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2 years ago

4x + 3 Y = 11 Me podrían ayudar por favor a resolverla

Mathematics

1 answer:

nydimaria [60]2 years ago

4 0

Answer:

x = 2

y = 1

Step-by-step explanation:

4 x 2 = 8

3 x 1 = 3

8 + 3 = 11

So therefore

x = 2

y = 1

Have a nice day :)

Spanish:

Responder:

x = 2

y = 1

Explicación paso a paso:

4 x 2 = 8

3 x 1 = 3

8 + 3 = 11

Asi que, por lo tanto

x = 2

y = 1

Que tenga un buen día :)

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Answer:

(0,0) and (7,10)

Step-by-step explanation:

recall that the slope - intercept form of a linear equation can be given as

y = mx + b

where m = slope and b = y intercept

in our case, we are given that slope, m = 10/7, hence our equation becomes

y = (10/7) x + b

since we are not given any information about the y-intercept, we can simply pick a value for b that is most convenient for us.

We pick b = 0, hence the equation simplifies to:

y = (10/7) x ----- eq 1

we can see immediately that if x = 0, y must also = 0

proof if x = 0:

y = (10/7)(0) =0

since 0 is an integer, then (0,0) must be the first point.

we can also observe that for y to be an integer, we must get rid of the denominator 7. We can do this by multiply the right side by 7. Hence we let x = 7:

y = (10/7) x 7

y = 10

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3 years ago

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2 years ago

Question 3 Solve for m. 4 over 3 m equals 4

Lorico [155]

Answer:

$m = 3$

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3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

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Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

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