Match each division expression to its quotient

Match Each Division Expression To Its Quotient

Answers

Answer 1

[tex]\frac{122}{10}*(-\frac{10}{61} )[/tex]Let's start by calculating their values one by one, and then we can match them.

Starting with [tex]-2\frac{2}{5} \div\frac{4}{5}[/tex], we can simplify this more by adding [tex]2*5[/tex] to the nominator. That gives us [tex]-\frac{12}{5} \div\frac{4}{5}[/tex]. Now we can apply the Keep-Change-Flip rule. Keep the first fraction as it is, change the division sign into multiplication, flip the second fraction. [tex]-\frac{12}{5} *\frac{5}{4}[/tex]. We apply fraction multiplication which is simply multiplying the first nominator by the first nominator and the same for the dominator.  and the result is [tex]-\frac{60}{20}[/tex] or simply -3.

[tex]-2\frac{2}{5} \div\frac{4}{5} = -3[/tex]

Now, we calculate the second one, [tex]-12.2\div(-6.1)[/tex]. This can be re-written as [tex]-\frac{122}{10}\div(-\frac{61}{10} )[/tex]. As we did in the previous part we apply the  Keep-Change-Flip, this will give us [tex]-\frac{122}{10}*(-\frac{10}{61} )[/tex]. Do the multiplication and the result will be [tex]\frac{1220}{610}[/tex], we can divide both the nominator and dominator by 10 which will result [tex]\frac{122}{61}[/tex] and finally we know that [tex]61*2=122[/tex] and we can divide both of them again by 61 which will result [tex]\frac{2}{1} =2[/tex]

[tex]-12.2\div(-6.1)=2[/tex]

You can try solving the rest by yourself but here's is the final answer for them both:

[tex]16\div(-8)=-2\\3\frac{3}{7} \div1\frac{1}{7} =3[/tex]


Related Questions

The smallest positive solution of tan bx = 2 is x = 0.3. Determine the general solution of tan bx = 2.

Answers

The general solution of [tex]\tan bx = 2[/tex] and [tex]x = 0.3[/tex] is [tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex].

From Trigonometry we remember that Tangent is a Transcendental Function that is positive both in 1st and 3rd Quadrants and have a periodicity of [tex]\pi[/tex] radians.  The procedure consists in using concepts of Direct and Inverse Trigonometric Functions as well as characteristics related to the behavior of the tangent function in order to derive a General Formula for every value of [tex]x[/tex], measured in radians.

First, we solve the following system of equations for [tex]b[/tex]:

[tex]\tan bx = 2[/tex] (1)

[tex]x = 0.3[/tex] (2)

Please notice that angles are measured in radians.

(2) in (1):

[tex]\tan 0.3b = 2[/tex]

[tex]0.3\cdot b = \tan^{-1} 2[/tex]

[tex]b = \frac{10}{3}\cdot \tan^{-1}2[/tex]

[tex]b\approx 3.690[/tex]

Under the assumption of periodicity, we know that:

[tex]y = \tan bx[/tex]

[tex]b\cdot x \pm \pi\cdot i = \tan^{-1} y[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]

[tex]b\cdot x = \tan^{-1}y \mp \pi\cdot i[/tex]

[tex]x = \frac{\tan^{-1}y \mp \pi\cdot i}{b}[/tex]

If we know that [tex]y = 2[/tex] and [tex]b \approx 3.690[/tex], then the general solution of this trigonometric function is:

[tex]x = \frac{0.352\pi \mp \pi\cdot i}{3.690}[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]

[tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex]

The general solution of [tex]\tan bx = 2[/tex] and [tex]x = 0.3[/tex] is [tex]x = 0.095\pi \mp 0.271\pi\cdot i[/tex], [tex]\forall \,i\in \mathbb{N}_{O}[/tex].

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the guys is wrong i checked

Answers

Step-by-step explanation:

the guys is wrong i checked

A car which was advertised for sale for 95000, was ultimately sold for 83600. Find the percent reduction in the price?

Answers

Answer: 12%

Step-by-step explanation:

95,000-83,600=11,400

(11,400/95000)(100) = 12%

The percentage reduction in the price of the car is 12%

What are percentages?

A percentage is a number or ratio that can be expressed as a fraction of 100. If we have to calculate the percent of a number, divide the number by the whole and multiply by 100. Hence, the percentage means, a part per hundred. The word percent means per 100. It is represented by the symbol “%”

Given here: Original price of car=95000 and Selling price=83600

Thus the reduction in price= 95000-83600

                                            =11400

Thus percentage reduction in the price of the car is

= 11400/95000 × 100

=12%

Hence, The percentage reduction in the price of the car is 12%

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Michael is 4 times as old as Brandon and is also 27 years older than Brandon.
How old is Brandon?

Answers

Answer:

9

Step-by-step explanation:

b = Brandon

4b=b+27

-b   -b

-------------

3b = 27

----   ----

3      3

b = 9

Brandon is 9 years old.

PLEASE HELP THIS IS DUE ASAP!!!!!!!!!!!!!!

Answers

the answer is 1/12

the first rolling a 4 has a 1/6 chance of happening and half of the numbers on the die are odd, so 1/6*1/2=1/12

Conan puts tennis balls into tubes after gym class. There are 17 tennis balls, and each tube holds 3 balls. How many tubes does Conan completely fill? How many tennis balls are left?

Answers

the answer is 5 tubes 2 left over

strontium-90 is a radioactive material that decays according to the function A(t)=A0e−0.0244t, where A0 is the initial amount present and A is the amount present at time t​ (in years). Assume that a scientist has a sample of 400 grams of​ strontium-90. ​
(a) What is the decay rate of​ strontium-90?
​(b) How much​ strontium-90 is left after 30 ​years?
​(c) When will only 100 grams of​ strontium-90 be​ left?
​(d) What is the​ half-life of​ strontium-90?

​(a) The decay rate of​ strontium-90 is nothing​%.
​(Type an integer or a decimal. Include the negative sign for the decay​ rate.)

Answers

Answer:

Step-by-step explanation:

The decay rate of strontium-90 is -.0244 as given.

For b., we have to use the formula to find out how much is left after 30 years. This will be important for part d.

[tex]A(t)=400e^{-.0244(30)}[/tex] which simplifies a bit to

A(t) = 400(.4809461353) so

A(t) = 192.4 g

For c., we have to find out how long it takes for the initial amount of 400 g to decay to 100:

[tex]100=400e^{-.0244t}[/tex]. Begin by dividing both sides by 400:

[tex].25=e^{-.0244t[/tex] and then take the natural log of both sides:

[tex]ln(.25)=lne^{-.0244t[/tex] . The natural log and the e cancel each other out since they are inverses of one another, leaving us with:

ln(.25) = -.0244t and divide by -.0244:

61.8 years = t

For d., we figured in b that after 30 years, 192.4 g of the element was left, so we can use that to solve for the half-life in a different formula:

[tex]A(t)=A_0(.5)^{\frac{t}{H}[/tex] and we are solving for H. Filling in:

[tex]192.4=400(.5)^{\frac{30}{H}[/tex] and begin by dividing both sides by 400:

[tex].481=(.5)^{\frac{30}{H}[/tex] and take the natural log of both sides, which allows us to pull the exponent out front. I'm going to include that step in with this one:

ln(.481) = [tex]\frac{30}{H}[/tex] ln(.5) and then divide both sides by ln(.5):

[tex]\frac{ln(.481)}{ln(.5)}=\frac{30}{H}[/tex] and cross multiply and isolate the H to get:

[tex]H=\frac{30ln(.5)}{ln(.481)}[/tex] and

H = 28.4 years

give the size of the letter figure below​

Answers

Answer: 150 degrees

Step-by-step explanation: 10+ 20 = 30

180-30 = 150 degrees.

[18].Simplify (TTE): x(2x+y+5) - 2(x²+xy+5) + y(x + y)

Answers

Answer:

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y) = 5x -10 + y\²[/tex]

Step-by-step explanation:

Given

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y)[/tex]

Required

Simplify

We have:

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y)[/tex]

Open brackets

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y) = 2x\²+xy+5x - 2x\²-2xy-10 + xy + y\²[/tex]

Collect like terms

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y) = 2x\²- 2x\²+xy-2xy+ xy+5x -10 + y\²[/tex]

[tex]x(2x+y+5) - 2(x\²+xy+5) + y(x + y) = 5x -10 + y\²[/tex]

4) The measure of the linear density at a point of a rod varies directly as the third power of the measure of the distance of the point from one end. The length of the rod is 4 ft and the linear density is 2 slugs/ft at the center, find the total mass of the given rod and the center of the mass​

Answers

Answer:

a. 16 slug b. 3.2 ft

Step-by-step explanation:

a. Total mass of the rod

Since the linear density at a point of the rod,λ varies directly as the third power of the measure of the distance of the point form the end, x

So, λ ∝ x³

λ = kx³

Since the linear density λ = 2 slug/ft at then center when x = L/2 where L is the length of the rod,

k = λ/x³ = λ/(L/2)³ = 8λ/L³

substituting the values of the variables into the equation, we have

k = 8λ/L³

k = 8 × 2/4³

k = 16/64

k = 1/4

So, λ = kx³ = x³/4

The mass of a small length element of the rod dx is dm = λdx

So, to find the total mass of the rod M = ∫dm = ∫λdx we integrate from x = 0 to x = L = 4 ft

M = ∫₀⁴dm

= ∫₀⁴λdx

= ∫₀⁴(x³/4)dx

= (1/4)∫₀⁴x³dx

= (1/4)[x⁴/4]₀⁴

= (1/16)[4⁴ - 0⁴]

= (256 - 0)/16

= 256/16

= 16 slug

b. The center of mass of the rod

Let x be the distance of the small mass element dm = λdx from the end of the rod. The moment of this mass element about the end of the rod is xdm =  λxdx = (x³/4)xdx = (x⁴/4)dx.

We integrate this through the length of the rod. That is from x = 0 to x = L = 4 ft

The center of mass of the rod x' = ∫₀⁴(x⁴/4)dx/M where M = mass of rod

= (1/4)∫₀⁴x⁴dx/M

= (1/4)[x⁵/5]₀⁴/M

= (1/20)[x⁵]₀⁴/M

= (1/20)[4⁵ - 0⁵]/M

= (1/20)[1024 - 0]/M

= (1/20)[1024]/M

Since M = 16, we have

x' =  (1/20)[1024]/16

x' = 64/20

x' = 3.2 ft

(x/4) + (2x/7 =135 solve it​

Answers

Answer:

the ans is 252................

A water trough is 9 m long and has a cross-section in the shape of an isosceles trapezoid that is 20 cm wide at the bottom, 70 cm wide at the top, and has height 50 cm. If the trough is being filled with water at the rate of 0.2 m3/min how fast is the water level rising when the water is 20 cm deep?

Answers

Answer:

dv = surface area * dh

so

dv/dt = surface area * dh/dt

width at surface = 40 + (80-40)(30/40)

= 40 + 30 = 70 cm = 0.70 m

so

surface area = 9 * .7 = 6.3 m^2

so

.3 m^3/min = 6.3 m^2 * dh/dt

and

dh/dt = .047 meters/min or 4.7 cm/min

Step-by-step explanation:

A coin is tossed and a die is rolled. Find the probability of getting a head and a number greater than 1.

___.
​(Type an integer or a simplified​ fraction.)

Answers

Answer:

5/12

Step-by-step explanation:

Heads: 1/2

Number greater than 1

A dice has 6 sides. 5 are greater than 1

The probability is 5/6

P(heads and a die greater than 1) = 1/2 * 5/6 = 5/12 or a little less than 1/2

e movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. How could you correctly rewrite the equation 4(10+5) = 6(12 - 2) using the distributive property​

Answers

9514 1404 393

Answer:

  4·10 +4·5 = 6·12 -6·2

Step-by-step explanation:

Each outside factor multiplies each inside term.

  4(10 +5) = 6(12 -2)

  4·10 +4·5 = 6·12 -6·2

lim(x-0) (sinx-1/x-1)

Answers

9514 1404 393

Answer:

as written: the limit does not existsin(x-1)/(x-1) has a limit of sin(1) ≈ 0.841 at x=0

Step-by-step explanation:

The expression written is interpreted according to the order of operations as ...

  sin(x) -(1/x) -1

As x approaches 0 from the left, this approaches +∞. As x approaches 0 from the right, this approaches -∞. These values are different, so the limit does not exist.

__

Maybe you intend ...

  sin(x -1)/(x -1)

This can be evaluated directly at x=0 to give sin(-1)/-1 = sin(1). The argument is interpreted to be radians, so sin(1) ≈ 0.84147098...

The limit is about 0.841 at x=0.

I am struggling and I would be so happy if any of you helped me. Can someone help me with the last two red boxes please? The rest of the question is for reference to help solve the problem. Thank you for your time!

Answers

Answer:

I think you can go with:

The margin of error is equal to half the width of the entire confidence interval.

so  try .74 ±   =   [ .724 , .756] as the confidence interval

Step-by-step explanation:

What is the phase of y= -3cos (3x-pi) +5

Answers

Answer:

[tex]- \frac{\pi}{3}[/tex]

Step-by-step explanation:

Given

[tex]y = -3\cos(3x - \pi) + 5[/tex]

Required

The phase

We have:

[tex]y = -3\cos(3x - \pi) + 5[/tex]

Rewrite as:

[tex]y = -3\cos(3(x - \frac{\pi}{3})) + 5[/tex]

A cosine function is represented as:

[tex]y = A\cos(B(x + C)) + D[/tex]

Where:

[tex]C \to[/tex] Phase

By comparison:

[tex]C = - \frac{\pi}{3}[/tex]

Hence, the phase is: [tex]- \frac{\pi}{3}[/tex]

Find the area of a triangle as a mixed number.

Answers

Answer:

I believe the answer is 4 37/50!

Use the order of operations to simplify 3/4+8(2.50-0.5).

Answers

Answer:

16[tex]\frac{3}{4}[/tex]

Step-by-step explanation:

Find the value of [(33.7)² - (15.3)²]^½ leaving your answer correct to 4 significant figures​

Answers

Answer:

30.03

Step-by-step explanation:

[(33.7)² - (15.3)²]^½

= [1135.69 - 234.09]^½

= [901.6]^½

= 30.02665483

= 30.03 (4sf)

PLZ ANSWER QUESTION IN PICTURE

Answers

Answer: y = 3x + 6

Step-by-step explanation:

(x-intercept of -2: (-2,0))

(slope = m)

y = mx + b, (-2,0), m = 3

[tex]y=mx+b\\0=3(-2)+b\\0=-6+b\\b=6\\y=3x+6[/tex]

Part b c and d please help

Answers

Answer:

b) Y =5.73X +4.36

C)  =5.73225*(21)X +4.359

    124.73625

D) 163.728 = 5.73X +4.36  

     X = (163.728 - 4.36)/5.73

     X = 27.81291449

  Year would be 2027

Step-by-step explanation:

x1 y1  x2 y2

4 27.288  16 96.075

   

(Y2-Y1) (96.075)-(27.288)=   68.787  ΔY 68.787

(X2-X1) (16)-(4)=    12  ΔX 12

   

slope= 5 41/56    

B= 4 14/39    

   

Y =5.73X +4.36      

If we decrease a dimension on a figure, how is the figure’s area affected?
The area decreases.
The area increases.
The area becomes 0.
The area remains the same.

Answers

The area decreases :)

35 + 3 x n with n = 7

Answers

56 would be the answer
It would just be 56 you can plug this in a Calcator

From September 1991 to September 1994 the enrollment at a particular school declined by 20 percent. If the number of students enrolled at that school in September 1994 was 720, what was the enrollment in September 1991

Answers

Answer:

900

Step-by-step explanation:

Given that :

Enrollment declined by 20% from between September 1991 to September 1994

This means there was a reduction in enrollment ;

Enrollment in September 1994 = 720

Enrollment in September 1991 = x

Hence,

Enrollment in 1994 = (1 - decline rate ) * enrollment in 1991

720 = (1 - 20%) * x

720 = (1 - 0.2) * x

720 = 0.8x

720 / 0.8 = 0.8x/0.8

900 = x

Hence, Enrollment in September 1991 = 900 enrollments

if my savings of $x grows 10 percent each year, how much will i have in 2 years?

