Step-by-step explanation:
Start Unlimited:
$70 for one line
$60 for two lines
$45 for three lines
$35 for four lines
Play More Unlimited:
$80 for one line
$70 for two lines
$55 for three lines
$45 for four lines
Do More Unlimited:
$80 for one line
$70 for two lines
$55 for three lines
$45 for four lines
Get More Unlimited:
$90 for one line
$80 for two lines
$65 for three lines
$55 for four lines
Answer:
4 months
As we show that ;
40 + (60x * 4) = 280
100 + (45x * 4) = 280
but in simultaneous equations Verizon must be set equal to 240 being 80 x 3
and T mobile must be equal to 45 x 4 = 180
so that 240+ 180 = 420 to find 4
This would be a method on distribution as "60 sign up is 1/3 more than 1st equation.
Step-by-step explanation:
This is a simultaneous equation but trial and error is below to prove all is true.
step 1 make all equations same
40s + (60x) * 3 = 180x LCM = 80 x 3 = 1 1/3 of 60
100s + (45x) * 4 = 180x LCM = 45 x 4 = 1 of 45 as verizon charges 1/3 more sign up.
100s- 40s + 180x = (180x) = 240 + 180
60s + 180x = 420
s = 60
so our equations must each end with 420
when we get 60s + 180x = 420 then
420 - 180 = 240
240/60 = 4
x = 4 months
Verizon = 1st and T mobile = 2nd
40 + (60x * 5) = 340
100 + 45x * 5 = 325
with $15 out after 5 months so we try 6 months
40 + (60x * 6) = 400
100 + 45x * 6 = 370
and see this is increasing in difference, so try a smaller value of months..
We try 4 months;
40 + (60x * 4) = 280
100 + (45x * 4) = 280
Which of the following theorems verifies that AABC - ASTU?
A. AA
B. HL
C. HA
D. LL
Answer:
AA
Step-by-step explanation:
See In Triangle ABC and Triangle STU
[tex]\because\begin{cases}\sf \angle A=\angle S=90° \\ \sf \angle B=\angle T=31°\end{cases}[/tex]
Hence
[tex]\sf \Delta ABC~\Delta STU(Angle-Angle)[/tex]
By AA similarity triangle ABC is similar to triangle SUT. Therefore, option A is the correct answer.
What are similar triangles?Two triangles are similar if the angles are the same size or the corresponding sides are in the same ratio. Either of these conditions will prove two triangles are similar.
In the given triangle ABC, ∠C=180°-90°-31°
∠C=59°
In the given triangle SUT, ∠U=180°-90°-31°
∠U=59°
Here, ∠B=∠T (Given)
∠C=∠U (Obtained using angle sum property of a triangle)
So, by AA similarity ΔABC is similar to ΔSUT.
Therefore, option A is the correct answer.
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After leaving an airport, a plane flies for 2 hours on a course of 60 degrees at a speed of 200 kilometers per hour. The plane then flies for 3 hours on a course of 210 degrees at a speed of 100 kilometers per hour What is the distance of the airport from the plane in kilometers? Round to the nearest tenth
Answer: 205.3
I suppose all measures of angles are done from the same axis (for example x-axis)
Step-by-step explanation:
You just have to use the theorem of Al'Kashi:
[tex]d^2=400^2+300^2-2*300*400*cos(30^o)\\\\d\approx{205.3(km)}[/tex]
Which of the following inequalities matches the graph?
10
6
-10
Oxs-1
Ox>-1
Oys-1
Oy 2-1
Answer:
y > -1
Step-by-step explanation:
the line is going across the y axis and is everything above -1
Plz help a beggar I don’t get it
Answer: 3
happy learning
Answer:
B.
Step-by-step explanation:
From the point (-1,0) the next point on the graph is up 3, right 1, making the slope a positive 3.
help pls! I need the answer quickly and pls explain. thank you!
