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Zarrin [17]
3 years ago
10

I really need help with this pls help me

Mathematics
2 answers:
nexus9112 [7]3 years ago
7 0

9514 1404 393

Answer:

  (a)  -2

Step-by-step explanation:

Put -1 where x is and do the arithmetic.

  f(x) = x³ +2x² -3

  f(-1) = (-1)³ +2(-1)² -3

  f(-1) = -1 +2 -3 = 1 -3

  f(-1) = -2

Agata [3.3K]3 years ago
6 0
I think it is the first one -2
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A cereal company is exploring new cylinder-shaped containers. Two possible containers are shown in the diagram.x is 6 and r=4.5
KatRina [158]

Answer:

Container \rm X will have less label area than container \rm Y by about 9\; \rm in.

Step-by-step explanation:

A rectangular sheet of paper can be rolled into a cylinder. Conversely, the lateral surface of a cylinder can be unrolled into a rectangle- without changing the area of that surface.

Indeed, the width of that rectangle will be the same as the height of the cylinder. On the other hand, the length of that rectangle should be exactly equal to the circumference of the circles on the top and the bottom of the cylinder. In other words, if a cylinder has a height of h and a radius of r at the top and the bottom, then its lateral surface can be unrolled into a rectangle of width h and length 2\,\pi\, r.

Apply this reasoning to both cylinder \mathrm{X} and \rm Y:

For cylinder \mathrm{X}, h = 6\; \rm in while r = 4.5\; \rm in. Therefore, when the lateral side of this cylinder is unrolled:

  • The width of the rectangle will be 6\; \rm in, while
  • The length of the rectangle will be 2 \, \pi \times 4.5\; \rm in = 9\, \pi\; \rm in.

That corresponds to a lateral surface area of 54\, \pi\; \rm in^2.

For cylinder \rm Y, h = 10.5\; \rm in while r = 3\; \rm in. Similarly, when the lateral side of this cylinder is unrolled:

  • The width of the rectangle will be 10.5\; \rm in, while
  • The length of the rectangle will be 2\pi\times 3\; \rm in = 6\,\pi \; \rm in.

That corresponds to a lateral surface area of 63\,\pi \; \rm in^2.

Therefore, the lateral surface area of cylinder \rm X is smaller than that of cylinder \rm Y by 9\,\pi\; \rm in^2.

5 0
3 years ago
100-point Question!!!
FrozenT [24]

Answer:

Two or more independent functions (say f(x) and g(x)) can be combined to generate a new function (say g(x)) using any of the following approach.

h(x) = f(x) + g(x)h(x)=f(x)+g(x) h(x) = f(x) - g(x)h(x)=f(x)−g(x)

h(x) = \frac{f(x)}{g(x)}h(x)=

g(x)

f(x)

h(x) = f(g(x))h(x)=f(g(x))

And many more.

The approach or formula to use depends on the question.

In this case, the combined function is:

f(x) = 75+ 10xf(x)=75+10x

The savings function is given as

s(x) = 85s(x)=85

The allowance function is given as:

a(x) = 10(x - 1)a(x)=10(x−1)

The new function that combined his savings and his allowances is calculated as:

f(x) = s(x) + a(x)f(x)=s(x)+a(x)

Substitute values for s(x) and a(x)

f(x) = 85 + 10(x - 1)f(x)=85+10(x−1)

Open bracket

f(x) = 85 + 10x - 10f(x)=85+10x−10

Collect like terms

mark as brainiest

f(x) = 85 - 10+ 10xf(x)=85−10+10x

f(x) = 75+ 10xf(x)=75+10x

4 0
2 years ago
What is the number of diagonals that intersect at a given vertex of a hexagon, heptagon, 30-gon and n-gon?
DENIUS [597]

Answer:

i. 9

ii. 14

iii. 405

iv. \frac{n(n-3)}{2}

Step-by-step explanation:

The number of diagonals in a polygon of n sides can be determined by:

\frac{n(n-3)}{2}

where n is the number of its sides.

i. For a hexagon which has 6 sides,

number of diagonals = \frac{6(6-3)}{2}

                                   = \frac{18}{2}

                                   = 9

The number of diagonals in a hexagon is 9.

ii. For a heptagon which has 7 sides,

number of diagonals = \frac{7(7-3)}{2}

                                   = \frac{28}{2}

                                   = 14

The number of diagonals in a heptagon is 14.

iii. For a 30-gon;

number of diagonals = \frac{30(30-3)}{2}

                                          = \frac{810}{2}

                                         = 405

The number of diagonals in a 30-gon is 405.

iv. For a n-gon,

number of diagonals = \frac{n(n-3)}{2}

The number of diagonals in a n-gon is \frac{n(n-3)}{2}

7 0
3 years ago
Please help!! Explain how you got the answer!
mash [69]

multiply 3.5 x 3 = 10.5

 add -5 +-10 = -15

 so you have 10.5 x 10^15

 then you need to move the decimal point 1 place to the left to get 1.05

 since you move the decimal point 1 place to the left you add 1 to the -15 to get -14

 so answer = 1.05 x 10^-14


8 0
3 years ago
Help me please I need to finish
IrinaK [193]
The awnser is c give me brainliest
8 0
2 years ago
Read 2 more answers
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