15 cm.

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Please help you will get 20 points and explain your answer please
sp2606 [1]

Answer:

Top prism = 262 in.² Bottom prism = 478 in.²

Step-by-step explanation:

top prism:

front + back: 5 x 3 = 15

sides: 19 x 4 x 2 = 152

bottom: 19 x 5 = 95

15 + 152 + 95 = 262

bottom prism:

front + back: 5 x 6 x 2 = 60

sides: 19 x 6 x 2 = 228

top + bottom: 19 x 5 x 2 = 190

60 + 228 + 190 = 478

7 0
4 years ago
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A local car dealer is advertising a 2-year lease
MAVERICK [17]

Answer:

c = 150m+3000

Step-by-step explanation:

So if 150 is the monthly payment and c is the total cost you'd be looking for c and 150 would be multiplied by m for how ever many months is takes to pay it off. You would then just add on the 3000 dollar deposit since its just a one time payment to start out with.

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3 years ago
What is the image of (-7,-1) after a dilation by a scale factor of 4 centered at the origin
nlexa [21]

Answer: (-28,-4)  

Step-by-step explanation:

Multiply the x and y coordinate values by the scale factor to find the image.

-7 * 4 = -28  

-1 * 4 = -4

(-28,-4)

7 0
3 years ago
There are 10 seats at each cafeteria table. How many ways could me and my nine friends sit at the table?
Andru [333]

Answer:

100

Step-by-step explanation:

5 0
3 years ago
Find a polynomial of the form
Sloan [31]

Answer:

f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3

Step-by-step explanation:

We are given that

f(x)=ax^3+bx^2+cx+d

f(0)=-3,f(1)=2,f(3)=5 and f(4)=0

We have to find the polynomial

Substitute the value x=0 then ,we get

f(0)=d=-3

Substitute x=1 then we get

a+b+c-3=2

a+b+c=2+3=5

a+b+c=5 (equation I)

Substitute x=3 then we get

a(3)^3+b(3)^2+c(3)-3=5

27a+9b+3c=5+3=8

27a+9b+3c=8  (Equation II)

Substitute x=4 then we get

a(4)^3+b(4)^2+c(4)-3=0

64a+16b+4c=3 (Equation III)

Equation I multiply by 3 then subtract from equation II

24a+6b=-7 (Equation IV)

Equation II multiply by 4 and equation III multiply by 3 and subtract equation II from III

84a+12b=-23 (Equation V)

Equation IV multiply by 2 and then subtract from equation V

36a=-9

a=-\frac{9}{36}

a=-\frac{1}{4}

Substitute the value of a in equation IV then we get

24(-\frac{1}{4})+6b=-7

-6+6b=-7

6b=-7+6=-1

b=-\frac{1}{6}

Substitute the value of b in equation I then we get

-\frac{1}{4}-\frac{1}{6}+ c=5

-\frac{5}{12}+c=5

c=5+\frac{5}{12}=\frac{65}{12}

Substitute the values then we get

f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3

5 0
4 years ago
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