Hey there!
We are given two functions - one is Exponential while the another one is Linear.
[tex] \large{ \begin{cases} f(x) = {4}^{x} - 8 \\ g(x) = 5x + 6 \end{cases}}[/tex]
1. Operation of Function
(f+g)(x) is a factored form of f(x)+g(x). We can common factor out x. Therefore:[tex] \large{(f + g)(x) = f(x) + g(x)}[/tex]
2. Substitution
Next, we substitute f(x) = 4^x+8 and g(x) = 5x+6.[tex] \large{(f + g)(x) = ( {4}^{x} - 8) + (5x + 6)}[/tex]
3. Evaluate/Simplify
Cancel out the brackets and combine like terms.[tex] \large{(f + g)(x) = {4}^{x} - 8 + 5x + 6} \\ \large{(f + g)(x) = {4}^{x} + 5x - 8 + 6} \\ \large{(f + g)(x) = {4}^{x} + 5x - 2}[/tex]
4. Final Answer
(f+g)(x) = 4^x+5x-2I need to find the distance B in the special counter sink shown
Answer:
Step-by-step explanation:
87°32' = 86°92'
(86°92')/2 = 43°46'
B = 13/(16cos(43°46')) = 1.125
Answer:
Step-by-step explanation:
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:
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]
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]
A rectangular tank 4 feet long, 3 feet wide, and 5 feet deep is full of oil with weight density 50 lb ft 3 lbft3 . Calculate the work required to pump all of the oil out over the top of the tank.
The work required for the given task of pumping all of the oil out over the top of the tank is 7,500 ft·lb
The known parameters;
The length of the rectangular tank, l = 4 feet
The width of the tank, w = 2 feet
The depth of the tank, h = 5 feet
The weight density of the oil with which the tank is filled, ρ × g = 50 lb/ft³
The unknown parameter
The work required to pump all of the oil out over the top of the tank
Method;
Calculate the force required to lift each slice (layer) of the oil to the top multiplied by the distance, y, the slice moves and summing the result as an integration as follows;
The volume of each slice, [tex]\mathbf{V_i}[/tex] = l × w × dy
The force required to move each slice, [tex]\mathbf{F_i}[/tex] = ρ × g × l × w × dy
The work done, [tex]\mathbf{W_i}[/tex], in moving the slice a distance, y, is given as follows;
[tex]\mathbf{W_i}[/tex] = ρ × g × l × w × y × dy
Therefore, the total work done, W, in pumping all the water located from y = 0, to y = 5, to the top of the tank, is given as follows;
[tex]\mathbf{W = \int\limits^5_0 {(\rho \times g \times l \times w \times y) } \, dy}[/tex]
Therefore;
W = (ρ × g × l × w × y²)/2
Plugging in the values, gives;
W = (50 lb/ft³ × 4 ft. × 3 ft. × (5 ft.)²)/2 = 7,500 ft·lb
The work required to pump all of the oil out over the top of the tank, W = 7,500 ft·lb.
Learn more about the use of integration to calculate the amount of work required for a given task here;
https://brainly.com/question/14318035
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
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.
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.
A student found the solution below for the given inequality.Which of the following explains whether the student is correct?The student is completely correct because the student correctly wrote and solved the compound inequality.The student is partially correct because only one part of the compound inequality is written correctly.The student is partially correct because the student should have written the statements using “or” instead of “and.”The student is completely incorrect because there is “ no solution “ to this inequality.
Answer:
The student is completely incorrect because there is no solution to this inequality.
Answer:
D on edge
Step-by-step explanation:
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
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 .
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]
Can someone help please
Step-by-step explanation:
a) The volume of the prism is
[tex]V = (n^2 - 1)×(n^2 - 1)×(5n)[/tex]
[tex]\:\:\:\:= (n^4 - 2n^2 + 1)(5n)[/tex]
[tex]\:\:\:\:=5n^5 - 10n^3 + 5n[/tex]
b) If the dimensions L of the prism are tripled, the new volume will be
[tex]V' = (3L)^3 = 27L^3 = 27V[/tex]
so it will increase by a factor of 27.
The graph of y= -2x + 10 is:
O A. a line that shows only one solution to the equation.
O B. a point that shows the y-intercept.
O C. a line that shows the set of all solutions to the equation.
O D. a point that shows one solution to the equation.
SUBM
9514 1404 393
Answer:
C. a line that shows the set of all solutions to the equation.
Step-by-step explanation:
Any graph shows the set of all solutions to the equation being graphed.
The graph of a linear function is a straight line.
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
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, ....
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 point on the number line shows the graph of 0.5?
Point A
Point B
Point C
Point D
Answer: D
Step-by-step explanation:
D
Answer:
D
Step-by-step explanation:
it is between 0 and 1
How to write -.04 as a fraction?
Answer:
[tex]0.04 = 4 \div 100 [/tex]
[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
4. Explain how the graphs of the functions are similar and how they are different
2x+3y=1470
And
2x+3y=1593
Answer:
they are parallel lines so have the same slope. Difference would be the where they intersect the x and y axis
Step-by-step explanation:
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 )
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
The height of a triangle is 5 yards greater than the base. The area of the triangle is 273 square yards. Find the length of the base and the height of the triangle.
Answer:
Base = 21 while Height = 16
What is the GCF of the expression 5x2y + 10xy2?
Answer:
[tex]5xy[/tex]
Step-by-step explanation:
[tex]\mathrm{Factor\:}:5x^2y[/tex]
[tex]5\cdot \:x\cdot \:x\cdot \:y[/tex]
[tex]\mathrm{Factor\:}:10xy^2[/tex]
[tex]2\cdot \:5\cdot \:x\cdot \:y\cdot \:y[/tex]
Common factor:-
[tex]5\cdot \:x\cdot \:y[/tex]
OAmalOHopeO
find the measure of d
forty-six times y is no more than 276.
