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4 years ago

Four Representations: Equations, Tables, Words,

Mathematics

1 answer:

andrey2020 [161]4 years ago

5 0

1) C

2) Graph 1

3) a= 3

4) b= 0

Youre welcome :)

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What are the coordinates of the midpoint of a segment AB , if: A(–3, 4), B( 1, 2)

Alik [6]

Hi there! :)

Answer:

$\huge\boxed{(-1, 3)}$

Use the midpoint formula to derive the midpoint of the segment:

$(x_{m} ,y_{m}) = (\frac{x_{1}+x_{2} }{2} , \frac{y_{1}+y_{2} }{2} )$

Substitute in the coordinates:

$(x_{m} ,y_{m}) = (\frac{-3+1 }{2} , \frac{4+2 }{2} )$

Simplify:

$(x_{m} ,y_{m}) = (-1 , 3 )$

The coordinates of the midpoint are:

$(-1, 3)$

3 0

3 years ago

Integrate dx/3sinx+4cosx

german

$\displaystyle\int\frac{\mathrm dx}{3\sin x+4\cos x}$

A standard approach would be the tangent half-angle substitution:

$t=\tan\dfrac x2\implies\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx$

Then

$\sin x=2\sin\dfrac x2\cos\dfrac x2\implies\sin x=\dfrac{2t}{1+t^2}$

$\cos x=\cos^2\dfrac x2-\sin^2\dfrac x2\implies\cos x=\dfrac{1-t^2}{1+t^2}$

from which we get

$\mathrm dx=\dfrac2{1+t^2}\,\mathrm dt$

So the integral becomes

$\displaystyle\int\frac{\frac2{1+t^2}}{\frac{6t}{1+t^2}+\frac{4(1-t^2)}{1+t^2}}\,\mathrm dt=\int\frac{\mathrm dt}{3t+2(1-t^2)}=-\int\frac{\mathrm dt}{2t^2-3t-2}$

Rewrite the denominator as

$2t^2-3t-2=(2t+1)(t-2)$

and expand the integrand into its partial fractions:

$\dfrac1{2t^2-3t-2}=\dfrac15\left(\dfrac1{t-2}-\dfrac2{2t+1}\right)$

We have

$\displaystyle-\frac15\int\frac1{t-2}-\frac2{2t+1}\,\mathrm dt=-\frac15(\ln|t-2|-\ln|2t+1|)+C$

$=\dfrac15\ln\left|\dfrac{2t+1}{t-2}\right|+C$

$=\dfrac15\ln\left|\dfrac{2\tan\frac x2+1}{\tan\frac x2-2}\right|+C$

6 0

3 years ago

This chart shows a bamboo plants growth over 10hours. The plant grew at a constant speed what is the rate of speed for this bamb

blagie [28]

Answer:

???

Step-by-step explanation:

???

5 0

3 years ago

Which one is it, I need help

Y_Kistochka [10]

Answer:

Their y-intercepts are equal

Step-by-step explanation:

The y-intercept is the y-value where the function crosses the y-axis. In this problem, functions are presented in 2 ways: algebraically and in a table.

1) Fortunately, the algebraic equation is written in slope-intercept form; this means that intercept is easy to find. The slope-intercept form is y=mx+b, where b is the y-intercept. In function 1, the b value is 10.

2) Another way to describe the y-intercept is the y-value when x=0. So, the y-intercept on a table is wherever the x-value is 0. In this case, the first row represents when x=0. The table says that when x=0, y=10. This means that the y-intercept for function 2 is 10.

Since the y-intercept for both of the functions is 10, it can be said that the 2 functions have equivalent y-intercepts.

7 0

2 years ago

Help me please :''(

ki77a [65]

1/4 2/5 83/100 100/100

4 0

3 years ago