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yanalaym [24]
3 years ago
11

1000 millimeters equals what =

Mathematics
1 answer:
scZoUnD [109]3 years ago
7 0
I don't know what you want 1000 milliliters to be equivalent to so ill give u a small list. Hope this helps.

1000 millimeters equals:
0.001 kilometers
1 meter
100 centimeters
1e+6 Micrometers
1e+9 Nanometers
0.000621371 Miles
1.09361 Yard
3.28084 Feet
39.3701 Inches
0.0005399570950324 Nautical Miles
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Margie's car can go 32 miles on a gallon of gas, and gas currently cost $4 per gallon. how many miles can Margie drive on $20 wo
Marysya12 [62]
To do this first we will divide the amount of dollars (20) by the amount of dollars 1 gallon is for (4). Then we will multiply the amount of times 4 goes into 20 by 32 and the answer will be the amount of miles she can drive worth $20. 

20 ÷ 4 = 3
3 × 32 = 96

So, Margie can drive 96 miles in $20. 

Hope I helped ya!! xD 
6 0
3 years ago
ABC has vertices A(–3, –4), B(1, 5), and C(4,1). ABC is translated so that B’ is located at point (–3, 2). State the coordinates
koban [17]
If B went from (1, 5) to (-3, 2), that means it was shifted down 4 units and left 3 units. So for each of the other points, we subtract 4 from the x value, and subtract 3 from the y value. Point A is (-3, -4), moving this by subtracting 4 from x and 3 from y, we get A'(-7, -7). Point C was (4, 1), subtract 4 and 3, leaving C'(0, -2).
So A'(-3, -4), B'(-3, 2) and C'(0, -2)
5 0
3 years ago
Read 2 more answers
A cylinder has a diameter of 5 centimeters and a height of 50 millimeters. Which of the following steps should be done
kolbaska11 [484]

Answer:

convert the diameter and the height to the same unit

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6
taurus [48]

Answer:

f(x) = \frac{6}{x^2} -8 or f(x) = -\frac{6}{x^2} + 4

Step-by-step explanation:

Given

(x,y) = (1,-2) --- Point

\int\limits^6_2 {(1 + 144x^{-6})} \, dx

The arc length of a function on interval [a,b]:  \int\limits^b_a {(1 + f'(x^2))} \, dx

By comparison:

f'(x)^2 = 144x^{-6}

f'(x)^2 = \frac{144}{x^6}

Take square root of both sides

f'(x) =\± \sqrt{\frac{144}{x^6}}

f'(x) = \±\frac{12}{x^3}

Split:

f'(x) = \frac{12}{x^3} or f'(x) = -\frac{12}{x^3}

To solve fo f(x), we make use of:

f(x) = \int {f'(x) } \, dx

For: f'(x) = \frac{12}{x^3}

f(x) = \int {\frac{12}{x^3} } \, dx

Integrate:

f(x) = \frac{12}{2x^2} + c

f(x) = \frac{6}{x^2} + c

We understand that it passes through (x,y) = (1,-2).

So, we have:

-2 = \frac{6}{1^2} + c

-2 = \frac{6}{1} + c

-2 = 6 + c

Make c the subject

c = -2-6

c = -8

f(x) = \frac{6}{x^2} + c becomes

f(x) = \frac{6}{x^2} -8

For: f'(x) = -\frac{12}{x^3}

f(x) = \int {-\frac{12}{x^3} } \, dx

Integrate:

f(x) = -\frac{12}{2x^2} + c

f(x) = -\frac{6}{x^2} + c

We understand that it passes through (x,y) = (1,-2).

So, we have:

-2 = -\frac{6}{1^2} + c

-2 = -\frac{6}{1} + c

-2 = -6 + c

Make c the subject

c = -2+6

c = 4

f(x) = -\frac{6}{x^2} + c becomes

f(x) = -\frac{6}{x^2} + 4

3 0
3 years ago
A father is twice as old as his son and the sum of their ages is 75. How old is the<br> father?
castortr0y [4]

Answer: 50

Let's first of all assume the son's age as x.

So,

Father's age will be 2x i.e. twice as old as the son

Given that,

Sum of their ages = 75 years

So,

x + 2x = 75

3x = 75

x = 75/3

x = 25

Hence,

Son's age = x = 25 years

Father's age = 2x = 2 * 25 = 50 years

8 0
2 years ago
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