x = (2, -4)
Explanation:Given:
[tex]\begin{gathered} 3|x+1|-9=0 \\ \\ 3|x+1|=9 \\ \\ |x+1|=3 \\ \\ x+1=3 \\ \Rightarrow x=2 \\ \\ OR \\ -(x+1)=3 \\ \\ -x-1=3 \\ \\ -x=3+1 \\ \\ -x=4 \\ \\ x=-4 \end{gathered}[/tex]x = (2, -4)
Does the following equation have a unique solution, no solution or infinitely manysolutions:3x + 9 = 3x - 9A. Unique SolutionB. No SolutionC. Infinitely Many Solutions
The given equation is:
[tex]3x+9=3x-9[/tex]Solve the equation:
[tex]\begin{gathered} \text{ Subtract }3x\text{ from both sides:} \\ 3x+9-3x=3x-9-3x \\ \Rightarrow9=-9 \end{gathered}[/tex]Notice that the equation results in a contradiction. Hence, the equation has no solution.
The answer is B.
can someone help me with this question explain
Given expression [tex]\frac{2x^3+11x^2-21x}{x^2+3x}[/tex] is equivalent to [tex]2x+5 -\frac{36}{x+3}[/tex].
What do you mean by algebraic expression?
The concept of algebraic expressions is the use of letters or alphabets to represent numbers without providing their precise values. We learned how to express an unknown value using letters like x, y, and z in the fundamentals of algebra. Here, we refer to these letters as variables.
Variables and constants can both be used in an algebraic expression.
There are 3 main types of algebraic expressions which include:
Monomial Expression
Binomial Expression
Polynomial Expression
Given expression:
[tex]\frac{2x^3+11x^2-21x}{x^2+3x}[/tex] for [tex]x[/tex] ≠ -3 or 0.
Using long division method and euclid lemma
On dividing [tex]2x^3+11x^2-21x[/tex] by [tex]x^2+3x[/tex] we get, (given in the snip)
As we know division can be written as
dividend = divisor × quotient + remainder
[tex]2x^3+11x^2-21x = (2x+5)(x^2+3x)-36x[/tex]
⇒ [tex]2x^3+11x^2-21x = 2x+5 -\frac{36x}{x^2+3x}[/tex]
⇒ [tex]2x^3+11x^2-21x = 2x+5 -\frac{36}{x+3}[/tex]
Therefore, given expression [tex]\frac{2x^3+11x^2-21x}{x^2+3x}[/tex] is equivalent to [tex]2x+5 -\frac{36}{x+3}[/tex].
To learn more about the algebraic expression from the given link.
https://brainly.com/question/28036476
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name a situation when a domain could not have a negative values
The domain of a function is the set of numbers that can be used as an input. For every case when we are dealing with real world objects the domain can't have negative numbers. For example: If a function is modeling the revenue of a parking lot as a function of the number of cars that park there in a month, there is no negative car, therefore we can't have negative values as inputs. The worst case scenario would be a situation where no cars visited the parking lot for the whole month, which would be an input of 0.
A movie with an aspect ratio of 1.25:1 is shown as a pillarboxed image on a 36-inch 4:3 television. Calculate the Areas of the TV, the Image and One Blackbar
Explanation
The television has a diagonal that measures 36 inches:
And the ratio is 4:3
[tex]\begin{gathered} \frac{w}{h}=\frac{4}{3} \\ w=\frac{4}{3}h \end{gathered}[/tex]We can use the Pythagorean theorem to find the height of the TV:
[tex]\begin{gathered} 36^2=h^2+w^2 \\ 36^2=h^2+(\frac{4}{3}h)^2 \\ 36^2=h^2(1+\frac{4^2}{3^{2}}) \\ 1296=h^2(1+\frac{16}{9}) \\ 1296=h^2\times\frac{25}{9} \\ h^2=1296\times\frac{9}{25} \\ h=\sqrt[]{1296\times\frac{9}{25}}=21.6 \end{gathered}[/tex]The height of the TV is 60 inches. It's width is:
[tex]w=\frac{4}{3}h=\frac{4}{3}\times21.6=28.8[/tex]w=80 inches
Therefore the area of the TV is
[tex]A_{TV}=w\times h=28.8\times21.6=622.08in^2[/tex]The move has an aspec ratio of 25:1 shown as a pillarboxed image. This means that this is what we see:
So we know that the image height is the same as the TV's, 21.6 inches.