Answer:
h = 6[tex]\sqrt{3}[/tex]
Step-by-step explanation:
The given is the special right triangle with angle measures : 90-60-30
and the side lengths for the given angles are represented by :
2a-a[tex]\sqrt{3}[/tex]-a
the side length that sees 60 degrees is represented by a[tex]\sqrt{3}[/tex] (h in this case)
the area of a triangle is calculated by multiplying height and base and that is divided by 2
a[tex]\sqrt{3}[/tex]*a/2 = 18[tex]\sqrt{3}[/tex] multiply both sides by 2
a^2[tex]\sqrt{3}[/tex] = 36[tex]\sqrt{3}[/tex] divide both sides by [tex]\sqrt{3}[/tex]
a^2 = 36 find the roots for both sides
a = 6
since h sees angle measure 60 and is represented by a[tex]\sqrt{3}[/tex]
h = 6[tex]\sqrt{3}[/tex]
f=((-1,1),(1,-2),(3,-4)) g=((5,0),(-3,4),(1,1),(-4,1)) find (fg)(1)
Answer:
f(g(1)) = - 2
Step-by-step explanation:
Find g(1) then use the value obtained to find f(x)
g(1) = 1 ← value of y when x = 1 (1, 1 ) , then
f(1) = - 2 ← value of y when x = 1 (1, - 2 )
If Bobby drinks 5 waters in 10 hours how many does he drink in 1 hour ?
Water drunk by Bobby in 10 hours = 5 units
So, water drink by Bobby in 1 hour
= 5/10 units
= 1/2 units
= 0.5 units
Answer:
1/2 water
Step-by-step explanation:
We can use a ratio to solve
5 waters x waters
----------- = ------------
10 hours 1 hours
Using cross products
5*1 = 10 *x
5 = 10x
Divide by 10
5/10 = x
1/2 waters =x
Given f(x) = 4x - 3 and g(x) = 9x + 2, solve for (f + g)(x).
[tex]\\ \sf\longmapsto (f+g)(x)[/tex]
[tex]\\ \sf\longmapsto f(x)+g(x)[/tex]
[tex]\\ \sf\longmapsto 4x-3+9x+2[/tex]
[tex]\\ \sf\longmapsto 4x+9x-3+2[/tex]
[tex]\\ \sf\longmapsto 13x-1[/tex]
Answer:
13x - 1
Step-by-step explanation:
f(x) + g(x) = 4x - 3 + 9x + 2
f(x) + g(x) = 4x+9x + 2 - 3
f(x) + g(x) = 13x - 1
on the same graph draw line 2y-x=10 and y=3x
Answer:
Step-by-step explanation:
If it takes 5 years for an animal population to double, how many years will it take until the population
triples?
9514 1404 393
Answer:
7.92 years
Step-by-step explanation:
We want to find t such that ...
3 = 2^(t/5)
where 2^(t/5) is the annual multiplier when doubling time is 5 years.
Taking logs, we have ...
log(3) = (t/5)log(2)
t = 5·log(3)/log(2) ≈ 7.92 . . . years
It will take about 7.92 years for the population to triple.
Suppose that 22 inches of wire costs 66 cents.
At the same rate, how much (in cents) will 17 inches of wire cost?
cents
Х
?
Answer:
51 cents for 17 inches of wire
Step-by-step explanation:
22 = 66
17 = x
22x = 66 * 17
22x = 1122
x = 51 cents
or
22 inches costs 66 cents
1 inch costs 3 cents (66 / 22 = 3 cents)
17 inches costs 51 cents (17 * 3 = 51 cents)
Assume the random variable X is normally distributed with mean μ = 50 and standard deviation σ = 7. Compute the probability
P(35 < X < 58)= ________
Answer:
Probability-Between .8574 = 85.74%
Step-by-step explanation:
Z1=-2.14 Z2=1.14
*x-1 35
*x-2 58
*µ 50
*σ 7
a) __m=10km 25m =___km
b) __m=__km__m=1.5 km
Example :
a) 7250m= 7km 250m = 7.250km
Please help me
Answer:
a) 10,025 m = 10km 25m = 10.025 km
b) 1,500 m = 1 km 500 m = 1.5 km
Answer:
a) 10025m = 10km 25m = 10.025km
b) 1500m = 1km 500m = 1.5km
Step-by-step explanation:
Concept:
Here, we need to know the idea of unit conversion.
