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pentagon [3]
3 years ago
9

Anyone good at composite functions??

Mathematics
1 answer:
sammy [17]3 years ago
6 0

\bf \begin{cases} ~\hfill f(x) = &-3x+4\\ ~\hfill g(x) = &x^2\\ (g \circ f)(0)=&g(~~f(0)~~) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ f(0) = -3(0)+4\implies \boxed{f(0) = 4} \\\\\\ g(~~f(0)~~)\implies g\left( \boxed{4} \right) = (4)^2\implies \stackrel{(g \circ f)(0)}{g(4)} = 16

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5 times 3 2/5 I need help pls k have school soon
timama [110]
5 times 3 is 15! i’m confused at the rest of the question!
8 0
2 years ago
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What problem makes 49
torisob [31]
45 + 4 makes 49 if that counts
6 0
3 years ago
Linear algebra is a fundamental component of continuum mechanics (and most engineering disciplines). Solve the following matrix
Mars2501 [29]

Answer:

A ) Not orthogonal to each other

B) 50i + 40j + 105k

C) The tensor product is attached below

D ) The value of X = F.X is attached below

Step-by-step explanation:

attached below is the detailed solution of the above problem

A) for the vectors ( u ) and ( v ) to be orthogonal to each other [ U.V has to be = 0 ] but in this scenario  U.V = 4 hence they are not orthogonal to each other

b) The vector normal to plane is gotten by : U x V

= 50i + 40j + 105k

4 0
3 years ago
SA = 2tr2 + 2trh
julsineya [31]

Given:

Radius of the cylinder = 5 in

Height of the cylinder = 30 in.

To find:

The surface area of the cylinder.

Solution:

The surface area of the cylinder is:

SA=2\pi r^2+2\pi rh

Where, r is the radius and h is the height of the cylinder.

Putting r=5,h=30,\pi=3.14 in the above formula, we get

SA=2(3.14)(5)^2+2(3.14)(5)(30)

SA=2(3.14)(25)+942

SA=157+942

SA=1099


Therefore, the surface area of the cylinder is 1099  square inches.

6 0
3 years ago
Calculate the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass of 1 m^3 of water to be 1000 kg
Lady bird [3.3K]

Answer:

1813.137 KJ

Step-by-step explanation:

potential energy of the body = mgH

where m is mass in Kg , g= 9.81 m/sec^2 and H= height in m

here m= 1000 kg, g= 9.81 m/s^2  and H= 607 feet = 607×0.305= 185.135 m

hence the potential energy p= 1000×9.81×185.135= 1813137.2 J

= 1813.137 KJ

hence the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass as 1000 kg is = 1813.137 KJ

4 0
3 years ago
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