Use the sequence below to complete each task. 34, 25, 16, 7, ... a. Identify the common difference (a). b. Write an equation to represent the sequence. c. Find the 20th term (azo)
Answers
Problem
Solution
We have the following sequence of terms 34,25,16,7,....
Part a
The common difference for this case would be:
25-34= -9
16-25=-9
7-16= -9
Then the answer for part a would be -9
Part b
We want to write the following form:
an = a1 + (n-1) d
For this case d=-9, a1= 34
And then we can write the genral expression like this:
an = 34 + (n-1 ) (-9)
With n = 1,2,3,4....
Part c
In order to find the 20 th term we can replace n =20 and we got:
a20= 34 + (20-1) (-9) = 34-171= -137
Related Questions
Which of the following are solutions to the following solutions to the following solutions?
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We have to find the solutions to the equation:
[tex]|x+4|=8[/tex]The absolute value function is in fact a piecewise function, so it may have two solutions.
We consider for the first solution that the argument inside the absolute function is positive, that is x + 4 > 0. Then, we will have:
[tex]\begin{gathered} x+4=8 \\ x=8-4 \\ x=4 \end{gathered}[/tex]Now, we consider that the the argument is negative and is made positive by the absolute value function (it will shift the sign, which can be represented by a multiplication by -1). This means that x + 4 < 0, and the solution will be:
[tex]\begin{gathered} -(x+4)=8 \\ -x-4=8 \\ -x=8+4 \\ -x=12 \\ x=-12 \end{gathered}[/tex]We can see it in a graph as:
Answer: the solutions are x = 4 and x = -12.
How do I get my answer?
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Answer:
[tex] \frac{2}{9 {d}^{14} } [/tex]
Step-by-step explanation:
[tex] \frac{ {4d}^{ - 5} }{18 {d}^{9} } = \frac{4}{18} \times \frac{ {d}^{ - 5} }{ {d}^{9} } = \frac{2}{9} {d}^{ - 14} = \frac{2}{ {9d}^{14} } [/tex]
5. What is the correlation coefficient for the given data?
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Let us plot the data on the graph to obtain its correlation coefficient.
5) 'r' is known to be the symbol for the correlation coefficient.
Hence, the correlation coefficient from the graph is
[tex]r=0.9741[/tex]6) There is a strong correlation for the data set because the r-value is larger than 0.7 which is close to +1.
Fifteen strips, 11/4" wide, are to be ripped from a sheet of plywood. If 1/8" is lost with each cut, how much of the plywood sheet is used in making the 15 strips? (Assume 15 cuts are necessary.)
Answers
The size of the plywood sheet used is;
[tex]\frac{37}{8}^{\doubleprime}[/tex]Here, we want to get the size of the part of the plywood sheet lost
From the question, we are told that 1/8 inches is lost
The size lost would be;
[tex]\frac{1}{8}\times\text{ 15 = }\frac{15}{8}[/tex]This is the size that was lost
To get the total part of the plywood used, we simply add the width of all the strips to the amount of the plywood lost
We have this as;
[tex]\frac{11}{4}\text{ + }\frac{15}{8}\text{ = }\frac{22+15}{8}\text{ = }\frac{37}{8}[/tex]Yoko plans to watch 2 movies each month. Write an equation to represent the total number of movies n that she will watch in m months.
Answers
Answer:
2m because 2 times the months will tell us how many she has watched for example in 2 months she will watch 4 because 2*2 is 4
A machine worked for 4hours and used 6kilowatts of electricity.What is the rate ofenergy consumed inkilowatts per hour?*Enter your answer as a decimal
Answers
4 hours ---> 6 kilowatts
1 hour -----> x kilowatts
[tex]\begin{gathered} 4\times x=1\times6 \\ 4x=6 \\ \frac{4x}{4}=\frac{6}{4} \\ x=\frac{3}{2}=1.5 \end{gathered}[/tex]answer:
1.5 kilowatts per hour
The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 170 cm2 , what is the length of the diagonal?The length of the diagonal is cm.Give your answer to 2 decimal places.Submit QuestionQuestion 25
Answers
The formula to find the area of a rectangle is:
[tex]\begin{gathered} A=l\cdot w \\ \text{ Where} \\ \text{ A is the area} \\ l\text{ is the length} \\ w\text{ is the width} \end{gathered}[/tex]Since the rectangle area is 170cm², we can write the following equation.
[tex]170=l\cdot w\Rightarrow\text{ Equation 1}[/tex]On the other hand, we know that the width of the rectangle is 6 less than twice its length. Then, we can write another equation.