Unit conversion is the conversion between different units of measurement for the same quantity.
1 km = 1000 m
Solve:
a)
10km 25m = 10×1000 + 25 = 10025 m10km 25m = 10 + 25/1000 = 10.025 kmb)
1.5km = 1 + 0.5 × 1000 = 1km 500m1.5km = 1.5 × 1000 = 1500mHope this helps!! :)
Please let me know if you have any questions
what is the volume of a cube with a length of a 10cm,
a width of 8cm and a height of 8cm
Answer:
640
Step-by-step explanation:
muntiply all the number length width height
[tex] \frac{3x - 2}{7} - \frac{5x - 8}{4} = \frac{1}{14} [/tex]
Answer:
[tex]x=2[/tex]
Step-by-step explanation:
[tex]\frac{3x-2}{7}-\frac{5x-8}{4}=\frac{1}{14}[/tex]
In order to factor an integer, we need to divide it by the ascending sequence of primes 2, 3, 5.
The number of times that each prime divides the original integer becomes its exponent in the final result.
In here, Prime number 2 to the power of 2 equals 4.
[tex]\frac{3x-2}{7}-\frac{5x-8}{2^{2} }=\frac{1}{14}[/tex]
First, We need to add fractions-
Rule:-
[tex]\frac{A}{B} +\frac{C}{D} =\frac{\frac{LCD}{B}+\frac{LCD}{D}C }{LCD}[/tex]
LCD = [tex]7 \cdot 2^{2}[/tex]
[tex]\frac{4(3x-2)+7(-(5x-8))}{7*2^{2} } =\frac{1}{14}[/tex]
[tex]x=2[/tex]
OAmalOHopeO
A telescope contains both a parabolic mirror and a hyperbolic mirror. They share focus , which is 46feet above the vertex of the parabola. The hyperbola's second focus is 6 ft above the parabola's vertex. The vertex of the hyperbolic mirror is 3 ft below . Find the equation of the hyperbola if the center is at the origin of a coordinate system and the foci are on the y-axis. Complete the equation.
the center is at the origin of a coordinate system and the foci are on the y-axis, then the foci are symmetric about the origin.
The hyperbola focus F1 is 46 feet above the vertex of the parabola and the hyperbola focus F2 is 6 ft above the parabola's vertex. Then the distance F1F2 is 46-6=40 ft.
In terms of hyperbola, F1F2=2c, c=20.
The vertex of the hyperba is 2 ft below focus F1, then in terms of hyperbola c-a=2 and a=c-2=18 ft.
Use formula c^2=a^2+b^2c
2
=a
2
+b
2
to find b:
\begin{gathered} (20)^2=(18)^2+b^2,\\ b^2=400-324=76 \end{gathered}
(20)
2
=(18)
2
+b
2
,
b
2
=400−324=76
.
The branches of hyperbola go in y-direction, so the equation of hyperbola is
\dfrac{y^2}{b^2}- \dfrac{x^2}{a^2}=1
b
2
y
2
−
a
2
x
2
=1 .
Substitute a and b:
\dfrac{y^2}{76}- \dfrac{x^2}{324}=1
76
y
2
−
324
x
2
=1 .
HELP ASAP!!
The equation (blank) has no solution
Answer:
Just to recap, an equation has no solution when it results in an incorrect "equation".
For example:
Equation: x+3 = x+4
Subtract x: 3 = 4???
