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dangina [55]
3 years ago
13

Sam is 5 years old. His older brother Tom, is three times as old as Sam. When Sam is 20, how old will Tom be?

Mathematics
1 answer:
GrogVix [38]3 years ago
6 0

Answer:

60

Step-by-step explanation:

Sam is 5, Tom is 3 times as old, 5 times three is 15, so tom is 15, 20 is 4 times 5, so you take 15, and multiply it by 4, and that's your answer

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If u, v, and w are nonzero vectors in r 2 , is w a linear combination of u and v?
Tju [1.3M]
Not necessarily. \mathbf u and \mathbf v may be linearly dependent, so that their span forms a subspace of \mathbb R^2 that does not contain every vector in \mathbb R^2.

For example, we could have \mathbf u=(0,1) and \mathbf v=(0,-1). Any vector \mathbf w of the form (r,0), where r\neq0, is impossible to obtain as a linear combination of these \mathbf u and \mathbf v, since

c_1\mathbf u+c_2\mathbf v=(0,c_1)+(0,-c_2)=(0,c_1-c_2)\neq(r,0)

unless r=0 and c_1=c_2.
8 0
3 years ago
HELP; PLZZZ GIVING 20 BRAINLIEST PLZZZ.
Andrew [12]

Answer:

Step-by-step explanation:

In the equation y=kx+1, k is the slope and 1 is the y-intercept.

To find the slope, use the formula k=y2-y1/x2-x1. To use this formula, you need two points. The first point, point M, is given to you. The second point can be the y-intercept, point (0, 1). Now, apply the formula to find the slope: k=3-1/1-0

k=2/1

k=2

We now have the equation y=2x+1.

The next step is to graph the line.

First, graph the y-intercept. Should look something like this: (star represents point. the point is on (0, 1).)

           |

           |

           *

------------------------------

           |

           |        

Then, use the slope to find your second point: (the point is on (1, 3)).

           |    *

           |

           *

------------------------------

           |

           |        

Draw a line through those two points and you have your answer!

6 0
3 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
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Karo-lina-s [1.5K]

Answer:

i can;t copy or paste either bro

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
give me answers for the blank parts please ive been trying this for at least 15 minutes and i swear i will get in so much troubl
DIA [1.3K]

Answer:

4 , 1 and 6

Step-by-step explanation:

now ur not in trouble :)

7 0
3 years ago
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