Segment EF is a mid segment of triangle ABC. If EF = 15 cm, what is the measure of segment BC?

Find the point slope equation for the line that passes through the points (7,-21) (-4,23)

zheka24 [161]

$\bf (\stackrel{x_1}{7}~,~\stackrel{y_1}{-21})\qquad (\stackrel{x_2}{-4}~,~\stackrel{y_2}{23}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{23-(-21)}{-4-7}\implies \cfrac{23+21}{-11}\\\\\\ \cfrac{44}{-11}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-21)=-4(x-7)\implies y+21=-4(x-7)$

5 0

3 years ago

Find the distance and midpoint for the points (5,5) and (1,2).

irinina [24]

Answer:

try and subtract the orgin is 0,0 so maybe it will help

Step-by-step explanation:

4 0

3 years ago

How do you determine the area under a curve in calculus using integrals or the limit definition of integrals?

RSB [31]

Answer:

Please check the explanation.

Step-by-step explanation:

Let us consider

$y = f(x)$

To find the area under the curve $y = f(x)$ between $x = a$ and $x = b$ , all we need is to integrate $y = f(x)$ between the limits of $a$ and $b$ .

For example, the area between the curve y = x² - 4 and the x-axis on an interval [2, -2] can be calculated as:

$A=\int _a^b|f\left(x\right)|dx$

= $\int _{-2}^2\left|x^2-4\right|dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=\int _{-2}^2x^2dx-\int _{-2}^24dx$

solving

$\int _{-2}^2x^2dx$

$\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=\left[\frac{x^{2+1}}{2+1}\right]^2_{-2}$

$=\left[\frac{x^3}{3}\right]^2_{-2}$

computing the boundaries

$=\frac{16}{3}$

Thus,

$\int _{-2}^2x^2dx=\frac{16}{3}$

similarly solving

$\int _{-2}^24dx$

$\mathrm{Integral\:of\:a\:constant}:\quad \int adx=ax$

$=\left[4x\right]^2_{-2}$

computing the boundaries

$=16$

Thus,

$\int _{-2}^24dx=16$

Therefore, the expression becomes

$A=\int _a^b|f\left(x\right)|dx=\int _{-2}^2x^2dx-\int _{-2}^24dx$

$=\frac{16}{3}-16$

$=-\frac{32}{3}$

$=-10.67$ square units

Thus, the area under a curve is -10.67 square units

The area is negative because it is below the x-axis. Please check the attached figure.

6 0

3 years ago

Graph the point (−1, 5/2) on the coordinate plane.<br><br> PLEASE HELP ASAP

drek231 [11]

Answer: screenshot below

Step-by-step explanation:

-1 means 1 space left from 0 on the x axis

5/2 means 5/2 spaces above 0 on the y axis

i converted 5/2 to 2 1/2 to make it easier and plotted the point by drawing a line from -1 and 2 1/2 (this is not needed for the showing of the answer)

the point is where the lines connect, aka (-1, 5/2)

4 0

3 years ago

The functions f(x) = (x + 1)2 − 2 and g(x) = -(x − 2)2 + 1 have been rewritten using the completing-the-square method. Is the ve

Ksenya-84 [330]

Answer:

The negative outside the parentheses indicates that the vertex is a maximum

f(x) has a minimum vertex

g(x) has a maximum

Step-by-step explanation:

7 0

3 years ago