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Marina CMI [18]
2 years ago
11

What is the product of -3y²(-4y³+y-9)​

Mathematics
1 answer:
ddd [48]2 years ago
7 0

Answer:

-3y²(-4y³+y-9)​

-9y(-63y-9)

567y-81y

4 8 6 y

You might be interested in
Is 5361 divisible by three how do you know just by looking at it
sleet_krkn [62]

Answer:

Yes!

Step-by-step explanation:

Here is a trick. If you add up all of the numbers in the number, and if that is divisible by three, the number is divisible by three. So...

5+3+6+1=15

15 is divisible by 3,

so 5361 is divisible by 3

(if you wanted to know, it is 1787)

8 0
3 years ago
A set of pulleys is used to lift a piano with a mass of 102 kg. The piano is lifted 3 meters in 1 minutes. How much power is use
agasfer [191]

Answer:

50.031 watts

Step-by-step explanation:

Formula

Power = Force * distance / time

The units are critical

distance = meters

time = seconds

force = newtons = kg * m / s^2

Solution

F = m * a

a = 9.81

m = 102 kg

F = 102 * 9.81 = 1000.62 Newtons

T =  minute = 60 sec/min * 1 min

T = 60 seconds

d = 3 meters

Power = 1000.62 * 3 / 60

Power = 50.031

3 0
3 years ago
1.64 as a fraction in simplest form
mrs_skeptik [129]
<span>The correct answer is 1 16/25.

Explanation<span>:
This is read as "one and sixty-four one-hundredths." This means the fraction would be 1 64/100.

Both numerator and denominator are even, so we will divide both by 2 and get 1 32/50.

Again, both are even, so we divide by 2 again and get 1 16/25.</span></span>
4 0
3 years ago
If (2x - 3) is a factor of mx^2 - 11x - 6, find the other factor.​
Travka [436]

Answer:

(2x - 3) ^2 - 11x - 6

Step-by-step explanation:

6 0
2 years ago
A huge ice glacier in the Himalayas initially covered an area of 454545 square kilometers. Because of changing weather patterns,
guajiro [1.7K]

We have been given that a huge ice glacier in the Himalayas initially covered an area of 45 square kilo-meters. The relationship between A, the area of the glacier in square kilo-meters, and t, the number of years the glacier has been melting, is modeled by the equation A=45e^{-0.05t}.

To find the time it will take for the area of the glacier to decrease to 15 square kilo-meters, we will equate A=15 and solve for t as:

15=45e^{-0.05t}

\frac{15}{45}=\frac{45e^{-0.05t}}{45}

\frac{1}{3}=e^{-0.05t}

Now we will switch sides:

e^{-0.05t}=\frac{1}{3}

Let us take natural log on both sides of equation.

\text{ln}(e^{-0.05t})=\text{ln}(\frac{1}{3})

Using natural log property \text{ln}(a^b)=b\cdot \text{ln}(a), we will get:

-0.05t\cdot \text{ln}(e)=\text{ln}(\frac{1}{3})

-0.05t\cdot (1)=\text{ln}(\frac{1}{3})

-0.05t=\text{ln}(\frac{1}{3})

t=\frac{\text{ln}(\frac{1}{3})}{-0.05}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-0.05\cdot 100}

t=\frac{\text{ln}(\frac{1}{3})\cdot 100}{-5}

t=-\text{ln}(\frac{1}{3})\cdot 20

t=-(\text{ln}(1)-\text{ln}(3))\cdot 20

t=-(0-\text{ln}(3))\cdot 20

t=20\text{ln}(3)

Therefore, it will take 20\text{ln}(3) years for area of the glacier to decrease to 15 square kilo-meters.

6 0
3 years ago
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