The relation between it's height and it's width is:
[tex]\begin{gathered} \frac{w}{h}=\frac{1.25}{1} \\ w=1.25h \\ \text{if h = 21.6 in} \\ w=27in \end{gathered}[/tex]The area of the image is:
[tex]A_{\text{image}}=w_{\text{image}}\times h=27\times21.6=583.2[/tex]The area of the two blackbars is the difference between the area of the TV and the area of the image:
[tex]A_{2-blackbars}=A_{TV}-A_{image}=622.08-583.2=38.88in^{2}[/tex]Since we need to find the area of just one blackbar, we just have to divide the area of both blackbars by 2:
[tex]A_{1-blackbar}=\frac{A_{2-blackbars}}{2}=\frac{38.88}{2}=19.44in^{2}[/tex]Answer
• Area of the TV: ,622.08 in²
,• Area of the image: ,583.2 in²
,• Area of one blackbar: ,19.44 in²
The price-demand and cost functions for the production of microwaves are given as p= 205 - q/70 and C(q) = 18000 + 20q,where q is the number of microwaves that can be sold at a price of p dollars per unit and C(q) is the total cost (in dollars) of producing q units.(A) Find the marginal cost as a function of q.C'(q)= (B) Find the revenue function in terms of q.R(q) =(C) Find the marginal revenue function in terms of q.R'(q)=
(A)
Find the derivative of C(q):
[tex]\begin{gathered} C^{\prime}(q)=0+20(1) \\ C^{\prime}(q)=20 \end{gathered}[/tex](B)
The revenue function is:
[tex]\begin{gathered} R(q)=q\cdot p \\ so: \\ R(q)=q(205-\frac{q}{70}) \\ R(q)=205q-\frac{q^2}{70} \end{gathered}[/tex](C)
The derivative of R(q) is:
[tex]\begin{gathered} R^{\prime}(q)=205(1)-\frac{1}{70}(2q) \\ so: \\ R^{\prime}(q)=205+\frac{q}{35} \end{gathered}[/tex]Hi are you a tutor for the HESI exam for nursing Maria can walk 3 1/2 miles in one hour. At this time how far can Maria walk in 1/2 hour?
Given that Maria can walk 3 1/2 miles in one hour.
[tex]\text{Speed}=3\text{ }\frac{1}{2}\text{ miles per hour}[/tex][tex]\text{Distance =sp}eed\times time[/tex]The distance that Maria can walk in 1/2 hour is
[tex]\text{Distance =3}\frac{1}{2}\times\frac{1}{2}\text{ miles}[/tex]Multiply the 3 1/2 miles by 1/2 to compute the distance covered in 1/2 hour.
[tex]3\frac{1}{2}\times\frac{1}{2}=\frac{3\times2+1}{2}\times\frac{1}{2}[/tex][tex]=\frac{7}{2}\times\frac{1}{2}=\frac{7}{4}[/tex][tex]=1\frac{3}{4}\text{ miles.}[/tex]Maria can walk 1 3/4 miles in 1/ 2 hour.
write the following in scientific notation:(5 • 10^13) (3 • 10^15)
Solution
Step 1
Obey the multiplication law of indices where
[tex]a^b\times a^{d\text{ }}=a^{b+d}[/tex]So that we will have
[tex]5\times3\times10^{13}\times10^{15}[/tex][tex]\begin{gathered} 15\times10^{13+15} \\ =15\times10^{28} \end{gathered}[/tex]I'm terrible at explaining answers and this is one of them...
We see that the sign is a triangle and the area of a triangle is given by the expression:
[tex]A=\frac{1}{2}\text{base}\times height[/tex]base=36
height=32
Then,
[tex]\begin{gathered} A=\frac{1}{2}36\times32 \\ A=\frac{1152}{2}=576\text{ square inches} \end{gathered}[/tex]in a 45-45-90 triangle, given the hypotenuse 9√2, find the leg of the triangle
Hypothenuse Z= 9√2
then
Z^2 = X^2 + X^2= 2X^2
(9√2)^2 = 2X^2
then
(9√2)/√2= X
cancel √2, then
9 = X
Then now find perimeter
Perimeter P= Z + X + X = 9√2 + 9 + 9 =
P = 9√2 + 18 = 9•(√2 + 2)
Answer is length of triangle= 9√2 + 18
Leg of triangle X= 9
List all zeros for the function f(x) = x^4 - 81. Be sure to include real and complex zeros.
