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lana [24]
2 years ago
9

Simplify (5b)(-3a). -35ab -15ab 2ab

Mathematics
2 answers:
Strike441 [17]2 years ago
5 0
(5b)(-3a) \\ \\ -15ab \\ \\ Answer: \fbox {-15ab}
Fofino [41]2 years ago
3 0
(5b)(-3a)
= (5*(-3))(b*a) (combine like terms)
= -15ab

The final answer is -15ab~
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Harry rolls 2 fair dice. Probability of both dice showing a factor of 12
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Answer:

P = 4/9

Step-by-step explanation:

Event A1: The first dice shows a factor of 12. P(A1) = 4/6 = 2/3

Event A2: The second dice shows a factor of 12. P(A2) = 4/6 = 2/3

P (A1*A2)=4/9

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A falcon flies 800,000 meters in 4 hours. Use the formula d = rt, where d represents distance, r represents rate, and t represen
photoshop1234 [79]

Answer:

a. r=d/t b. 800000/4=200,000 c.800/2=200 d.not sure km are used more often as theyre simpler to use but there is not much that i am basing this off of

Step-by-step explanation:

5 0
2 years ago
If x and y satisfy both 9x+2y=8 and 7x+2y=4, then y=?
Alex

Answer:

The value of y is -5

Step-by-step explanation:

we have

9x+2y=8 ------> equation A

7x+2y=4 ------> equation B

we know that

If x and y satisfy both equations, then (x,y) is the solution of the system of equations

Using a graphing tool

Remember that

The solution of the systems of equations is the intersection point both graphs

The intersection point is (2,-5)

therefore

The solution of the system of equations is the point (2,-5)

The value of y is -5

7 0
2 years ago
Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times and did 20 mi
atroni [7]

Answer:

10 calories were burned for each minute of jogging  and  10 calories were burned for each minute of cycling.

Step-by-step explanation:

Let be "j" the amount of calories Shelly burned for each minute of jogging and "c" the amount of calories Shelly burned for each minute of cycling.

Set up a system of equations:

\left \{ {{10j+20c=300} \atop {20j+10c=300}} \right.

You can apply the Elimination method: multiply the first equation by -2, then add the equations and  solve for "c":

\left \{ {{-20j-40c=-600} \atop {20j+10c=300}} \right.\\...............................\\-30c-300\\\\c=10

Now, substitute the value of "c" into any original equation and solve for "j":

10j+20(10)=300\\\\10j=300-200\\\\10j=100\\\\j=10

7 0
2 years ago
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