But clearly, 3 is not equal to 4, so this equation has NO SOLUTION.
Now onto our problem:
13y+2-2y = 10y+3-y
11y+2 = 9y+3
2y=1
y=1/2
9(3y+7)-2 = 3(-9y+9)
27y+61 = -27y+27
54y = -34
y = -34/54
32.1y+3.1+2.4y-8.2=34.5y-5.1
34.5-5.1=34.5y-5.1
5.1=5.1
infinite solutions
5(2.2y+3.4) = 5(y-2)+6y
11y+17 = 11y-10
17 = -10??
That's not true, so the option "5(2.2y+3.4) = 5(y-2)+6y" has no solution.
Let me know if this helps
The legend on a map states that 1 inch is 20 miles. If you measure 5 inches on the map, how many miles would the actual distance be? Actual distance = [ ? ] miles
Answer:
1 inch= 20 miles. 5*20=100 miles. The answer is 100 miles.
Step-by-step explanation:
On these types of questions just do that every time, then you don't need to ask, for example:
1 foot = 50 miles
If it measures 3 feet.
3*50=150 miles.
If you have any questions regarding my answer, tell me in the comments, and I will answer them.
Consider an x distribution with standard deviation o = 34.
(a) If specifications for a research project require the standard error of the corresponding distribution to be 2, how
large does the sample size need to be?
B) If specifications for a research project require the standard error of the corresponding x distribution to be 1, how large does the sample size need to be?
Part (a)
The standard error (SE) formula is
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\[/tex]
where n is the sample size. We're given SE = 2 and sigma = 34, so,
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\2 = \frac{34}{\sqrt{n}}\\\\2\sqrt{n} = 34\\\\\sqrt{n} = \frac{34}{2}\\\\\sqrt{n} = 17\\\\n = 17^2\\\\n = 289\\\\[/tex]
So we need a sample size of n = 289 to have an SE value of 2.
Answer: 289========================================================
Part (b)
We'll use SE = 1 this time
[tex]\text{SE} = \frac{\sigma}{\sqrt{n}}\\\\1 = \frac{34}{\sqrt{n}}\\\\1*\sqrt{n} = 34\\\\\sqrt{n} = 34\\\\n = 34^2\\\\n = 1156\\\\[/tex]
Because we require greater precision (i.e. a smaller SE value), the sample size must be larger to account for this. In other words, as SE goes down, then n must go up, and vice versa.
Answer: 1156The sum of three numbers is 124
The first number is 10 more than the third.
The second number is 4 times the third. What are the numbers?
Answer:
182/3,3 8/3, 152/3
Step-by-step explanation:
a+b+c=124
a trừ c= 10
4b=c
Answer:
a=29,b=79,c=19
Step-by-step explanation:
a=c+10
b=4c
=> a+b+c=c+10+4c+c=124
=> c=19
=> a= 29, b=79
find the measure of d
Each course at college X is worth either 2 or 3 credits. The members of the men's swim team are taking a total of 48 courses that are worth a total of 107 credits. How many 2-credit courses and how many 3-credit courses are being taken?
Answer:
Let the number of courses that are worth 3 credits each be x and those worth 4 credits be y. With the given information, you can write the following equations:
x + y = 48
3x + 4y = 155
You can solve the above equations by method of elimination/substitution
x + y = 48 ⇒ x = 48 - y (Now, substitution this equation into 3x + 4y = 155)
3(48 - y) + 4y = 155
144 -3y + 4y = 155
y + 144 = 155
y = 11
Now plug this solution back into x = 48 - y
x = 48 - 11 = 37
Check work (by plugging the solutions back into the 3x + 4y and see if it's equal to 155):
3(37) + 4(11) = 155
Answer: There are 37 of the 3-credit course and 11 of the 4-credit course
In an assembly-line production of industrial robots, gearbox assemblies can be installed in one minute each if holes have been properly drilled in the boxes and in ten minutes if the holes must be redrilled. Twenty-four gearboxes are in stock, 6 with improperly drilled holes. Five gearboxes must be selected from the 24 that are available for installation in the next five robots. (Round your answers to four decimal places.) (a) Find the probability that all 5 gearboxes will fit properly. (b) Find the mean, variance, and standard deviation of the time it takes to install these 5 gearboxes.
