Please help! Will give Brainliest!
Steps 1-4 in attachment (#4 below)
Step 4: Use the equation you wrote in Step 3. Write the equation for the graph of g(x) that has also been shifted right 1 unit.

Answer:

156 minutes

Step-by-step explanation:

So we need to create an equation to represent how Frank's phone company bills him

So the phone company charges an $8 monthly fee, so this value remains constant and will be our "y-intercept"

They then charge $0.06 for every minute he talks, this will be our "slope"

Combining everything into an equation, we have: y = 0.06x + 8

Now since we were given Franks phone bill total and want to figure out how many minutes he used, we just need to solve the equation for x and plug in our known y value

Frank used up a total of 156 minutes

Answer:

b

Step-by-step explanation:

In a parralel graph, the slopes would always be the same. The intercept in the answer is 2, showing that the coordinate points are (0,2)

Hope this helps!:)

Answer:

B) y = 2x + 2

Step-by-step explanation:

Firstly, you have to know that parallel lines have congruent slopes. That means that the slope of this line will be 2.

Next, make a point slope form of the equation:

y - y1 = m(x - x1)

y - 2 = 2(x - 0)

y - 2 = 2x - 0

Now, we can make it into slope intercept form.

y - 2 = 2x

y = 2x + 2

Hope this helps :)

Solution,

Radius=2 m

Area =pi r^2

= 3.142*(2)^2

=12.568 m^2

hope it helps

Good luck on your assignment

Answer:

Step-by-step explanation:

The general form representing limit of a function is expressed as shown below;

[tex]\lim_{h \to a} f(h)[/tex] where a is the value that h will take and use in the function f(h). It can be expressed in words as limit of function f as h tends to a. Comparing the genaral form of the limit to the limit given in question [tex]\lim_{h \to 0} \frac{\sqrt{9+h} - 3}{h}[/tex], it can be seen that a = 0 and f(h) = [tex]\frac{\sqrt{9+h} - 3}{h}[/tex]

Taking the limit of the function

[tex]\lim_{h \to 0} \frac{\sqrt{9+h} -3}{h}\\= \frac{\sqrt{9+0}-3 }{0}\\= \frac{0}{0}(indeterminate)[/tex]

Applying l'hopital rule

[tex]\lim_{h \to 0} \frac{\frac{d}{dh} (\sqrt{9+h} - 3)} {\frac{d}{dh} (h)}\\= \lim_{h \to 0} \frac{1}{2} (9+h)^{-1/2} /1\\=\frac{1}{2} (9+0)^{-1/2}\\= \frac{1}{2} * \frac{1}{\sqrt{9} } \\= 1/2 * 1/3\\= 1/6[/tex]

Answer:

a) The velocity at which the water leaves the gun = 37.66 m/s

b) The volume rate of flow = (1.183 × 10⁻⁴) m³/s

c) The water hits the ground 18.64 m from the point where the water gun was shot.

Step-by-step explanation:

a) Using Bernoulli's equation, an equation that is based on the conservation of energy.

P₁ + ½ρv₁² + ρgh₁ = P₂ + ½ρv₂² + ρgh₂

The two levels we are considering is just inside the water reservoir and just outside it.

ρgh is an extension of potential energy and since the two levels are at the same height,

ρgh₁ = ρgh₂

Bernoulli's equation becomes

P₁ + ½ρv₁² = P₂ + ½ρv₂²

P₁ = Pressure inside the water reservoir = 8 atm = 8 × 101325 = 810,600 Pa

ρ = density of water = 1000 kg/m³

v₁ = velocity iof f water in the reservoir = 0 m/s

P₂ = Pressure outside the water reservoir = atmospheric pressure = 1 atm = 1 × 101325 = 101,325 Pa

v₂ = velocity outside the reservoir = ?

810,600 + 0 = 101,325 + 0.5×1000×v₂²

500v₂² = 810,600 - 101,325 = 709,275

v₂² = (709,275/500) = 1,418.55

v₂ = √(1418.55) = 37.66 m/s

b) Volumetric flowrate is given as

Q = Av

A = Cross sectional Area of the channel of flow = πr² = π×(0.001)² = 0.0000031416 m²

v = velocity = 37.66 m/s

Q = 0.0000031416 × 37.66 = 0.0001183123 m³/s = (1.183 × 10⁻⁴) m³/s

c) If the height of gun above the ground is 1.2 m. Where does the water hit the ground?

The range of trajectory motion is given as

R = vT

v = horizontal component of the velocity = 37.66 m/s

T = time of flight = ?

But time of flight is given as

T = √(2H/g) (Since the initial vertical component of the velocity = 0 m/s

H = 1.2 m

g = acceleration due to gravity = 9.8 m/s²

T = √(2×1.2/9.8) = 0.495 s

Range = vT = 37.66 × 0.495 = 18.64 m

Hope this Helps!!!

Answer:

option 2

Step-by-step explanation:

4^2=16/8=2.  4^2=16/16=1.  2-1=1

Answer:

The probability that the next customer will request mid-grade gas and fill the tank is 0.1800

Step-by-step explanation:

In order to calculate the probability that the next customer will request mid-grade gas and fill the tank we would have to make the following calculation:

probability that the next customer will request mid-grade gas and fill the tank= percentage of the people using mid-grade gas* percentage of the people using mid-grade gas that fill their tanks

probability that the next customer will request mid-grade gas and fill the tank=  30%*60%

probability that the next customer will request mid-grade gas and fill the tank= 0.1800

The probability that the next customer will request mid-grade gas and fill the tank is 0.1800

Answer:

the volume of the box is increasing

dV = +310,000 ft^3/s

Step-by-step explanation:

Volume of a rectangular box with side a,b and c can be expressed as;

V = abc

The change in volume dV can be expressed as;

dV = d(abc)/da + d(abc)/db + d(abc)/dc

dV = bc.da + ac.db + ab.dc ......1

Given:

a= 150 feet,

b= 80 feet, and

c= 50 feet

ais increasing at 100 feet/sec,bis decreasing 20 feet/sec, andcisincreasing at 5 feet/sec

da = +100 feet/s

db = -20 feet/s

dc = +5 feet/s

Substituting the values into the equation 1;

dV = (80×50×+100) + (150×50×-20) + (150×80×+5)

dV = +400000 - 150000 + 60000 ft^3/s

dV = +310,000 ft^3/s

Since dV is positive, the volume of the box is increasing at that instant.

Answer:

[tex]=\left(x+4\right)\left(x-9\right)[/tex]

Step-by-step explanation:

[tex]x^2-5x-36\\\mathrm{Break\:the\:expression\:into\:groups}\\=\left(x^2+4x\right)+\left(-9x-36\right)\\\mathrm{Factor\:out\:}x\mathrm{\:from\:}x^2+4x\mathrm{:\quad }x\left(x+4\right)\\\mathrm{Factor\:out\:}-9\mathrm{\:from\:}-9x-36\mathrm{:\quad }-9\left(x+4\right)\\=x\left(x+4\right)-9\left(x+4\right)\\\mathrm{Factor\:out\:common\:term\:}x+4\\=\left(x+4\right)\left(x-9\right)[/tex]