The roots can be found as,
[tex]\begin{gathered} x^4-81=0 \\ (x^2+9)(x^2-9)=0 \\ (x^2+9)(x+3)(x-3)=0 \\ x=\pm3i,3\&-3 \end{gathered}[/tex]Thus, the roots of the equations are 3i,-3i,3 and -3.
The formula used to calculate the value of a savings accounty =(1+)120What does theafter t years is A(t)=0.04= 1500 1+120.04fraction represent?12y=a(1)aeAthe daily interest rateB how long the money has been in the accountCthe monthly interest rateD the starting balance in the account
We have here the formula for Compound Interest:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Where:
• A is the accrued amount.
,• P is the Principal (the original amount of money, the starting amount of money).
,• r is the interest rate.
,• n is the number of times per year compounded.
,• t is the time in years.
When we have that n is equal to 12, we are talking here about that the amount of money is being compounded monthly (we have 12 months in a year, 12 periods, n = 12). Therefore, we are dividing the rate, r, by the number of compoundings per year, n, and this is the rate per each new compounding period of time, r/n, and, in this case, n = 12 (monthly interest rate).
Therefore, in few words, the fraction (0.04/12) is the monthly interest rate (option C).
[If we see the other options, we have:
• The daily interest rate would be given by 0.04/365.
,• How long the money has been in the account is time, t.
,• The starting balance in the account is the Principal, P. ]
If x = 8 units and y = 24 units, then what is the volume of the square pyramid shown above?
In this problem, we want to find the volume of a pyramid. In general, the formula for the volume of a pyramid is
[tex]V=\frac{1}{3}Bh[/tex]where B represents the base shape's area, and h represents the height.
From the image, we can see the base shape is a square, and we can use the formula:
[tex]V=\frac{1}{3}x^2y[/tex]Note: the area of a square is the side-length squared, and since we know the side length is labeled x, we can update the formula as we did above.
We are given x = 8 and y = 24, so we can substitute and simplify to find the volume:
[tex]\begin{gathered} V=\frac{1}{3}(8)^2(24) \\ \\ V=\frac{1}{3}(64)(24) \\ \\ V=512 \end{gathered}[/tex]The final volume is 512 cubic units.
insert the values given in the problem then scale up or down to find the missing value
Given the statements in the image, 2% of the fans implies that 2/100 of the fans in attendance equals to 120 teenagers. i.e,
[tex]\frac{2}{100}\text{ of fans }=120[/tex]Let the total number of fans equal to x. We have the following ratio:
[tex]\frac{2}{100}=\frac{120}{x}[/tex]To get x by scaling up our ratios, we multiply both numerator and denominator by a common factor of 60 to have:
It can be seen that:
[tex]\begin{gathered} 2\times60=120 \\ 100\times60=x \\ 6000=x \\ x=6000 \end{gathered}[/tex]Hence, the total number of people at the stadium is 6000.
Use an explicit formula to find the 10th term of the geometric sequence. 2,8, 32, 128, ...
To find the explicit formula of a geometric sequence you use the next:
[tex]a_n=a_1\cdot r^{n-1}[/tex]a1 is the first term in the sequence
r is the ratio between each pair of terms
2,8,32,128,...
Find r:
[tex]\begin{gathered} \frac{8}{2}=4 \\ \\ \frac{32}{8}=4 \\ \\ \frac{128}{32}=4 \end{gathered}[/tex]Find the explicit formula:
[tex]a_n=2\cdot4^{n-1}[/tex]To find the 10th term you substitute the n in the formula for 10:
[tex]\begin{gathered} a_{10}=2\cdot4^{10-1} \\ \\ a_{10}=2\cdot4^9_{}_{} \\ \\ a_{10}=2\cdot262144 \\ \\ a_{10}=524288 \end{gathered}[/tex]Then, the 10th term is 524,288If Erik checks his pulse for 8 minutes, what is his rate if he counts 600 beats? beats per minute.
