All categories

2 years ago

PLEASE HELP

Mathematics

1 answer:

Lena [83]2 years ago

3 0

Answer:

11.4 cm

Step-by-step explanation:

Using pythagoras theorem,

You might be interested in

Suppose a population of 250 crickets doubles in size every 3 months. How many crickets will there be after 2 years?

aivan3 [116]

The equation is:
f ( t )=

6 0

3 years ago

Read 2 more answers

“Help me get my feet back on the ground”

finlep [7]

Answer:

Angle 2

Step-by-step explanation:

Remember the X, whatever is opposite to the given angle.

6 0

3 years ago

Read 2 more answers

If 6=2x+4y, what is the value of x+2y?

Tamiku [17]

Answer:

Step-by-step explanation:

Take 2 as a common then divide by 2

6÷2 =3

6 0

3 years ago

Use implicit differentiation to find the points where the parabola defined by x2−2xy+y2+4x−8y+20=0 has horizontal and vertical t

Komok [63]

Answer:

The parabola has a horizontal tangent line at the point (2,4)

The parabola has a vertical tangent line at the point (1,5)

Step-by-step explanation:

Ir order to perform the implicit differentiation, you have to differentiate with respect to x. Then, you have to use the conditions for horizontal and vertical tangent lines.

-To obtain horizontal tangent lines, the condition is:

$\frac{dy}{dx}=0$ (The slope is zero)

--To obtain vertical tangent lines, the condition is:

$\frac{dy}{dx}=\frac{1}{0}$ (The slope is undefined, therefore the denominator is set to zero)

Derivating respect to x:

$\frac{d(x^{2}-2xy+y^{2}+4x-8y+20)}{dx} = \frac{d(x^{2})}{dx}-2\frac{d(xy)}{dx}+\frac{d(y^{2})}{dx}+4\frac{dx}{dx}-8\frac{dy}{dx}+\frac{d(20)}{dx}$ = $2x -2(y+x\frac{dy}{dx})+2y\frac{dy}{dx}+4-8\frac{dy}{dx}= 0$

Solving for dy/dx:

$\frac{dy}{dx}(-2x+2y-8)=-2x+2y-4\\\frac{dy}{dx}=\frac{2y-2x-4}{2y-2x-8}$

Applying the first conditon (slope is zero)

$\frac{2y-2x-4}{2y-2x-8}=0\\2y-2x-4=0$

Solving for y (Adding 2x+4, dividing by 2)

y=x+2 (I)

Replacing (I) in the given equation:

$x^{2}-2x(x+2)+(x+2)^{2}+4x-8(x+2)+20=0\\x^{2}-2x^{2}-4x+x^{2} +4x+4+4x-8x-16+20=0\\-4x+8=0\\x=2$

Replacing it in (I)

y=(2)+2

y=4

Therefore, the parabola has a horizontal tangent line at the point (2,4)

Applying the second condition (slope is undefined where denominator is zero)

2y-2x-8=0

Adding 2x+8 both sides and dividing by 2:

y=x+4(II)

Replacing (II) in the given equation:

$x^{2}-2x(x+4)+(x+4)^{2}+4x-8(x+4)+20=0\\x^{2}-2x^{2}-8x+x^{2}+8x+16+4x-8x-32+20=0\\-4x+4=0\\x=1$

Replacing it in (II)

y=1+4

y=5

The parabola has vertical tangent lines at the point (1,5)

4 0

4 years ago

Graph the quadratic functions y=-2xandy=-2x2 + 4 on a separate piece of paper. Using those graphs, compare and contrast the shap

dedylja [7]

Y = -2x^2

h = Xv = -B/2A = 0/-4 = 0.

k = -2*0^2 = 0.

V(0,0).

Use the following points for graphing:

Y = -2x^2.

(x,y)

(-2,-8)

(-1,-2)

V(0,0)

(1,-2)

(2,-8)

Y = -2x^2 + 4.

h = Xv = -B/2A = 0/-4 = 0.

k = Yv = -2*0^2 + 4 = 4.

V(0,4).

(-2,-4)

(-1,2)

V(0,4)

(1,2)

(2,-4)

8 0

2 years ago