ABCD is an isosceles trapezoid. What is the m∠DAB?

(y − 3)^2 = 4(x − 5), (6, 1). Find an equation of the tangent line to the graph at the given point.

Mandarinka [93]

Answer:

y=x−1. Explanation: (y−3)2=4(x −5). Expanding: y2−6y+9=4x−20. y2−6y−4x=− 29. Differentiating implicitly: 2ydydx−6dydx−4=0.

y=x−1 Explanation: (y−3)2=4(x−5) Expanding: y2−6y+9=4x−20 y2−6y−4x=−29 Differentiating implicitly: 2ydydx−6dydx−4=0 Factor: ... More

4 0

3 years ago

Pls ill give brainelist

Ivenika [448]

Answer: A, negative nonlinear

Step-by-step explanation:

The line goes down, and the points aren't in a straight line; they're scattered. as the x-axis increases, the y-axis decreases.

I didn't really explain too much but like-- i hope it helps lol

6 0

2 years ago

Read 2 more answers

Michelle throws a frisbee into the air. The height of the frisbee at a given time can be modeled by the equation h(t)= -2t

viva [34]

Answer:

(a) The ball will hit the ground after 3 seconds

(b) The maximum height is 6.125

Step-by-step explanation:

Given

$h(t) = -2t^2 +5t +3$

Solving (a): When the frisbee will hit the ground?

To do this, we set h(t) to 0

So, we have:

$h(t) = -2t^2 +5t +3$

$-2t^2 +5t +3=0$

Expand

$-2t^2 +6t-t +3=0$

Factorize

$-2t(t - 3) -1(t - 3) = 0$

Factor out t - 3

$(-2t -1)(t - 3) = 0$

Split:

$-2t -1= 0\ or\ t - 3 = 0$

Solve for t in both equations

$-2t =1\ or\ t = 3$

$t =-\frac{1}{2}\ or\ t = 3$

Time can't be negative; So:

$t = 3$

Solving (b): How height the frisbee will go?

First, we calculate time to reach the maximum height

$t = -\frac{b}{2a}$

Where:

$h(t) = at^2 + bt + c$

By comparison:

$a = -2,\ b =5,\ c =3$

So:

$t = -\frac{b}{2a}$

$t = -\frac{5}{2*-2}$

$t = \frac{5}{4}$

$t = 1.25$

So, the maximum height is:

$h_{max} = -2 * 1.25^2 + 5 * 1.25 + 3$

$h_{max} = 6.125$

3 0

3 years ago

Tyrell exercised this week both by walking and by biking. He walked at a rate of 4 mi/h and biked at a rate of 12 mi/h. The tota

Flura [38]

Answer:

A. The equation is 4(t + 1) + 12t = 36

B. He spent 2 hours on biking and 3 hours on walking

Step-by-step explanation:

* Lets explain how to solve the problem

- Tyrell walked at a rate 4 miles per hour

∴ His speed on walking is 4 miles/hour

- Tyrell biked at rate 12 miles per hour

∴ His speed on biking is 12 miles/hour

- The total distance he covered both walking and biking was

36 miles

- Assume that he walked x and biked y

∴ x + y = 36 ⇒ (1)

- Tyrell spent one more hour walking than biking

- Assume that he biked for t hours

∵ He walked one more hour than he biked

∵ He biked for t hours

∴ He walked for t + 1 hours

A.

∵ Distance = speed × time

∴ x = 4 × (t + 1)

∴ x = 4(t + 1)

∴ y = 12 × t

∴ y = 12t

- Substitute x and y in equation (1)

∴ 4(t + 1) + 12t = 36 ⇒ the equation

B.

* Lets solve the equation

- Multiply the bracket by 4

∴ 4t + 4 + 12t = 36

- Add like terms in left hand side

∴ (4t + 12t) + 4 = 36

∴ 16t + 4 = 36

- Subtract 4 from both sides

∴ 16t = 32

- Divide both sides by 16

∴ t = 2

∵ t represents the time of biking

∴ He biked for 2 hours

∵ t + 1 represents the time of walking

∵ t + 1 = 2 + 1 = 3

∴ He walked for 3 hours

5 0

3 years ago

Use a trigonometric ratio to find the distance EF.

leonid [27]

Answer:

The height of the building is approximately <u>73 m</u>.

Step-by-step explanation:

The triangle EFG is shown below.

Given:

Length of shadow of building is, $FG=34\ m$

Angle of inclination of Sun with the vertical is 25°.

Therefore, the angle of inclination of Sun with the horizontal is given as:

$\angle G=90-25=65\°$

Now, in triangle EFG as shown below, we use the tangent ratio to determine the height of building EF. Therefore,

$\tan (\angle G)=\frac{EF}{FG}\\\tan 65\°=\frac{EF}{34}\\EF=34\times \tan 65\°=72.9\ m\approx 73\ m$

Therefore, the height of the building is approximately 73 m.

4 0

3 years ago

ABCD is an isosceles trapezoid. What is the m∠DAB?​

ABCD is an isosceles trapezoid. What is the m∠DAB?