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3 years ago

I don't know how to do this and need help please 21y^5+7y^3/7y^6

Mathematics

1 answer:

Rasek [7]3 years ago

7 0

Answer:

Step-by-step explanation:

I am assuming that there are 2 terms in the numerator,

(21y^5+7y^3)/7y^6

= 21y^5/7y^6 + 7y^3/7y^6

= 3/y + 1/y^3.

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Solve 2 ÷ x =5

Volgvan

So to help with the first one
2 / x = 5
multiply the x on both sides
2 = 5x
divide by 5 to isolate the x
2/5 = x

For the second one
We will use the diamond to help us find the common factor
\ 1 /
\ /
\ /
2 / \ 4
/ 3 \
/ \

1) the product
2) and 4) the two numbers
and 3) is sum

10 is the product and -7 is the sum
so what two numbers (factors of 10)
will equal -7 when added

so we have these numbers that will equal the product of +10 and we will need to find the ones that will equal -7 as the sum
10*1, 2*5, -1*-10, -2*-5
if we add the two numbers we will find respectively
11, 7, -11, -7
As you can see that -2 + -5 = +10 and -2+-5= +10
So we have found the two numbers
now before we factor the expression looks like
( x + a) (x + b)
and when factored looks like
x^2 + (a+b)x + (a*b)
Now we can plug in the numbers and solve to see if -2 and -5 are right
(x + -2) (x + -5)
we will factor it
x^2 +-5x + -2x + 10
x^2 + -7x + 10
so a = -2 and b = -5

Hope this helps :)

5 0

3 years ago

Bill drove 871 miles in 13 hours.

wariber [46]

Answer:

536 miles

Step-by-step explanation:

871/13 = 67

67*8 = 536

8 0

3 years ago

Read 2 more answers

A tank contains 60 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

MissTica

Answer:

(a) 60 kg; (b) 21.6 kg; (c) 0 kg/L

Step-by-step explanation:

(a) Initial amount of salt in tank

The tank initially contains 60 kg of salt.

(b) Amount of salt after 4.5 h

$\text{Let A = mass of salt after t min}\\\text{and }r_{i} = \text{rate of salt coming into tank}\\\text{and }r_{0} =\text{rate of salt going out of tank}$

(i) Set up an expression for the rate of change of salt concentration.

$\dfrac{\text{d}A}{\text{d}t} = r_{i} - r_{o}\\\\\text{The fresh water is entering with no salt, so}\\ r_{i} = 0\\r_{o} = \dfrac{\text{3 L}}{\text{1 min}} \times \dfrac {A\text{ kg}}{\text{1000 L}} =\dfrac{3A}{1000}\text{ kg/min}\\\\\dfrac{\text{d}A}{\text{d}t} = -0.003A \text{ kg/min}$

(ii) Integrate the expression

$\dfrac{\text{d}A}{\text{d}t} = -0.003A\\\\\dfrac{\text{d}A}{A} = -0.003\text{d}t\\\\\int \dfrac{\text{d}A}{A} = -\int 0.003\text{d}t\\\\\ln A = -0.003t + C$

(iii) Find the constant of integration

$\ln A = -0.003t + C\\\text{At t = 0, A = 60 kg/1000 L = 0.060 kg/L} \\\ln (0.060) = -0.003\times0 + C\\C = \ln(0.060)$

(iv) Solve for A as a function of time.

$\text{The integrated rate expression is}\\\ln A = -0.003t + \ln(0.060)\\\text{Solve for } A\\A = 0.060e^{-0.003t}$

(v) Calculate the amount of salt after 4.5 h

a. Convert hours to minutes

$\text{Time} = \text{4.5 h} \times \dfrac{\text{60 min}}{\text{1h}} = \text{270 min}$

b.Calculate the concentration

$A = 0.060e^{-0.003t} = 0.060e^{-0.003\times270} = 0.060e^{-0.81} = 0.060 \times 0.445 = \text{0.0267 kg/L}$

c. Calculate the volume

The tank has been filling at 6 L/min and draining at 3 L/min, so it is filling at a net rate of 3 L/min.

The volume added in 4.5 h is

$\text{Volume added} = \text{270 min} \times \dfrac{\text{3 L}}{\text{1 min}} = \text{810 L}$

Total volume in tank = 1000 L + 810 L = 1810 L

d. Calculate the mass of salt in the tank

$\text{Mass of salt in tank } = \text{1810 L} \times \dfrac{\text{0.0267 kg}}{\text{1 L}} = \textbf{21.6 kg}$

(c) Concentration at infinite time

$\text{As t $\longrightarrow \, -\infty,\, e^{-\infty} \longrightarrow \, 0$, so A $\longrightarrow \, 0$.}$

This makes sense, because the salt is continuously being flushed out by the fresh water coming in.

The graph below shows how the concentration of salt varies with time.

3 0

3 years ago

Determine the slope and y -intercept for each line shown. Then write the equation of the line (will give Brainliest)

Pavlova-9 [17]

Answer:

$y=\frac{1}{4}x+2$

m= $\frac{1}{4}$

b= 2

$y=-\frac{1}{3}x+1$

m= $-\frac{1}{3}$

b= 1

$y=3x+4$

m= 3

b=4

Step-by-step explanation:

1. The line intersects the y-axis at the point (0,2) therefore its y-intercept is b=2.

The line rises up 1 unit on the y-axis for every 4 units on the x-axis therefore the line has a slope of m=1/4.

Considering the equation of a line (y=mx+b), we plug in the variables we have found into the formula to find that $y=\frac{1}{4}x+2$

2. The line intersects the y-axis at the point (0,1) therefore its y-intercept is b=1.

The line down up 1 unit on the y-axis for every 3 units on the x-axis therefore the line has a slope of m= -1/3.

Considering the equation of a line (y=mx+b), we plug in the variables we have found into the formula to find that $y=-\frac{1}{3}x+1$

3. The line intersects the y-axis at the point (0,4) therefore its y-intercept is b=4.

The line rises up 3 units on the y-axis for every 1 unit on the x-axis therefore the line has a slope of m=3.

Considering the equation of a line (y=mx+b), we plug in the variables we have found into the formula to find that $y=3x+4$

3 0

2 years ago

. Mr. Perez owns a house on a lot shaped like a square. The perimeter of the lot is 60 feet. What is the length and the width of

pantera1 [17]

Hi there!

If we know the perimeter of the lot, it's pretty easy to find the length and width. Since we know that the lot is a square, we know that all sides are the same. And there are four sides on a square. Using this information, we can get an equation: 60 = x + x + x + x. Then, we need to simplify and solve the equation: 60 = 4x | x = 15 feet. Therefore, the length and width of the lot is 15 feet.

Hope this helps!! :)
If there's anything else that I can help you with, please let me know!

7 0

3 years ago