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4 years ago

Can somebody please help me with this?

Mathematics

1 answer:

Semmy [17]4 years ago

3 0

(X^2 -5x+6)(x-5) = x^3- 10x^2 +31x -30

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Melissa has 885 cards and she wants to store them in envelopes. She has 35 envelopes. How many envelopes does she need

Verizon [17]

Answer:

I don't know

Step-by-step explanation:

I don't understand the point of this probem. It says that she has 35 envelopes best it doen't say how much can fit into each one. Maybe you ftogot some details pls reply in the comments.

8 0

3 years ago

What is the length of the missing side?

Lina20 [59]

Answer:

seventeen [17]

Step-by-step explanation:

8*8=64+15*15=225 so 225+64=289 and √289=17

6 0

3 years ago

1 point) Are the functions f,g, and h given below linearly independent? f(x)=e3x+cos(5x), g(x)=e3x−cos(5x), h(x)=cos(5x). If the

Fynjy0 [20]

Answer:

Functions are linearly dependent (are not linearly independent.)

Step-by-step explanation:

Remember that two functions f(x), g(x) and h(x) are said linearly independent on an interval I if the <em>only solution</em> to the equation

$\alpha f(x)+\beta g(x)+\omega h(x)=0, \ \text{for all } x\in I$

is the trivial one: α = 0, β = 0, ω = 0. If they are not linearly independent, they are called linearly dependent.

Now, let f(x), g(x) and h(x) be the functions:

$f(x)=e^{3x}+\cos(5x),$

$g(x)=e^{3x}-\cos(5x),$

$h(x)=\cos(5x).$

Then, letting α = 1, β= -1 and ω = -2, we see that:

$\alpha f(x)+\beta g(x)+ \omega h(x)=e^{3x}+\cos(5x)-e^{3x}+\cos(5x)+2\cos(5x)=0.$

Hence, the functions f(x), g(x) and h(x) are not linearly independent, or equivalently, are linearly dependent.

8 0

3 years ago

For exercises 16-18 , trace each figure and draw its rotated image

Artemon [7]

The answer is r(b x p) d

6 0

3 years ago

The line shown in the graph passes through the points (2, 4) and (-1, 6). Which is the

Shkiper50 [21]

Y2 -y 1 / x2 - x1
6 - 4 / -1 - 2
2 / - 3
m = -2/3 x (slope)

6 0

3 years ago