All categories

3 years ago

What is the equation of the line shown in this graph? The points are (-2, 2) and (-2, -3)

Mathematics

1 answer:

KengaRu [80]3 years ago

5 0

$\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-3-2}{-2-(-2)}\implies \cfrac{-3-2}{-2+2}\implies \stackrel{und efined}{\cfrac{-5}{0}}$

so our slope is undefined, that means a vertical line, check the picture below

You might be interested in

What is the value of x? ______ * i will give you a brainliest answer!!!!!!

mamaluj [8]

Two ways of solving it:

The angle beside the 71 is equal to

180 - 71 = 109

so a triangle angles sums up to 180

so the reminaing is

x = 180 - 109 - 28 = 43

solution (2):

the exterior angle = sum of interior angles except the one near it.

28 + x = 71
x = 43

8 0

3 years ago

Read 2 more answers

I NEED PLEASE HELP ASAP

azamat

Answer:

Step-by-step explanation:

the lat answer for the first picture correct just plug it in the hihger number for x and then try a larger number y decreases. The same answer for the second one 2

7 0

2 years ago

Convert 1 cal/(m^2 * sec * °C) into BTU/(ft^2 * hr * °F)

Crazy boy [7]

Answer:

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Step-by-step explanation:

To find : Convert $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}$ into $\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Solution :

We convert units one by one,

$1\text{ m}^2=10.7639\text{ ft}^2$

$1\text{ sec}=\frac{1}{3600}\text{ hour}$

$1\text{ cal}=0.003968\text{ BTU}$

Converting temperature unit,

$^\circ C\times \frac{9}{5}+32=^\circ F$

$1^\circ C\times \frac{9}{5}+32=33.8^\circ F$

So, $1^\circ C=33.8^\circ F$

Substitute all the values in the unit conversion,

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{10.7639\times \frac{1}{3600}\times 33.8}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=\frac{0.003968}{0.101061}\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

$1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

Therefore, The conversion of unit is $1\ \frac{\text{cal}}{m^2\times sec\times ^\circ C}=0.03926\frac{\text{BTU}}{ft^2\times hr\times ^\circ F}$

3 0

2 years ago

A typical running track is an oval with 74-m-diameter half circles at each end. a runner going once around the track covers a di

denpristay [2]

Answer:

Her centripetal acceleration during the turn at each end of the track is $0.432\, \frac{m}{s^{2}}$

Step-by-step explanation:

Total distance covered in one round , D= 400 m

Time taken to cover one round , T = 1 min 40 s = 100 sec

Speed of runner , $V=\frac{D}{T}=\frac{400}{100}\, \frac{m}{s}= 4.0\, \frac{m}{s}$

Now centripetal acceleration is given by

$a_c=\frac{V^{2}}{R}$

where $V= 4.0\frac{m}{s}$

$Radius, R= \frac{Diameter}{2}=\frac{74}{2}m=37\, m$

$\therefore a_c=\frac{4^{2}}{37}\, \frac{m}{s^{2}}=0.432\, \frac{m}{s^{2}}$

Thus her centripetal acceleration during the turn at each end of the track is $0.432\, \frac{m}{s^{2}}$

4 0

3 years ago

Read 2 more answers

Find the complex zeros of <img src="https://tex.z-dn.net/?f=%20f%28x%29%20%3D%20%7Bx%7D%5E%7B3%20%7D%20%20-%2013%20%7Bx%7D%5

timama [110]

$f(x) = 0 \\\\\implies x^3-13x^2+59x -87 =0\\\\\implies x^3 -3x^2 -10x^2 +30x +29x - 87=0\\\\\implies x^2(x-3) - 10x(x-3) + 29(x-3) =0\\\\\implies (x-3)(x^2 -10x +29) =0\\\\\implies x - 3 = 0 ~~ \text{or}~~x^2 -10x +29 = 0\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{-(-10) \pm \sqrt{(-10)^2 -4 \cdot 1 \cdot 29}}{2(1)}\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{10\pm \sqrt{-16}}{2}\\\\\implies x =3~~ \text{ or} ~~ x = \dfrac{10\pm 4i }{2}\\$

$\\\implies x =3~~ \text{ or} ~~ x =5 \pm 2i\\\\\text{Hence the complex roots are,}~ 5- 2i ~~ \text{and} ~~5 + 2i$

6 0

2 years ago