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3 years ago

Julissa walked 8 miles.

Mathematics

2 answers:

djyliett [7]3 years ago

8 0

Answer:

she walked another 2 miles.

Step-by-step explanation:

topjm [15]3 years ago

5 0

Answer:8 miles+2 miles=10 miles

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Simplify............

Kay [80]

$= \frac{6x + 8}{4x + 4} \div \frac{4x + 4}{4x + 4} \\ = \frac{2x + 4}{1} \\ = 2x + 4$
Hope this helped!!

3 0

3 years ago

Read 2 more answers

(50 POINTS!!!) find the total surface area and volume, then multiply by $0.004. please show your work!!

mrs_skeptik [129]

The Total surface area of the square pyramid = 115.2 in.²

The Volume of the square pyramid = 1,024 in.³.

<h3>What is the Total Surface Area of a Square Pyramid?</h3>

The total surface are of a square pyramid is given as: TSA = SA = 1/2(P × l), where:

P = perimeter of the square base

l = slant height of the pyramid

<h3>What is the Volume of a Square Pyramid?</h3>

The volume of a square pyramid = 1/3(a³ × h), where:

a = edge of the square base

h = height of the pyramid

Find the Total surface area of the Pyramid:

P = 4(a) = 4(8) = 32 in.

l = 7.2 in.

Total surface area of the square pyramid = 1/2(32 × 7.2)

Total surface area of the square pyramid = 115.2 in.²

Volume of the square pyramid:

a = 8 in.

h = 6 in.

Volume of the square pyramid = 1/3(8³ × 6)

Volume of the square pyramid = 1,024 in.³

Learn more about the square pyramid on:

brainly.com/question/3688277

#SPJ1

6 0

2 years ago

Fedrick salary varies directly with hour he works last week he earned 328.30 and worked 33.5 hours if he works this week what wi

stiks02 [169]

Answer:

Step-by-step explanation:

8 0

4 years ago

A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of

belka [17]

Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min

so that the net rate at which salt is exchanged through the tank is

ds(t)/dt = 12 - s(t)/150 … (lb/min)

Solve for s(t). The DE is separable, so we have

ds/dt = 12 - s/150

150 ds/dt = 1800 - s

150/(1800 - s) ds = dt

Integrate both sides to get

-150 ln|1800 - s| = t + C

Solve for s :

ln|1800 - s| = -t/150 + C

1800 - s = exp(-t/150 + C )

1800 - s = C exp(-t/150)

s = 1800 - C exp(-t/150)

Now given that s(0) = 50, we solve for C :

50 = 1800 - C exp(-0/150)

50 = 1800 - C

C = 1750

Then the amount of salt in the tank at any time t ≥ 0 is

s(t) = 1800 - 1750 exp(-t/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for t in

100 = 1800 - 1750 exp(-t/150)

1700 = 1750 exp(-t/150)

34/35 = exp(-t/150)

ln(34/35) = -t/150

t = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for s(t), we could have instead stopped with

-150 ln|1800 - s| = t + C

Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so

-150 ln|1800 - 50| = 0 + C

C = -150 ln|1750|

==> t = 150 ln(1750) - 150 ln|1800 - s|

Then s(t) = 100 lb when

t = 150 ln(1750) - 150 ln(1700)

t = 150 ln(1750/1700)

t = 150 ln(35/34)

t ≈ 4.348

5 0

3 years ago

What is the length of arc PQ? What is the area of the shaded region?

Liula [17]

Comment
I'm assuming you mean the minor arc. That's the part of the circumference that is in the shaded part.

Part One
The length of the arc is connected to the central angle which you are given as 155o
r = 9 in
The arc length = $\theta*r * 2\pi /360$
Since $\theta$ = 155 degrees

The arc length = 2*3.14 * 9*155 / 360
The arc length = 24.3 inches

The sector area = (central angle/360) * pi * r^2 .
The sector area = (155/360) * 3.14 * 9^2
The sector area = 109.5 in^2

Answers
Arc length = 24.3 <<<<<<< answer
Arc area = 109.5 <<<<<<<<answer

7 0

3 years ago