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3 years ago

5

Need help asap.pls and thanks

1 answer:

love history [14]3 years ago

6 0

Answer:

your speed is 135 mph

Step-by-step explanation:

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Henry spent 1 hour 30 minutes less than Joey reading last week. Joey spent 50 minutes less than Pete. Pete spent 3 hours reading

Nookie1986 [14]

Answer: Henry read for a total of 40 minutes

Step-by-step explanation: 3:00 - 0:50 is 2:10 then 2:10 - 1:30= 40 minutes

3 0

3 years ago

Read 2 more answers

5. A rectangular pyramid measures three feet by four feet at the base, and has a height of seven feet. Find its volume.

den301095 [7]

Answer: 28 ft

Step-by-step explanation:

Multiply 3,4, and 7 to get 84 ft cubed. To find the volume of a pyramid, divide by 3 to get 28 ft.

4 0

4 years ago

Solve for t and find the value for t if I= 115, P=650, r=5%

Brrunno [24]

P=650
R= 5% annually= 0.05
T=1 month=1/12 year

Sl=Prt=650*0.05*1/12= $2.70
So the answer would be 2.7 or 2.70

5 0

3 years ago

What is the distributive property of 2 times 8

malfutka [58]

2*8 using <span>distributive property
= 2(3+5)
= 6+ 10
= 16</span>

5 0

4 years ago

What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5?

telo118 [61]

Answer:

$f(x)=-\frac{1}{8}(x-3)^2+3$

Step-by-step explanation:

We want to find the equation of the parabola with a focus of $(3,1)$ and directrix $y=5$ .

Considering the directrix, the quadratic graph must open downwards.

The equation of this parabola is given by the formula,

$(x-h)^2=4p(y-k)$ , where $(h,k)$ is the vertex of the parabola.

The axis of this parabola meets the directrix at $(3,5)$ .

Since the vertex is the midpoint of the focus and the point of intersection of the axis of the parabola and the directrix,

$h=\frac{3+3}{2}=3$ and $k=\frac{5+1}{2}=3$ .

The equation of the parabola now becomes,

$(x-3)^2=4p(y-3)$ .

Also $|p|$ is the distance between the vertex and the directrix.

$|p|=2$

This implies that $p=-2\:or\:2$ .

Since the parabola turns downwards,

$p=-2$ .

Our equation now becomes,

$(x-3)^2=4(-2)(y-3)$ .

$(x-3)^2=-8(y-3)$ .

We make y the subject to get,

$y=-\frac{1}{8}(x-3)^2+3)$ .

This is the same as

$f(x)=-\frac{1}{8}(x-3)^2+3)$ .

4 0

3 years ago