Find the values of k for which y = kx + 1 is a tangent to the curve y = 2x^2 + x +3

sladkih [1.3K]

Answer:

A. If the side lengths are the same, then a triangle is not scalene.

Step-by-step explanation:

A triangle can be defined as a two-dimensional shape that comprises three (3) sides, three (3) vertices and three (3) angles.

Simply stated, any polygon with three (3) lengths of sides is a triangle.

In Geometry, a triangle is considered to be the most important shape.

Generally, there are three (3) main types of triangle based on the length of their sides and these include;

I. Equilateral triangle: it has all of its three (3) sides and interior angles equal.

II. Isosceles triangle: it has two (2) of its sides equal in length and two (2) equal angles.

III. Scalene triangle: it has all of its three (3) sides and interior angles different in length and size respectively.

6 0

3 years ago

Four lawn sprinkler heads are fed by a 1.9-cm-diameter pipe. The water comes out of the heads at an angle of 35° above the horiz

klemol [59]

Answer:

Step-by-step explanation:

Given:

Angle, θ = 35°

Vertical distance, Δx = 6 m

Diameter, d = 1.9 cm

= 0.019 m

A.

When the water leaves the sprinkler, it does so at a projectile motion.

Therefore,

Using equation of motion,

(t × Vox) = 2Vo²(sin θ × cos θ)/g

= Δx = 2Vo²(sin 35 × cos 35)/g

Vo² = (6 × 9.8)/(2 × sin 35 × cos 35)

= 62.57

Vo = 7.91 m/s

B.

Area of sprinkler, As = πD²/4

Diameter, D = 3 × 10^-3 m

As = π × (3 × 10^-3)²/4

= 7.069 × 10^-6 m²

V_ = volume rate of the sprinkler

= area, As × velocity, Vo

= (7.069 × 10^-6) × 7.91

= 5.59 × 10^-5 m³/s

Remember,

1 m³ = 1000 liters

= 5.59 × 10^-5 m³/s × 1000 liters/1 m³

= 5.59 × 10^-2 liters/s

= 0.0559 liters/s.

For the 4 sprinklers,

The rate at which volume is flowing in the 4 sprinklers = 4 × 0.0559

= 0.224 liters/s

C.

Area of 1.9 cm pipe, Ap = πD²/4

= π × (0.019)²/4

= 2.84 × 10^-4 m²

Volumetric flowrate of the four sprinklers = 4 × 5.59 × 10^-5 m³/s

= 2.24 × 10^-4 m³/s

Velocity of the water, Vw = volumetric flowrate/area

= 2.24 × 10^-4/2.84 × 10^-4

= 0.787 m/s

7 0

3 years ago

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How do u multiply monimials

krek1111 [17]

Two multiply 2 monomials, you have to multiply the coefficients and add the degrees. For example

2x^2 × 3x^3 =

6x^5

6 0

3 years ago

A plane flying horizontally at an altitude of 1 mi and a speed of 470 mi/h passes directly over a radar station. Find the rate a

kiruha [24]

Answer:

407 mi/h

Step-by-step explanation:

Given:

Speed of plane (s) = 470 mi/h

Height of plane above radar station (h) = 1 mi

Let the distance of plane from station be 'D' at any time 't' and let 'x' be the horizontal distance traveled in time 't' by the plane.

Consider a right angled triangle representing the above scenario.

We can see that, the height 'h' remains fixed as the plane is flying horizontally.

Speed of the plane is nothing but the rate of change of horizontal distance of plane. So, $s=\frac{dx}{dt}=470\ mi/h$

Now, applying Pythagoras theorem to the triangle, we have:

$D^2=h^2+x^2\\\\D^2=1+x^2$

Differentiating with respect to time 't', we get:

$2D\frac{dD}{dt}=0+2x\frac{dx}{dt}\\\\\frac{dD}{dt}=\frac{x}{D}(s)$

Now, when the plane is 2 miles away from radar station, then D = 2 mi

Also, the horizontal distance traveled can be calculated using the value of 'D' in equation (1). This gives,

$2^2=1+x^2\\\\x^2=4-1\\\\x=\sqrt3=1.732\ mi$

Now, plug in all the given values and solve for $\frac{dD}{dt}$ . This gives,

$\frac{dD}{dt}=\frac{1.732\times 470}{2}=407.02\approx 407\ mi/h$

Therefore, the distance from the plane to the station is increasing at a rate of 407 mi/h.

6 0

3 years ago

Ok so umm last question for tonight i hope but um please help

amid [387]

Answer:

fish=3,5,15

snails=1,4,4

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers