Answer:
The exponent of y is 1
Step-by-step explanation:
Given
[tex]6xy^2 - 5x^{[]}y^{[]} + 9x^{[]}[/tex]
[tex]degree = 3[/tex]
Required
The exponent of y (second term)
Since the polynomial has a degree of 3, the exponents of y will decrease from left to right (i.e. 2,1,0) while x will increase from left to right (i.e. 1,2,3)
So, we have:
[tex]6xy^2 - 5x^{[]}y^{[]} + 9x^{[]} = 6xy^2 - 5x^{[2]}y^{[1]} + 9x^{[3]}[/tex]
Remove square brackets
[tex]6xy^2 - 5x^{[]}y^{[]} + 9x^{[]} = 6xy^2 - 5x^2y + 9x^3[/tex]
The second term is:
[tex]T_2 =5x^2y[/tex]
The exponent of y is 1
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
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
PLZ ANSWER QUESTION IN PICTURE
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]
[18].Simplify (TTE): x(2x+y+5) - 2(x²+xy+5) + y(x + y)
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]
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?
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
Can someone do #4 and #5
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
What is the phase of y= -3cos (3x-pi) +5
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]
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.
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
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
write your answer in simplest radical form
Answer:
z = √3
Step-by-step explanation:
sin (30°) = z / 2√3
z = sin (30°) 2√3
z = √3
(x/4) + (2x/7 =135 solve it
Answer:
the ans is 252................
lim(x-0) (sinx-1/x-1)
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=0Step-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.
Find the value of [(33.7)² - (15.3)²]^½ leaving your answer correct to 4 significant figures
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)
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?
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.
In a class of students, the following data table summarizes how many students have a cat or a dog. What is the probability that a student who has a cat also has a dog?
Has a cat Does not have a cat
Has a dog 7 6
Does not have a dog 8 2
Part b c and d please help
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 my savings of $x grows 10 percent each year, how much will i have in 2 years?
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
in each figure below find m<1 and m < 2 if a||b
please help i don't have a lot of time I will give brainliest if you help
Answer:
m∠1 = 105°
m∠2 = 75°
Step-by-step explanation:
From the picture attached,
Two lines 'a' and 'b' are parallel and a transversal 't' is intersecting these lines at two distinct lines.
Therefore, m∠2 = 75° [Corresponding angles measure the same]
m∠1 + m∠2 = 180° [Linear pair of angles are supplementary]
m∠1 + 75° = 180°
m∠1 = 105°
Use the order of operations to simplify 3/4+8(2.50-0.5).
Answer:
16[tex]\frac{3}{4}[/tex]
Step-by-step explanation:
I need help please can not figure out this problem
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?
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:
Find the area of a triangle as a mixed number.
Answer:
I believe the answer is 4 37/50!
What is the first step to solve the equation 16x-21 = 52?
1 Add 52 to both sides
2 Add 21 to both sides
3 Subtract 21 from both sides
4 Subtract 52 from both sides
Answer:
2) Add 21 to both sides
Step-by-step explanation:
When solving [tex]16x-21=52[/tex] for [tex]x[/tex], our goal to isolate [tex]x[/tex] such that we have [tex]x[/tex] set equal to something.
Therefore, we want to start by adding 21 to both sides. This leaves us with [tex]16x=73[/tex] and we are one step closer to isolating [tex]x[/tex].
The smallest positive solution of tan bx = 2 is x = 0.3. Determine the general solution of tan bx = 2.
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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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
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
35 + 3 x n with n = 7
the guys is wrong i checked
Step-by-step explanation:
the guys is wrong i checked
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?
PLEASE HELP THIS IS DUE ASAP!!!!!!!!!!!!!!
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
give the size of the letter figure below
Answer: 150 degrees
Step-by-step explanation: 10+ 20 = 30
180-30 = 150 degrees.
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.)
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
