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3 years ago

What is the area of a triangle with the coordinates (-4,14) (10,0) (-14,-10)

1 answer:

3 0

You can use Heron Formula in the calculator to output the result, but i still show you the steps on paper thro

lengths of the 3 lines: \sqrt{392} , \sqrt{676} , \sqrt{676}

s= (\sqrt{392} + \sqrt{676} + \sqrt{676})/2=35.89949494

area= \sqrt{s(s- \sqrt{392} )(s- \sqrt{676})^2 } =238

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YALL HURRYY I NED HELP!

elena-s [515]

Answer:

A = correct equation

x = 60

Step-by-step explanation:

__+ 65 = 125

x + 65 = 125

x = 60

60 + 65 = 125

Hope it helps!

3 0

3 years ago

Carla is going to draw a marble from the bag shown below, replace it, and then draw another marble. What is the probability that

Contact [7]

Answer:

7/80

Step-by-step explanation:

(3 green, 4 yellow, 5 blue, 8 pink) = 20 marbles

P ( green or yellow ) = number of green or yellow / total

= (3+4)/20 = 7/20

Replace

so we have (3 green, 4 yellow, 5 blue, 8 pink) = 20 marbles

P (blue) = number of blue/ total

= 5/20 = 1/4

P ( green or yellow, replace, blue) = 7/20 * 1/4 = 7/80

8 0

3 years ago

Find cos(2115°) and sin(2115°). Identify the measure of the reference angle.

ivolga24 [154]

Answer:

Cos(2115°) =1/√2

Sin(2115°) = -1/√2

Step-by-step explanation:

We have to find the values of Cos (2115°) and Sin (2115°).

Now, 2115° can be written as (23×90°+ 45°).

Therefore, the angle 2115° lies in the 4th quadrant where Cos values are positive and Sin values are negative.

Hence, Cos (2115°) = Cos(23×90° +45°) =Sin 45° {Since 23 is an odd number, so the CosФ sign will be changed to SinФ} =1/√2 (Answer)

Again, Sin (2115°) = Sin(23×90° +45°) = -Cos 45° {Since 23 is an odd number, so the SinФ sign will be changed to CosФ} = -1/√2 (Answer)

Now, the required reference angle is 45°. (Answer)

8 0

3 years ago

(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

3 years ago

Pls help giving brainlyist

dangina [55]

Answer:

x = 0

Step-by-step explanation:

when y is -3, x is 0

5 0

3 years ago