Given
Time taken by Erik to check his pulse = 8minutes
Count = 600 beats
To get his rate in beats per minute, you will use the formula:
Beat rate = Count (in beat)/Time taken (in minutes)
Sustitute the given paremeters into the formula given as shown:
Beat rate = 600beats/8minutes
Beat rate = 75beats per minutes
Hence his rate is 75 beats per minute.
form
Kellen earned $52.92 for waitressing at a burger restaurant on Friday night.if she waitressed for 4.5 hours,on average how much did she earn per hour?
Given data:
The amount Kellen earned in 4.5 hours is $52.92.
The expression for the given statement is,
[tex]\begin{gathered} 4.5\text{ hours=\$52.92} \\ 1\text{ hour= \$11.76} \end{gathered}[/tex]Thus, Kellen earned $11.76 per hour.
write an equation in point slope form that passes through (-4,-6) and is parallel to y= -7/2x +6. I added the pic for better information
as the line is parallel to the other line. They have the same slope. So the equation is:
[tex]y+6=-\frac{7}{2}(x+4)[/tex]hi Mr or Ms i need help with this problem please guide me step by step because I don't understand this. the part with the Hj=7x-27 do i bring that down and make an equation? or do i leave that there and make an equation with 3x-5 and x-1?
Let's begin by listing out the information given to us:
HJ = 7x - 27
HI = 3x - 5
IJ = x - 1
The key to solving this is to bear in mind that HJJ = HI + IJ
7x - 27 = 3x - 5 + x - 1
7x - 27 = 3x + x - 5 - 1
7x - 27 = 4x - 6
Subtract 4x from each side, we have:
7x - 4x - 27 = 4x - 4x - 6
3x - 27 = - 6
Add 27 to each side, we have:
3x - 27 + 27 = 27 - 6
3x = 21
Divide each side by 3, we have:
x = 7
The equation and graph of a polynomial are shown below. The graph reaches its maximum when the value of x is 3. What is the y-value of this maximum? y=-x+6x-8
The maximum value of y is
[tex]Y=x^2+6x-8[/tex][tex]y=(3)^2+6(3)-8[/tex][tex]\begin{gathered} y=\text{ 9+18-8} \\ y=19 \end{gathered}[/tex]So, the maximum value of y when x=3 is 19
is the point (1,3) a solution to the linear equation 5x - 9y = 32?
We are given the following equation:
[tex]5x-9y=32[/tex]To determine if the point (1, 3) is a solution to this equation we will replace the given values since the point (1, 3) means that when x = 1, y = 3. Replacing in the equation:
[tex]5(1)-9(3)=32[/tex]Solving the operation we should get the same value on both sides of the equation:
[tex]\begin{gathered} 5-27=32 \\ -22\ne32 \end{gathered}[/tex]Since both sides are different this means that the point (1, 3) is not a solution to the equation.
Answer:
89
Step-by-step explanation:
The product of the polynomials (2ab + b) and (a^2 - b^2) is 2(a^3b)-2a(b^3) + (a^2)b - b^3.If this product is multiplied by (2a + b), the result is a polynomial with_____ terms.
Ok, so
We know that the product of the polynomials (2ab + b) and (a^2 - b^2) is 2(a^3b)-2a(b^3) + (a^2)b - b^3.
Now, we have to multiply the last result per (2a+b)
If me multiply term by term, we get a new polynomial that will has 8 terms.
That's because we have four different terms and we're multiplying each term by 2 different ones. So, there's 8.
Instructions: Determine the word or words that appropriately complete the sentence.
Okay, here we have this:
Considering the provided statement, we are going to identify wich is the correct word, so we obtain the following:
Remember that if two lines intersect, it means that there is a unique point (x, y) that satisfies both equations. According to this we have:
A system of linear equation will have one solution when the equation intersect.
How many years will it take for an initial investment of $10,000 to grow to $25,000? Assume a rate of interest of 3% compounded daily.
The formula for compounded interest is the following:
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]Where A is the final amount, P is the principal, the initial investment, r is the annual rate of interest, n is how many times it is compounded per year and t is the time in years.