Answers

Answer:

$240

Step-by-step explanation:

A year has 12 month in it so lets multiply the 10 by 12 which is $120,Mean a year is $120 so 2years will be $120×2 which is $240

Answer:

x+1/5x

Step-by-step explanation:

Because the eqaution would be x+10%=x+1/10+10%=1/5+x

Then the equation equals x+1/5

write your answer in simplest radical form​

Answers

Answer:

z = √3

Step-by-step explanation:

sin (30°) = z / 2√3

z = sin (30°) 2√3

z = √3

Can someone do #4 and #5

Answers

Answer:

First, find two points on the graph:

(x₁, y₁) = (0, 2)(x₂, y₂) = (2, 8)

Slope = [tex]\frac{y_{2}-y_{1} }{x_{2}-x_{1}} = \frac{8-2}{2-0} =\frac{6}{2}=3[/tex]

16 + (-3) = 16 - 3 = 13

A scientist has two solutions, which she has labeled Solution A and Solution B. Each contains salt. She knows that Solution A is 55% salt and Solution B is 70% salt. She wants to obtain 30 ounces of a mixture that is 60% salt. How many ounces of each solution should she use?

Answers

Answer:

Let x = the number of ounces of Solution A

Let y = the number of ounces of Solution B

x + y = 180        y = 180 - x

.60x + .85y = .75(180)

.60x + .85y = 135      Multiply both sides of the equation by 100 to remove the decimal points.

60x + 85y = 13500

60x + 85(180 - x) = 13500

60x + 15300 - 85x = 13500

-25x = -1800

x = 72ounces

y = 180 - 72

y = 108 ounces          

Step-by-step explanation:

Wyzant (ask an expert) solution on their website.

If a tank holds 6000 gallons of water, which drains from the bottom of the tank in 50 minutes, then Toricelli's Law gives the volume V of water remaining in the tank after t minutes as

V=5000 (1-1/50*t)^2 0⤠t ⤠50.

1. Find the rate at which water is draining from the tank after the following amount of time. (Remember that the rate must be negative because the amount of water in the tank is decreasing.)

a. 5 min
b. 10 min
c. 20 min
d. 50 min

2. At what time is the water flowing out the fastest?
3. At what time is the water flowing out the slowest?

Answers

Answer: hello from the question the volume of tank = 6000 gallons while the value in the Torricelli's equation = 5000 hence I resolved your question using the Torricelli's law equation

answer:

1) a) -180  gallons/minute ,

  b)  -160 gallons/minute

  c)  -120 gallons/minute

  d) 0

2) The water is flowing out fastest when t = 5 min

3) The water is flowing out slowest after t = 20 mins

Step-by-step explanation:

Volume of tank = 5000 gallons

Time to drain = 50 minutes

Volume of water remaining after t minutes by Torricelli's law

V = 5000 ( 1 - [tex]\frac{1}{50}t[/tex] )^2 ----- ( 1 )

1) Determine the rate at which water is draining from the tank

First step : differentiate equation 1 using the chain rule to determine the rate at which water is draining from the tank

V' = [tex]-10000[ ( 1 - \frac{1}{50}t ) (\frac{1}{50}) ][/tex]

a) After t = 5minutes

V' = - 10000[ ( 1 - 0.1 ) * ( 0.02 ) ]

   = -180  gallons/minute

b) After t = 10 minutes

V' = - 10000[ ( 1 - 0.2 ) * ( 0.02 ) ]

   = - 160 gallons/minute

c) After t = 20 minutes

V' = - 10000 [ ( 1 - 0.4 ) * ( 0.02 ) ]

   = -120 gallons/minute

d) After t = 50 minutes

V' = - 10000 [ ( 1 - 1 ) * ( 0.02 ) ]

   = 0 gallons/minute

2) The water is flowing out fastest when t = 5 min

3) The water is flowing out slowest after t = 20 mins  because no water flows out after 50 minutes

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