Answer:
The right answer is:
(a) 0.1456
(b) 18.125, 69.1202, 8.3139
Step-by-step explanation:
Given:
N = 24
n = 5
r = 7
The improperly drilled gearboxes "X".
then,
⇒ [tex]P(X) = \frac{\binom{7}{x} \binom {17}{5-x}}{\binom{24}{5}}[/tex]
(a)
P (all gearboxes fit properly) = [tex]P(x=0)[/tex]
= [tex]\frac{\binom{7}{0} \binom{17}{5}}{\binom{24}{5}}[/tex]
= [tex]0.1456[/tex]
(b)
According to the question,
[tex]X = 91+5[/tex]
Mean will be:
⇒ [tex]\mu = E(x)[/tex]
[tex]=E(91+5)[/tex]
[tex]=9E(1)+5[/tex]
[tex]=9.\frac{nr}{N}+5[/tex]
[tex]=9.\frac{5.7}{24} +5[/tex]
[tex]=18.125[/tex]
Variance will be:
⇒ [tex]\sigma^2=Var(X)[/tex]
[tex]=V(9Y+5)[/tex]
[tex]=81.V(Y)[/tex]
[tex]=81.n.\frac{r}{N}.\frac{N-r}{N}.\frac{N-n}{N-1}[/tex]
[tex]=81.5.\frac{7}{24}.\frac{24-7}{24}.\frac{24-5}{24-1}[/tex]
[tex]=69.1202[/tex]
Standard deviation will be:
⇒ [tex]\sigma = \sqrt{69.1202}[/tex]
[tex]=8.3139[/tex]
The weekly amount of money spent on maintenance and repairs by a company was observed, over a long period of time, to be approximately normally distributed with mean $440 and standard deviation $20. How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1
Answer:
$465.6 should be budgeted.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Normally distributed with mean $440 and standard deviation $20.
This means that [tex]\mu = 440, \sigma = 20[/tex]
How much should be budgeted for weekly repairs and maintenance so that the probability the budgeted amount will be exceeded in a given week is only 0.1?
The 100 - 10 = 90th percentile should be budgeted, which is X when Z has a p-value of 0.9, so X when Z = 1.28. Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 440}{20}[/tex]
[tex]X - 440 = 1.28*20[/tex]
[tex]X = 465.6[/tex]
$465.6 should be budgeted.
Which figure always has exactly one line of symmetry?
A. rectangle
B. trapezoid
C. isosceles right triangle
D. circle
A shape of a trapezoid has exactly one line of symmetry. The correct option is B.
What is a trapezoid?An open, flat object with four straight sides and one pair of parallel sides is referred to as a trapezoid or trapezium.
A balanced and proportionate likeness between an object's two halves is referred to as symmetry in geometry. It implies that one half is the other's mirror image.
A trapezium's non-parallel sides are referred to as the legs, while its parallel sides are referred to as the bases. The legs of a trapezium can also be parallel. The parallel sides may be vertical, horizontal, or angled.
Therefore, the shape of a trapezoid has exactly one line of symmetry. The correct option is B.
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It is estimated that 75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.
(a) On average, how many young adults do not own a landline in a random sample of 100?
(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?
(c) What is the proportion of young adults who do not own a landline?
(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?
(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?
(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?
(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?
Answer:
a) 75
b) 4.33
c) 0.75
d) [tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline
e) [tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.
f) Binomial, with [tex]n = 100, p = 0.75[/tex]
g) [tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.
