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Oksi-84 [34.3K]

3 years ago

6

Sara has already saved $6. She earns $3 per hour. She needs a minimum of $94 to buy her new bike. What is the minimum number of

hours she must work?

1 answer:

r-ruslan [8.4K]3 years ago

4 0

Answer:

30

Step-by-step explanation:

3 times 30 = 90 - 2(for the six she already had) so 30

i think

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A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will <u>plug $t=3$ into the above equation</u>. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then <u>we will plug $H= 400$ </u>

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find <u>the time taken by the rocket to reach the ground</u>. When it reaches the ground, then $H=0$ . So.....

$0=-16t^2+500t+20\\ \\ -4(4t^2-125t-5)=0\\ \\ 4t^2-125t-5=0$

Using <u>quadratic formula</u>, we will get.....

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(-5)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{15705}}{8}\\ \\ t=31.2899...\\ and\\ t= -0.0399...$

<em>(Negative value is ignored as time can't be in negative)</em>

So, the rocket will remain in the air for 31.2899... seconds.

5 0

4 years ago

A 20 foot ladder is leaning up against the side of a house the distance between the house and the base of the ladder is 4 feet f

Pie

Answer:

78.5°

Step-by-step explanation:

We solve for the above question, using the formula for the Trigonometric function of Cosine

cos θ = Adjacent/Hypotenuse

Adjacent = The distance between the house and the base of the ladder = 4 feet

Hypotenuse = Length of the Ladder = 20 feet

Hence,

cos θ = 4/20

θ = arc cos(4/20)

θ = 78.463040967°

Approximately = 78.5°

Therefore, the angle that the ladder makes with the ground is 78.5°

6 0

3 years ago

Write the equation of each Line in slope-intercept form

Neko [114]

y = 30x + 20 is the equation

8 0

4 years ago

List the side of triangle XYZ in order from shortest to longest if m

zlopas [31]

I think it would be c, although I don't really get the question

5 0

3 years ago

P and Q are two points on the line x - y + 1 = 0 and are at distance 5 from the origin. Find the area of the triangle OPQ.

Alecsey [184]

Answer:

P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.

Step-by-step explanation:

P and Q are the intersection points of

x-y+1 = 0 and the circle x^2 + y^2 = 25

sub y = x+1 into the circle

x^2 + (x+1)^2 = 25

x^2 + x^2 + 2x + 1 - 25 = 0

x^2 + x - 12 = 0

(x+4)(x-3) = 0

x = 3 or x = -4

y = 4 or y = -3

so P(3,4) and Q(-4,3) are our two points

Height of triangle.

h = |0 - 0 + 1|/√2 = 1/√2

PQ = √( (-7)^2 + 1^2) = √50 = 5√2

area POQ = (1/2)(1/√2)(5√2) = 5/2 square units

hope this helped

7 0

2 years ago