What is the slope of the line described by the equation below?

if Mr hansons motorcycle drives 432 miles on 6 gallons of gas how much gas will he need to drive 738

lozanna [386]

Answer:

10.25 gallons

Step-by-step explanation:

432 miles / 6 gallons=72 miles to gallon

738 miles / 72 miles per gallon=10.25 gallons

6 0

3 years ago

You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

vladimir1956 [14]

Answer:

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

$m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$m = -\frac{\cos 2\theta}{\sin \theta}$

The tangent line is horizontal when $m = 0$ . Then:

$\cos 2\theta = 0$

$2\theta = \cos^{-1}0$

$\theta = \frac{1}{2}\cdot \cos^{-1} 0$

$\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)$ , for all $i \in \mathbb{N}_{O}$

$\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}$ , for all $i \in \mathbb{N}_{O}$

The first four solutions are:

x: $\sqrt{2}$ $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$

y: 1 -1 1 -1

The solutions listed from the smallest to the greatest are:

x: $-\sqrt{2}$ $-\sqrt{2}$ $\sqrt{2}$ $\sqrt{2}$

y: -1 1 -1 1

6 0

3 years ago

Ginny is studying a population of frogs. She determines that the population is decreasing at an average rate of 3% per year. Whe

svp [43]

Answer:

Step-by-step explanation:

This is exponential, of the form

$y=a(b)^x$ where a is the initial amount and b is the growth/decay rate. If Ginny begins the study with 1200 frogs in the population, our a value is 1200.

If the population is decreasing 3% per year, it is decreasing 100% - 3% = 97% which, as a decimal, is .97. Our exponential function, then, is

$y=1200(.97)^x$

6 0

3 years ago

What is the relationship between segment ED and segment AC?

Bas_tet [7]

D midpoint of EC -----------------> FD parallel to AC and FD=AC/2=14/2=7
2-EB=EA E midpoint of AB 
DB=DC D midpoint BC ...............> ED=AC/2=2 
3-T midpoint of SR 
U midpoint of QR ---------> TU = QS/2 
QS=2 TU = 4.4 
4- The same steps SR=2 UV=9 
5-N midpoint of KM 
O midpoint of ML 
* NO parallel to Kl

6 0

3 years ago

If ABCD is an A4 sheet and BCPO is the square, prove that △OCD is an isosceles triangle. And find the angles marked as 1 to 8 wi

Dmitry [639]

Answer:

The diagram for the question is missing, but I found an appropriate diagram fo the question:

Proof:

since OC = CD = 297mm Therefore, Δ OCD is an isoscless triangle

∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

∠DOP = 22.5°

∠PDO = 67.5°

∠ADO = 22.5°

∠AOD = 67.5°

Step-by-step explanation:

Given:

AB = CD = 297 mm

AD = BC = 210 mm

BCPO is a square

∴ BC = OP = CP = OB = 210mm

Solving for OC

OCB is a right anlgled triangle

using Pythagoras theorem

(Hypotenuse)² = Sum of square of the other two sides

(OC)² = (OB)² + (BC)²

(OC)² = 210² + 210²

(OC)² = 44100 + 44100

OC = √(88200

OC = 296.98 = 297

OC = 297mm

An isosceless tringle is a triangle that has two equal sides

Therefore for △OCD

CD = OC = 297mm; Hence, △OCD is an isosceless triangle.

The marked angles are not given in the diagram, but I am assuming it is all the angles other than the 90° angles

Since BC = OB = 210mm

∠BCO = ∠BOC

since sum of angles in a triangle = 180°

∠BCO + ∠BOC + 90 = 180

(∠BCO + ∠BOC) = 180 - 90

(∠BCO + ∠BOC) = 90°

since ∠BCO = ∠BOC

∴ ∠BCO = ∠BOC = 90/2 = 45

∴ ∠BCO = 45°

∠BOC = 45°

∠PCO = 45°

∠POC = 45°

For ΔOPD

$Tan\ \theta = \frac{opposite}{adjacent}\\ Tan\ (\angle DOP) = \frac{87}{210} \\(\angle DOP) = Tan^-1(0.414)\\(\angle DOP) = 22.5 ^{\circ}$

Note that DP = 297 - 210 = 87mm

∠PDO + ∠DOP + 90 = 180

∠PDO + 22.5 + 90 = 180

∠PDO = 180 - 90 - 22.5

∠PDO = 67.5°

∠ADO = 22.5° (alternate to ∠DOP)

∠AOD = 67.5° (Alternate to ∠PDO)

3 0

3 years ago