So, assuming the given rate of interest is annual, we have:
[tex]\begin{gathered} A=25000 \\ P=10000 \\ r=3\%=0.03 \\ n=365 \\ t=? \end{gathered}[/tex]Where we got n = 365 because it is compounded daily and there are 365 days in an year.
So let's start by solving for t:
[tex]\begin{gathered} A=P(1+\frac{r}{n})^{nt} \\ \frac{A}{P}=(1+\frac{r}{n})^{nt} \\ \log \frac{A}{P}=\log (1+\frac{r}{n})^{nt} \\ \log \frac{A}{P}=nt\log (1+\frac{r}{n}) \\ \frac{\log\frac{A}{P}}{n\log(1+\frac{r}{n})}=t \\ t=\frac{\log\frac{A}{P}}{n\log(1+\frac{r}{n})} \end{gathered}[/tex]Where the log base can be anyone, but it has to be the sme for both log.
Let's calculate the numerator and denominator separately first, using base 10:
[tex]\log _{}\frac{A}{P}=\log \frac{25000}{10000}=_{}\log 2.5=0.39794\ldots[/tex][tex]\begin{gathered} n\log (1+\frac{r}{n})=365\log (1+\frac{0.03}{365})=365\log (1+0.0000821918\ldots)= \\ =365\log (1.0000821918\ldots)=365\cdot0.000035694\ldots=0.013028\ldots \end{gathered}[/tex]Putting them together, we have:
[tex]t=\frac{\log\frac{A}{P}}{n\log(1+\frac{r}{n})}=\frac{0.39794\ldots}{0.013028\ldots.}=30.54\ldots\approx31[/tex]So, it will take between 30 and 31 years, closer to 31 years for it to grow to $25,000.
Construct a circle through pointsX, Y, and Z.
When you need to construct a circle, the major factor to consider is the radius.
The radius is the same distance from any point around the circumference of the circle to the centre. Since the radius is not given, you however need to look for clues.
You start by joining the points to arrive at two lines, for example, join points X and Y and then join points Y and Z.
Next you bisect each of the two lines one after the other (bisect along the perpendicular)
You will observe that both perpendicular bisectors would touch at a point. That point where they touch or "cross each other" is the center of your circle.
Next you place the sharp tip of your compass on the center of your circle, adjust its distance to the pencil end (that is your radius) and as soon as it touches one of the three points, you draw your circle.
Mason was practicing free throws at basketball practice he made 5 throws every 2 he missed
Mason made 3 correct throws as every second he missed
describe the formations between f(x) = x-5 to g(x)=-6x+2
The given function is,
f(x) = x- 5
The transferred equation is,
g(x) = -6x + 2
So the transformation is,
[tex]g(x)=-6(f(x))-28[/tex]find the order pairs by following the tablegiven:y=x^2 -12x+36table of x : ?,5,9,4y : 0,?,1,?,?
x = ? , 5 , 9 , 4
y= 0, ? , 1 , ?
To find the missing x value, replace the matching value of y (0) in the equation and solve for x:
0 = x^2-12x +36
Apply the quadratic formula
[tex]\frac{-b\pm\sqrt[]{b^2-4\cdot A\cdot c}}{2\cdot a}=\frac{12\pm\sqrt[]{(-12)^2-4\cdot1\cdot36}}{2\cdot1}[/tex][tex]\frac{12\pm\sqrt[]{144-144}}{2}=\frac{12}{2}=6[/tex]For x = 5:
y= (5)^2-12 (5) +36 = 25-60+36 = 1
For y=1
1 =x^2-12x+36
0 = x^2-12x+36-1
0= x^2-12x+35
[tex]\frac{12\pm\sqrt[]{(12)^2-4\cdot1\cdot35}}{2\cdot1}=\frac{12\pm\sqrt[]{144-140}}{2}=\frac{12\pm2}{2}=\frac{14}{2}=7\text{ }[/tex]x =7
For x=9
y= (9)^2-12 (9)+36 = 81-108+36=9
For x=4
y= (4)^2-12(4)+35 = 16-48+36=4
a diver stands on a platform 15ft above a lake. he doesn't dive off the platform and lands in the water below. his height (H) above the lake after X seconds is shown on the graph below. what is the reasonable domain for the scenario?
