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mote1985 [20]
3 years ago
7

What is the definition of rate?

Mathematics
1 answer:
RUDIKE [14]3 years ago
5 0
A measure, quantity, or frequency, typically one measured against some other quantity or measure.
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How to get X or y alone
Trava [24]
To get X and Y by itself, move everything else to the other side of the equal sign. Do this by doing the opposite operation.

Example: x-2y=11 
             add 2y to both sides to get x by itself 
           x=2y+11
7 0
3 years ago
1. Find a markdown of 1/9 on $450?
vovangra [49]

Step-by-step explanation:

1. 1/9 of 450

of is multiply

1/9 × 450

= ???

2. 15/60 = decimal number

change number to percentage = ???

I dont know what 3 is

put answers to 1 and 2 in comments i will mark if u don't understand comment

please mark brainiest

5 0
2 years ago
On the Oregon Trail, each person over the age of 10 was given 1.5 cups of beans per day. One family who traveled the trail in 18
devlian [24]
1 Member = 1.5 Cup (Over the age of 10) for one day

7 Member (Over age of 10) = 1.5*7=10.5 cups in total.

Total of 10.5 per day.

In a week (Equivalent to 7 days):

10.5*7=73.5 Cups of beans in total

6 0
3 years ago
Two quantities are related, as shown in the table:
Rashid [163]

Answer:

y = \frac{1}{2} x + 2.

Step-by-step explanation:

From the given table it is clear that x and y are linearly related.

Now, any two points that satisfy the relation between x and y are sufficient to express the equation.

The first two given points are (2,3), and (4,4).

Therefore, the equation is \frac{y - 4}{4 - 3} = \frac{x - 4}{4 - 2}

⇒ y - 4 = \frac{1}{2} (x - 4)

⇒ y - 4 = \frac{1}{2} x - 2

⇒ y = \frac{1}{2} x + 2. (Answer)

6 0
3 years ago
Suppose that a particle following the given path c(t) flies off on a tangent at t = t0. Compute the position of the particle at
Svetach [21]

Answer:

(30,35,0)

Step-by-step explanation:

We are given that

C(t)=(2t^2,t^3-4t,0)

t_0=3,t_1=4

The velocity of particle,v(t)=c'(t)=(4t,3t^2-4,0)

Substitute t_0=3

C(3)=(2(3)^2,(3)^3-4(3),0))=(18,15,0)

v(3)=(4(3),3(3)^2-4,0)=(12,23,0)

The tangent line at t=3 is given by

l(t)=c(2)+(t-3)v(3)=(18,15,0)+(t-3)(12,23,0)

Now, substitute t=4

l(4)=(18,15,0)+(4-3)+(12,23,0)=(18,12,0)+(12,23,0)

l(4)=(30,35,0)

Hence, the position of particle  at t_1=4=(30,35,0)

4 0
3 years ago
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