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Advocard [28]
3 years ago
13

Clark typed 720 words in 18 minutes. How many words can he type in one minute?​

Mathematics
2 answers:
viva [34]3 years ago
5 0

Answer:

Clark can type 40 words in one minute

Step-by-step explanation:

  1. Set up a proportion: \frac{720}{18} = \frac{x}{1}  
  2. Cross multiply, then divide: 720 × 1 = 720, 720 ÷ 18 = 40
  3. So, x = 40

I hope this helps!

lbvjy [14]3 years ago
3 0
720/18 = 40

He can type 40 words a minute
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If the two lines below are perpendicular and the slope of the red line is -5/2, what is the slope of the green line
inna [77]
-2
----
5

Because perpendicular is the opposite and the opposite of -5/2 is -2/5
5 0
3 years ago
One box of strawberries cost $5. Which equation can be used to calculate the most number of boxes a person can buy with $307 30
alexgriva [62]

Answer:

30 = 5n

n = 6

Step-by-step explanation:

A box of strawberry cost = $5

A person has = $30

Number of boxes a person can buy = n

Equation to calculate the most number of boxes a person can buy with $30

Amount with the person = cost of a strawberry × number of strawberry

30 = 5 * n

30 = 5n

n = 30/5

n = 6

7 0
3 years ago
A stone is thrown horizontally with an initial speed of 8 m/s from the edge of a cliff. a stop watch measures the stone's trajec
Tamiku [17]

Answer:

90.6m

Step-by-step explanation:

Initial speed = 8m/s

Time = 4.3 s

In studying the motion of an object, we often make use of kinematic equation

s = ut + 1/2at^2

v^2 = u^2 + 2as

a = (v - u) / t

v = final velocity

u = initial velocity

s = distance or height

t = time

a = acceleration due to gravity

Given time to hit the ground as 4.3s, we calculate the height by using

s = ut + 1/2at^2

Since the stone is moving against gravity, a is negative

u = 0

s = 0(4.3) +1/2(-9.8)(4.3)^2

s = -90.6m

The height of the cliff is 90.6m

7 0
3 years ago
A bag of sand originally weighing 144 lb is lifted at a constant rate. As it rises, sand also leaks out at a constant rate. At t
frosja888 [35]

Answer:

Step-by-step explanation:

The bag of sand wave is

W=144lb at x=0

The bag of sand is lifted at a constant rate, i.e the it is not accelerating, so a=0m/s²

Let W be the mass of the sand,

Sand leaks out at a constant rate

dW/dx= C

At a height=18ft, half of the original mass.

i.e x=18ft, M=72lb

We are asked to find the work at 18ft

Work is given as

Work=∫F•dx,

Where F is the force and also the weight of the sand, F=W

The Weight of the bag is a linear function,

dW/dx= C

Then, using variable separation

dW=Cdx

Integrating both sides

∫dW=∫Cdt

W=Cx + B

So, at x=0, W=144

144=B

B=144

W=Cx+144

Also at x=18, W=72lb

72=C×18+144

72-144=18C

-72=18C

C=-4

Then, the weight function becomes

W=-4x+144

W=144-4x

Then applying work formula

Work=∫W•dx

Work=∫(144-4x)dx. x=0 to x=18

Work = 144x-4x²/2. x=0 to x=18

Work=144x-2x² x=0 to x=18

Work=144(18)-2(18²) -0 -0

Work =2592-648

Work =1944 ft lbs

The work done in lifting the sand to 18ft is 1944 ft lbs

7 0
3 years ago
Im confused and i need some help
lesantik [10]

Answer:

.2, .4, .44, .12, I'm too lazy to do the rest

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
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