Step-by-step explanation:
For each young adult, there are only two possible outcomes. Either they do not own a landline, or they do. The probability of an young adult not having a landline is independent of any other adult, which means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
75% of all young adults between the ages of 18-35 do not have a landline in their homes and only use a cell phone at home.
This means that [tex]p = 0.75[/tex]
(a) On average, how many young adults do not own a landline in a random sample of 100?
Sample of 100, so [tex]n = 100[/tex]
[tex]E(X) = np = 100(0.75) = 75[/tex]
(b) What is the standard deviation of probability of young adults who do not own a landline in a simple random sample of 100?
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100(0.75)(0.25)} = 4.33[/tex]
(c) What is the proportion of young adults who do not own a landline?
The estimation, of 75% = 0.75.
(d) What is the probability that no one in a simple random sample of 100 young adults owns a landline?
This is P(X = 100), that is, all do not own. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 100) = C_{100,100}.(0.75)^{100}.(0.25)^{0} = 3.2 \times 10^{-13}[/tex]
[tex]3.2 \times 10^{-13}[/tex] probability that no one in a simple random sample of 100 young adults owns a landline.
(e) What is the probability that everyone in a simple random sample of 100 young adults owns a landline?
This is P(X = 0), that is, all own. So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{100,0}.(0.75)^{0}.(0.25)^{100} = 6.2 \times 10^{-61}[/tex]
[tex]6.2 \times 10^{-61}[/tex] probability that everyone in a simple random sample of 100 young adults owns a landline.
(f) What is the distribution of the number of young adults in a sample of 100 who do not own a landline?
Binomial, with [tex]n = 100, p = 0.75[/tex]
(g) What is the probability that exactly half the young adults in a simple random sample of 100 do not own a landline?
This is P(X = 50). So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 50) = C_{100,50}.(0.75)^{50}.(0.25)^{50} = 4.5 \times 10^{-8}[/tex]
[tex]4.5 \times 10^{-8}[/tex] probability that exactly half the young adults in a simple random sample of 100 do not own a landline.
Given the central angle, name the arc formed.
Major arc for ∠EQD
A. EQDˆ
B. GDFˆ
C. EGDˆ
D. EDˆ
9514 1404 393
Answer:
C. EGD
Step-by-step explanation:
A major arc is typically named using the end points and a point on the arc. Here, the end points are E and D, and points on the major arc include C, G, and F. The major arc ED could be named any of
arc ECDarc EGD . . . . choice Carc EFDOf course, the reverse of any of these names could also be used: DCE, DGE, DFE.
The sum of the 3rd and 7th terms of an A.P. is 38, and the 9th term is 37. Find the A.P.
Answer:
The AP is 1, 11/2, 10, 29/2, 19, ....
Step-by-step explanation:
Let the first term be a and d be the common difference of the arithmetic progression.
ATQ, a+2d+a+6d=38, 2a+8d=38 and a+8d=37. Solving this, we will get a=1 and d=9/2. The AP is 1, 11/2, 10, 29/2, 19, ....
Team A scored 30 points less than four times the number of points that Team B scored. Team C scored 61 points more than half of the number of points that Team B scored. If Team A and Team C shared in the victory, having earned the same number of points, how many more points did each team have than Team B?
Answer:
team a and team c scored 74 points which is 48 points more than team b, scoring 26 points.
Step-by-step explanation:
Question
(X-5y/y3)-1=
Answer:
[tex]x = y^3+5y[/tex]
Step-by-step explanation:
Complete question
[tex]\frac{x - 5y}{y^3} - 1=0\\[/tex]
Required
Solve for x
We have:
[tex]\frac{x - 5y}{y^3} - 1=0[/tex]
Collect like terms
[tex]\frac{x - 5y}{y^3} = 1[/tex]
Multiply through by [tex]y^3[/tex]
[tex]x - 5y = y^3[/tex]
Make x the subject
[tex]x = y^3+5y[/tex]