The reasonable domain is when the time starts at 0 seconds and when the height is equal to 0 meters. Then, the domain is
[tex]0\le x\le3[/tex]which corresponds to the first option
In the diagram below of circle A. diameter MP = 26. m_GAI = 30° and radiGA and Al are drawn.MАP30°G1if MG IP, find the area of the sector MAG in terms of me and approximateto the nearest hundredth.The area of the sector in terms of u isTTThe area of the sector rounded to the nearest hundredth isunitssquared.
To find the area of a sector of a circle in terms of π having the angle in degrees you use the next formula:
[tex]A=\frac{\theta}{360}\cdot\pi\cdot r^2[/tex]r is the radius
To find area of sector MAG:
1. Find the angle of the sector MAG.
The semicircle has an angle of 180° and it is divided into 3 sectors MAG, GAI, and IAP.
As the arcs MG and IP are congruents (have the same measure) the angles of the sectors MAG and IAP are also congruent.
[tex]\begin{gathered} m\angle\text{MAG}+m\angle\text{GAI}+m\angle\text{IAP}=180 \\ \\ m\angle MAG=m\angle IAP \\ m\angle GAI=30 \\ \\ 2m\angle MAG+m\angle GAI=180 \\ 2m\angle MAG+30=180 \end{gathered}[/tex]Use the equation above to find the measure of angle MAG:
[tex]\begin{gathered} 2m\angle MAG=180-30_{} \\ 2m\angle MAG=150 \\ m\angle MAG=\frac{150}{2} \\ \\ m\angle MAG=75 \end{gathered}[/tex]2. Find the area of sector MAG:
Angle 75°
radius= half of the diameter (26/2 = 13)
r=13
[tex]\begin{gathered} A=\frac{75}{360}\cdot\pi\cdot(13)^2 \\ \\ A=\frac{75}{360}\cdot\pi\cdot169 \\ \\ A=\frac{12675}{360}\pi \\ \\ A=\frac{845}{24}\pi \\ \\ A\approx35.21\pi \\ \\ A\approx110.61 \end{gathered}[/tex]The exact area of the sector MAG is 845/24 π units squared.
Rounded to the nearest hundredth 35.21 π units squared or 110.61 units squared
e = radians. Identify the terminal point and tan e.O A. Terminal point: (33) tan = 13B. Terminal point: (1, 1); tan 6 = 73(1,1)tan 0 = 2C. Terminal point:; tane3D. Terminal point:
The correct answer is Option D
This following are the steps to take:
Step1: Convert the angle from radians to degrees
[tex]\begin{gathered} 1\pi radians=180^o \\ \text{Thus }\frac{\pi}{6}\text{ radians = }\frac{180^o}{6} \\ \text{ }\frac{\pi}{6}\text{ radians =}30^o \end{gathered}[/tex]Step 2: Draw a unit circle (with a radius of 1 unit), and show the line which forms angle 30 degrees with the x -axis
Step 3: Compute the values of the terminal points:
[tex]\begin{gathered} Th\text{e x-coordinate of the terminal point = 1 }\times cos30^0\text{ = }\frac{\sqrt[]{3}}{2} \\ Th\text{e y-coordinate of the terminal point = 1 }\times\sin 30^0\text{ = }\frac{1}{2} \\ \text{Thus the coordinates of ther terminal point = }(x,y)\text{ = (}\frac{\sqrt[]{3}}{2},\text{ }\frac{1}{2}\text{)} \end{gathered}[/tex]Step 4: Compute the values of the tangent of the angle:
[tex]\begin{gathered} \tan 30^0\text{ = }\frac{y}{x}=\frac{\frac{1}{2}}{\frac{\sqrt[]{3}}{2}}=\frac{1}{\sqrt[]{3}}\text{ } \\ \\ \tan 30^o=\frac{1}{\sqrt[]{3}}\text{ }\times\frac{\sqrt[]{3}}{\sqrt[]{3}}\text{ =}\frac{\sqrt[]{3}}{3} \\ \\ \tan 30^{o\text{ }}=\text{ }\frac{\sqrt[]{3}}{3} \end{gathered}[/tex]