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3 years ago

Please Help! I will give you a lot of points!

Mathematics

2 answers:

Alexandra [31]3 years ago

7 0

Answer:i can see it

Step-by-step explanation:

Eddi Din [679]3 years ago

7 0

Answer:

Hello! I noticed you are answering questions with answers that are too incorrect to say it was an accident. Please stop :)

p.s this was my revenge for the points you took from me

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Which is the equation of the line?

DIA [1.3K]

The answer is B, the x-axis crosses over to -8 the way the line goes also means it is a negative. I don't know for sure because i haven't done this in a while but I hope it helps.

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3 years ago

Find f(g(x)) and g(f(x)) for the following functions f(x) = 2x g(x) = x + 5

Helga [31]

F(x)=2x
G(x)=x+5
Now
F(g(x))=2(x+5)
=2x+10
G(f(x))=2x+5

4 0

3 years ago

Translate the figure 2 units left and 7 units down.

STALIN [3.7K]

The translation of the given figure will have vertices, A'(1,2), B'(3,2), C'(3,-5), and D(-1,-3) from the initial vertices A(3,9), B(5,9), C(5,2), and D(1,4). The figure is attached.

In the question, we are asked to translate the given figure 2 units left and 7 units down.

In the translation, the left-right movement implies the change in the x-coordinate, whereas the up-down movement implies the change in the y-coordinate.

For any point P(x, y) the given rule says that after translation,

P(x, y) becomes P'(x - 2, y - 7).

The coordinates of the given figure are:

A(3,9), B(5,9), C(5,2), and D(1,4).

After translation, these points can be shown as:

A(3,9) becomes A'(1,2).

B(5,9) becomes B'(3,2).

C(5,2) becomes C'(3,-5).

D(1,4) becomes D'(-1,-3).

Thus, the translation of the given figure will have vertices, A'(1,2), B'(3,2), C'(3,-5), and D(-1,-3) from the initial vertices A(3,9), B(5,9), C(5,2), and D(1,4). The figure is attached.

Learn more about translations at

brainly.com/question/1400185

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2 years ago

What is the value of m? 14 2m=4m-16

gogolik [260]

M=15...................................................................

4 0

3 years ago

Read 2 more answers

1. (a) Solve the differential equation (x + 1)Dy/dx= xy, = given that y = 2 when x = 0. (b) Find the area between the two curves

erastova [34]

(a) The differential equation is separable, so we separate the variables and integrate:

$(x+1)\dfrac{dy}{dx} = xy \implies \dfrac{dy}y = \dfrac x{x+1} \, dx = \left(1-\dfrac1{x+1}\right) \, dx$

$\displaystyle \frac{dy}y = \int \left(1-\frac1{x+1}\right) \, dx$

$\ln|y| = x - \ln|x+1| + C$

When x = 0, we have y = 2, so we solve for the constant C :

$\ln|2| = 0 - \ln|0 + 1| + C \implies C = \ln(2)$

Then the particular solution to the DE is

$\ln|y| = x - \ln|x+1| + \ln(2)$

We can go on to solve explicitly for y in terms of x :

$e^{\ln|y|} = e^{x - \ln|x+1| + \ln(2)} \implies \boxed{y = \dfrac{2e^x}{x+1}}$

(b) The curves y = x² and y = 2x - x² intersect for

$x^2 = 2x - x^2 \implies 2x^2 - 2x = 2x (x - 1) = 0 \implies x = 0 \text{ or } x = 1$

and the bounded region is the set

$\left\{(x,y) ~:~ 0 \le x \le 1 \text{ and } x^2 \le y \le 2x - x^2\right\}$

The area of this region is

$\displaystyle \int_0^1 ((2x-x^2)-x^2) \, dx = 2 \int_0^1 (x-x^2) \, dx = 2 \left(\frac{x^2}2 - \frac{x^3}3\right)\bigg|_0^1 = 2\left(\frac12 - \frac13\right) = \boxed{\frac13}$

7 0

2 years ago