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ludmilkaskok [199]
3 years ago
15

3(x-1)=2x+9 how many solutions are there

Mathematics
1 answer:
Nostrana [21]3 years ago
5 0
There are 4 solutions
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A circle has its center at (0,0) and passes through the point (0,9). What is the standard equation of the circle?
Alexxandr [17]

Answer:

x^2+y^2=9^2

Step-by-step explanation:

The standard equation of a circle is:

(x-h)^2+(y-k)^2=r^2

Where the center of a circle is (h,k) and r is the radius of the circle.  In this case because a circle is equidistant from the center and we have a point where it passes through 9 that means that the radius is 9.  However, since the standard equation states that we must write r in the form of r^2 this means that r^2=9^2.  Therefore by plugging in the values we have:

(x-0)^2+(y-0)^2=9^2\\\\x^2+y^2=9^2

7 0
3 years ago
7th grade math
Elza [17]

Answer:

1. b and g    2.a end e    3. d and f    4. g and h

Step-by-step explanation:

4 0
2 years ago
Arianna went running for 0.75 hours and ran 5.55 miles. What was her speed, in miles per hour, if Arianna
Nataliya [291]

Answer:

7.4 miles per hour

Step-by-step explanation:

To find her speed in miles per hour, divide the amount of miles she ran by the number of hours:

5.55/0.75

= 7.4

So, her speed is 7.4 miles per hour

6 0
3 years ago
A ball weighs 15lb on earth and 5.7 pounds on mars. What is the constant proportionality between weight of an object on mars and
krok68 [10]
So its asking you to find the gravity on earth relative to the gravity on mars.

Set up an equation to figure out the difference between the two gravities 

15x=5.7
\\x=5.7/15
x=.38

I don't know what constant of proportionality is in this context but a ball on mars would weigh .38 the amount that it does on earth.
5 0
3 years ago
A NASA spacecraft measures the rate R of at which atmospheric pressure on Mars decreases with altitude. The result at a certain
AysviL [449]

Answer:

R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}

Step-by-step explanation:

For this case we have the following value:

R = 0.0498 \frac{Kpa}{Km}

We can convert this first to \frac{Kpa}{m} like this:

R=0.0498 \frac{Kpa}{Km} *\frac{1km}{1000m}=0.0000498 \frac{Kpa}{m}

Now we use the fact the the pressure is defined as P =\frac{F}{A}, whre P is the pressure, F the force and A the area, so then Kpa= \frac{KN}{m^2} and then we can replace this:

R=0.0000498 \frac{KN}{m^3}

Now from definition of work we know that W= Fd where W is the work, F the force and d the distance, so then is equivalent KN =\frac{KJ}{m}

And if we replace this into the equation we got:

R = 0.0000498 \frac{KJ}{ m m^3}=0.0000498 \frac{KJ}{m^4}=4.98x10^{-5}\frac{KJ}{m^4}

4 0
3 years ago
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