All categories

2 years ago

Find the value of k if it is known that the graph of y=kx+2 goes through the point. a(14,-40)

Mathematics

2 answers:

Alex_Xolod [135]2 years ago

8 0

Answer:

k=-3

Explanation:

Plug in x and y values

-40=k(14)+2

-2 both sidea

-42=k(14)

divide by 14 both sides

k=-3

zimovet [89]2 years ago

8 0

k = -3

Step-by-step explanation:

y = kx + 2 through the point a(14, -40)

Subtitute…

y = kx + 2

-40 = k(14) + 2

-40 = 14k + 2

-40 - 2 = 14k

-42 = 14k

(-42)/14 = k

<u>-3 = k</u>

You might be interested in

Can someone please help me with the slope?

nalin [4]

Answer:

1/2

Step-by-step explanation:

For each two units that this line moves to the right, it rises one. This means that it has a slope of 1/2. Hope this helps!

3 0

3 years ago

-2/7x=4/7 how to check your answer

Ymorist [56]

Answer:

1)Evaluate the left-hand side expression at the given value to get a number.

2)Evaluate the right-hand side expression at the given value to get a number.

3)See if the numbers match.

Sorry if wrong

7 0

2 years ago

I need help with this

adoni [48]

Answer is D
Mark me the brainliest

6 0

2 years ago

Es 18 de junio de 1815, las tropas napoleónicas se encuentran justo adelante, tu eres el ingeniero en balística y el general Wel

Airida [17]

Answer:

89.1° or -1.4°

Step-by-step explanation:

1. Location:

You are on the Mont-Saint-Jean escarpment, near the Belgian town of Waterloo.

The French troops are about 50 m below you and 1.2 km distant.

2. Finding the firing angle

Data:

R = 1200 m

u = 600 m/s

h = -50 m (the height of the target)

a = 9.8 m/s²

We have two conditions.

Horizontal distance

(1) 1200 = 600t cosθ

Vertical distance

(2) -50 = 600t sinθ - 4.9t²

Divide each side of (1) by 600cosθ.

$(3) \, t =\dfrac{2}{\cos \theta}$

Substitute (3) into (2)

$-50 = 600t \sin \theta - 4.9t^{2} = 600 \left( \dfrac{2}{\cos \theta} \right ) \sin \theta - 4.9 \left( \dfrac{2}{\cos \theta} \right )^{2}\\\\(4) \, -50 = 1200 \tan \theta - \dfrac{19.6}{\cos^{2} \theta}$

Recall that

(5) sec²θ = 1/cos²θ = tan²θ + 1

Substitute (5) into (4)

$-50 = 1200 \tan \theta - 19.6 \left(\tan^{2} \theta}+ 1\right )$

Set up a quadratic equation

$\begin{array}{rcl}-50 & = & 1200 \tan \theta - 19.6\tan^{2} \theta -19.6 \\0 & = & 1200 \tan \theta - 19.6\tan^{2} \theta + 30.4\\0 & =&19.6\tan^{2} \theta - 1200 \tan \theta - 30.4\\0 & =&\tan^{2} \theta - 61.224 \tan \theta - 1.551\\\end{array}$

Solve for θ

Use the quadratic formula.

tanθ = 61.249 or -0.025

θ = arctan(61.249) = 89.1° or

θ = arctan(-0.025) = -1.4°

3 0

3 years ago

Solve the given initial-value problem. x^2y'' + xy' + y = 0, y(1) = 1, y'(1) = 8

Kitty [74]

Substitute $z=\ln x$ , so that

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)$

Then the ODE becomes

$x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0$
$\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0$

which has the characteristic equation $r^2+1=0$ with roots at $r=\pm i$ . This means the characteristic solution for $y(z)$ is

$y_C(z)=C_1\cos z+C_2\sin z$

and in terms of $y(x)$ , this is

$y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)$

From the given initial conditions, we find

$y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1$
$y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8$

so the particular solution to the IVP is

$y(x)=\cos(\ln x)+8\sin(\ln x)$

4 0

2 years ago