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3 years ago

Simplify the expression: The picture is down below

Mathematics

1 answer:

sertanlavr [38]3 years ago

6 0

Answer:

$\frac{4x^{21}}{y^{11}}$

Step-by-step explanation:

You are going to need to know two properties of exponents.

$x^a * x^b = x^{a+b}\\\frac{x^a}{x^b}=x^{a-b}$

Also know that a*b = b*a, so youc an change the order of multiplication.

Since everything is being multiplied in the numerator, and there is only multiplication, it's actually super easy to start. Just multiply everything.

$8x^9y^8(-2)x^5y^{-9} = -16x^{14}y^{-1}$

so now you have $\frac{-16x^14y^{-1}}{-4x^{-7}y^6}$

Again, since the numerator and denominator only have multiplication you can divide part by part. so the number divides the number, the x term divides the term and the y term divides the y term

-16/-4 = 4

x^14 / x^-7 = x^21

y^-1 / y^6 = y^-7

So all together that makes $4x^{21}y^{-7}$

The final step in simplifying is making all negative powers into fractions. Sometimes non whole numbers can be made into roots. Definitely make negative powers fractions though

$\frac{4x^{21}}{y^{-7}}$

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a fruit delivers its fruit in two types of boxes: large and small. a delivery of 3 large boxes and 5 small boxes has a total wei

Masteriza [31]

Answer:

The weight of a small box = 13.5 kg

The weight of 1 large box = 18.5 kg

Step-by-step explanation:

Let us assume the weight of a small box = m kg

And the weight of 1 large box = n kg

Now, the weight of 5 small box = 5 x (weight of 1 small box) = 5 m

Also, the weight of 3 large box = 3 x (weight of 1 large box) = 3 n

Here, 3 large boxes + 5 small boxes = of 123 kilograms

⇒ 5 m + 3 n = 123 .... (1)

Again, the weight of 2 small box = 2 x (weight of 1 small box) = 2 m

Also, the weight of 12 large box = 12 x (weight of 1 large box) = 12 n

Here, 12 large boxes + 2 small boxes = 249 kilograms

⇒ 2 m+ 12 n = 249 .... (2)

Now, solving both the given equations by ELIMINATION, we get:

5 m + 3 n = 123 x (2)

2 m+ 12 n = 249 x (-5)

we get the new set of equitation as:

10 m + 6 n = 246

- 10 m - 60 n = -1245

Adding both equation, we get

-54 n = 999

or, n = 18.5

Now, 5 m + 3 n = 123

So, 5 m = 123 -3 (18.5) = 123 - 55.5 = 67.5

⇒ m = 13.5

Hence, the weight of a small box = m kg = 13.5 kg

And the weight of 1 large box = n kg = 18.5 kg

8 0

3 years ago

At soccer practice on Saturday morning, Winston's team practiced dribbling for 25 minutes and practiced shooting for 15 minutes.

Hatshy [7]

Answer:

10h20m AM

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Please anyone help question 3a(i) and (ii) I will mark brainliest answer

Alex73 [517]

Answer:

i) is correct answer i think

7 0

3 years ago

The graph of y=ax^2+bx+c passes through points (0,5), (1,10), and (2,19). Find a+b+c.

nataly862011 [7]

Answer: a = 2 b = 3 c = 5

Literally a+ b+c = 10, but I don't think that's what the question means to ask.

the equation is 2x² +3x + 5

Step-by-step explanation: Substitute the values of the given coordinates into the formula and use algebra (solve a system of equations by substitution and elimination) to get the values of a,b & c

y=ax²+bx+c using (0,5), (1,10), and (2,19)

5=a(0)²+b(0)+c becomes 5= 0+0 + c .so we find c = 5

10=a(1)²+b(1)+c becomes 10 = a + b + c

19=a(2)²+b(2)+c . becomes 19 = 4a + 2b + c

Substitute the 5 for c. Then rewrite

10 = a + b + 5 becomes 5 = a+ b to eliminate b multiply this be (-2)

(-2)( 5 = a+ b) .to get -10 = -2a - 2b

19 = 4a + 2b + 5 becomes 14 = 4a + 2b

_______________ Add these two

14 = 4a + 2b

<u>-10 = -2a - 2b</u><u> </u> this will eliminate the "b" term

4 = 2a divide both sides by 2 so you have a = 2

substitute in 10=a(1)²+b(1)+5 to find b

10=2(1)²+b(1)+5 .becomes 5 = 2 + b Subtract 2 from both sides

b = 3

So now you have all three values:

a = 2 b= 3 c = 5

I am attaching the graph. I hope this deserves Brainiest.

Best wishes in your studies!

4 0

3 years ago

HI PLEASE HELP ME WITH MY CALCULUS 1 HW? I AM REALLY STUCK. I need help with parts d,e,g.

asambeis [7]

(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when $t=0$ and $t=3$ , and because the velocity function is continuous, you need only check the sign of $v(t)$ for values on the intervals (0, 3) and (3, 6).

We have, for instance $v(1)\approx-0.91$ and $v(4)\approx0.91>0$ , which means the particle is moving the positive direction for $3$ , or the interval (3, 6).

(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

$\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt$

which follows from the definition of absolute value. In particular, if $x$ is negative, then $|x|=-x$ .

The total distance traveled is then 4 ft.

(g) Acceleration is the rate of change of velocity, so $a(t)$ is the derivative of $v(t)$ :

$a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)$

Compute the acceleration at $t=4$ seconds:

$a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}$

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)

6 0

3 years ago