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stepladder [879]
3 years ago
6

Which is true about the polynomial –8m3 + 11m?

Mathematics
2 answers:
algol133 years ago
5 0

Answer:

B

Step-by-step explanation:

Veseljchak [2.6K]3 years ago
4 0

Answer:

stepbinomial with a degree of 3

explanation:

Step-by-step

:

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Correct answer gets brainliest!!!!!
Aliun [14]
Hello!

You can use the Pythagorean Theorem to solve this

Separate the rectangle into two triangles

a^{2} + b^{2} = c^{2}

c is the diagonal
a and b are the side lengths

Put in the values you know

5^{2} + 4^{2} = c^{2}

Square the numbers

25 + 16 =  c^{2}

Add

41 =  c^{2}

Take the square root of both sides

6.4 = c

The answer is B) 6.4 units

Hope this helps!
5 0
3 years ago
Read 2 more answers
To find the answer to the questions ask?
timurjin [86]

The answer is 6.

5*6-0=30

30=30

7 0
4 years ago
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A man makes a profit of 15%6y selling an article for Rs 2500. For what price should he sellituto make a profit of 38%​
dusya [7]

Find the price before the 15% profit:

2500 / 1.15 = 2173.91

Now multiply the original cost by 1 + 38%:

2173.91 x 1.38 = 2999.99

Answer: 2.999.99 ( may need to round to 3,000)

5 0
3 years ago
HELPPP PLEASE .☆........
tiny-mole [99]

Answer:

I think its 1/3/ and 4. But I dont know tho. just make sure with other.

Step-by-step explanation:

7 0
3 years ago
A man travels 20 km by car from Town P to Town Q at an average speed of x km/h. He finds that the time of the journey would be s
yuradex [85]

Answer:

x = 20.

Step-by-step explanation:

First, you should remember the relation:

Distance = Speed*Time.

First, we know that a man travels a distance of 20km at a speed of x km/h, in a time T.

We can write this as:

20km = (x km/h)*T

We know that the time is shortened by 12 minutes if the speed is increased by 5km/h

Rewriting these 12 minutes in hours (remember that 60min = 1 hour)

12 min = (12/60) hours = 0.2 hours

Then from this, he can travel the same distance of 20km in a time T minus 0.2 hours if the speed is increased by 5 km/h

We can write this as:

20km = (x + 5 km/h)*(T - 0.2 h)

Then we have a system of two equations, and we want to find the value of x:

20km = (x km/h)*T

20km = (x + 5 km/h)*(T - 0.2 h)

First, we should isolate the variable T in one of the equations, if we isolate it in the first one, we will get:

20km/(x km/h) = T

Replacing that in the other equation we get:

20km = (x + 5 km/h)*(T - 0.2 h)

20km = (x + 5 km/h)*( 20km/(x km/h) - 0.2 h)

Now we can solve this for x.

Removing the units (that we know that are correct) so the math is easier to read, we get:

20 = (x + 5)*(20/x - 0.2)

We only want to solve this for x.

20 = x*20/x - x*0.2 + 5*20/x - 5*0.2

20 = 20 - 0.2*x + 100/x - 1

subtracting 20 in both sides we get:

20 - 20 = 20 - 0.2*x + 100/x - 1 - 20

0 = -0.2*x + 100/x - 1

If we multiply both sides by x we get:

0 = -0.2*x^2 + 100 - x

-0.2*x^2 - x + 100 = 0

This is just a quadratic equation, we can solve it using the Bhaskara's equation, the solutions are:

x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*(-0.2)*100} }{2*-0.2}  = \frac{1 \pm 9 }{-0.4}

Then the two solutions are:

x = (1 + 9)/-0.4 = -25

x = (1 - 9)/-0.4 = 20

As x is used to represent a speed, the negative solution does not make sense, so we should use the positive one.

x = 20

then the average speed initially is 20 km/h

3 0
3 years ago
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