Answer:
yip that's all
Step-by-step explanation:
not more than 276 means less or equal to 276,
46 × y ≤ 276
y ≤ 6
Answer:
46y<276
Step-by-step explanation:
no more than means less than or equal to.
Place the labels in the chart
If you can draw this out for me or describe were they are that will be very helpful:)
Answer:
Check the image
find the n^th root of z = -2i, n = 6
Answer:
2^(1/6) (cos(-pi/12)+i sin(-pi/12))
2^(1/6) (cos(3pi/12)+i sin(3pi/12))
2^(1/6) (cos(7pi/12)+i sin(7pi/12))
2^(1/6) (cos(11pi/12)+i sin(11pi/12))
2^(1/6) (cos(5pi/4)+i sin(5pi/4))
2^(1/6) (cos(19pi/12)+i sin(19pi/12))
Step-by-step explanation:
Let's convert to polar form.
-2i=2(cos(A)+i sin(A) )
There is no real part so cos(A) has to be zero and since we want -2 and we already have 2 then we need sin(A)=-1 so let's choose A=-pi/2.
So z=2(cos(-pi/2)+i sin(-pi/2)).
There are actually infinitely many ways we can write this polar form which we will need.
z=2(cos(-pi/2+2pi k)+i sin(-pi/2+2pi k))
where k is an integer
Now let's find the 6 6th roots or z.
2^(1/6) (cos(-pi/12+2pi k/6)+i sin(-pi/12+2pi k/6))
Reducing
2^(1/6) (cos(-pi/12+pi k/3)+i sin(-pi/12+pi k/3))
Plug in k=0,1,2,3,4,5 to find the 6 6th roots.
k=0:
2^(1/6) (cos(-pi/12+pi (0)/3)+i sin(-pi/12+pi (0)/3))
=2^(1/6) (cos(-pi/12)+i sin(-pi/12))
k=1:
2^(1/6) (cos(-pi/12+pi/3)+i sin(-pi/12+pi/3))
2^(1/6) (cos(3pi/12)+i sin(3pi/12))
k=2:
2^(1/6) (cos(-pi/12+2pi/3)+i sin(-pi/12+2pi/3))
2^(1/6) (cos(7pi/12)+i sin(7pi/12))
k=3:
2^(1/6) (cos(-pi/12+3pi/3)+i sin(-pi/12+3pi/3))
2^(1/6) (cos(11pi/12)+i sin(11pi/12))
k=4:
2^(1/6) (cos(-pi/12+4pi/3)+i sin(-pi/12+4pi/3))
2^(1/6) (cos(15pi/12)+i sin(15pi/12))
2^(1/6) (cos(5pi/4)+i sin(5pi/4))
k=5:
2^(1/6) (cos(-pi/12+5pi/3)+i sin(-pi/12+5pi/3))
2^(1/6) (cos(19pi/12)+i sin(19pi/12))
Find two power series solutions of the given differential equation about the ordinary point x = 0. Compare the series solutions with the solutions of the differential equation obtained using the method of Section 4.3. Try to explain any differences between the two forms of the solution. y'' − y' = 0 y1 = 1 − x2 2! + x4 4! − x6 6! + and y2 = x − x3 3! + x5 5! − x7 7! + y1 = x and y2 = 1 + x + x2 2! + x3 3! + y1 = 1 + x2 2! + x4 4! + x6 6! + and y2 = x + x3 3! + x5 5! + x7 7! + y1 = 1 + x and y2 = x2 2! + x3 3! + x4 4! + x5 5! + y1 = 1 and y2 = x + x2 2! + x3 3! + x4 4! +
You're looking for a solution in the form
[tex]y(x) = \displaystyle \sum_{n=0}^\infty a_nx^n[/tex]
Differentiating, we get
[tex]y'(x) = \displaystyle \sum_{n=0}^\infty na_nx^{n-1} = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1)a_{n+1}x^n[/tex]
[tex]y''(x) = \displaystyle \sum_{n=0}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=1}^\infty (n+1)na_{n+1}x^{n-1} = \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n[/tex]
Substitute these for y' and y'' in the differential equation:
[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n - \sum_{n=0}^\infty (n+1)a_{n+1}x^n = 0[/tex]
[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1)a_{n+2}-(n+1)a_{n+1}\bigg)x^n = 0[/tex]
Then the coefficients of y are given by the recurrence
[tex]\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\frac{a_{n+1}}{n+2}&\text{for }n\ge0\end{cases}[/tex]
or
[tex]a_n = \dfrac{a_{n-1}}n[/tex]
But we cannot assume that [tex]a_0[/tex] and [tex]a_1[/tex] depend on each other; we can only guarantee that the recurrence holds for n ≥ 1, so that
[tex]a_2=\dfrac{a_1}2 \\\\ a_3=\dfrac{a_2}3=\dfrac{a_1}{3\times2} \\\\ a_4=\dfrac{a_3}4=\dfrac{a_1}{4\times3\times2} \\\\ \vdots \\\\ a_n=\dfrac{a_1}{n!}[/tex]
So in the power series solution, we split off the constant term and we're left with
[tex]y(x) = a_0 + a_1 \displaystyle \sum_{n=1}^\infty \frac{x^n}{n!}[/tex]
so that the fundamental solutions are
[tex]y_1=1[/tex]
and
[tex]y_2=x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\cdots[/tex]