[tex]\begin{gathered} w=2l-6\Rightarrow\text{ Equation 2} \\ \text{ Because} \\ 2l\Rightarrow\text{ Twice length} \\ 2l-6\Rightarrow\text{ 6 less than twice length} \end{gathered}[/tex]Now, we solve the found system of equations.
[tex]\begin{cases}170=l\cdot w\Rightarrow\text{ Equation 1} \\ w=2l-6\Rightarrow\text{ Equation 2}\end{cases}[/tex]For this, we can use the substitution method.
Step 1: we replace the value of w from Equation 2 into Equation 1. Then, we solve for l.
[tex]\begin{gathered} 170=l(2l-6) \\ \text{Apply the distributive property} \\ 170=l\cdot2l-l\cdot6 \\ 170=2l^2-6l \\ \text{ Subtract 170 from both sides} \\ 0=2l^2-6l-170 \end{gathered}[/tex]We can use the quadratic formula to solve the above equation.
[tex]\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow\text{ Quadratic formula} \\ \text{ For }ax^2+bx+c=0 \end{gathered}[/tex]Then, we have:
[tex]\begin{gathered} a=2 \\ b=-6 \\ c=-170 \\ l=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ l=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(2)(-170)}}{2(2)} \\ l=\frac{6\pm\sqrt[]{1396}}{4} \\ \end{gathered}[/tex]There are two solutions for l.
[tex]\begin{gathered} l_1=\frac{6+\sqrt[]{1396}}{4}\approx10.84 \\ l_2=\frac{6-\sqrt[]{1396}}{4}\approx-7.84 \\ \text{ The symbol }\approx\text{ is read 'approximately'.} \end{gathered}[/tex]Since the value of l can not be negative, the value of l is 10.84.
Step 2: We replace the value of l into any of the equations of the system to find the value of w. For example, in Equation 1.
[tex]\begin{gathered} 170=l\cdot w\Rightarrow\text{ Equation 1} \\ 170=10.84\cdot w \\ \text{ Divide by 10.84 from both sides} \\ \frac{170}{10.84}=\frac{10.84\cdot w}{10.84} \\ 15.68\approx w \end{gathered}[/tex]Now, the long side, the wide side and the diagonal of the rectangle form a right triangle.
Then, we can use the Pythagorean theorem formula to find the length of the diagonal.
[tex]\begin{gathered} a^2+b^2=c^2 \\ \text{ Where} \\ a\text{ and }b\text{ are the legs} \\ c\text{ is the hypotenuse} \end{gathered}[/tex]In this case, we have:
[tex]\begin{gathered} a=10.84 \\ b=15.68 \\ a^2+b^2=c^2 \\ (10.84)^2+(15.68)^2=c^2 \\ 117.51+245.86=c^2 \\ 363.37=c^2 \\ \text{ Apply square root to both sides of the equation} \\ \sqrt[]{363.37}=\sqrt[]{c^2} \\ 19.06=c \end{gathered}[/tex]Therefore, the length of the diagonal of the given rectangle is 19.06 cm rounded to 2 decimal places.
Find the value of each variable.All answers must be in simplest radical form
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Radical
x = √10 • tan 45° = √10• 1 = √10
then
x= √10
y= √x^2 + 10 = √ 10 +10 = √20
Then answer is
x=√10
y= √20
cell phone company A charges a fee of $50 per month plus an additional $0.10 for every minute talked. cell phone company B computes its monthly charge by using the equation y=$0.05 + $75 where y is the total cost and X is the number of minutes talked.
Answers
We will first write A equation
Let x be the number of minutes
y = 0.10x + 50
Comparing the above with y=mx + b where m is the rate of change
m = 0.10
Company B
y = 0.05x + 75
comparing with y =mx + b
rate of change (m) = 0.05
Hence, company A has a higher rate of change at $0.10
what's the answer for proportions 7/9=b/b-10
Answers
Answer:
-35
Step-by-step explanation:
[tex]\frac{7}{9}[/tex] = [tex]\frac{b}{b - 10}[/tex] multiply both sides by 9(b -10)
[tex]\frac{9(b - 10)}{1}[/tex] [tex](\frac{7}{9})[/tex] = [tex]\frac{9(b -10)}{1}[/tex] [tex](\frac{b}{b-10})[/tex] On the right side of the equation, the 9's cancel out and on the right side of the equation the (b -10) cancels out to leave
7(b -10) = 9b Distribute the 7
7b - 70 = 9b Subtract 7b from both sides
-70 = 2b Divide both sides by 2
-35 = b
Number 14. Directions in pic. And also when you graph do the main function in red and the inverse in blue
Answers
Question 14.
Given the function:
[tex]f(x)=-\frac{2}{3}x-4[/tex]Let's find the inverse of the function.
To find the inverse, take the following steps.
Step 1.
Rewrite f(x) for y
[tex]y=-\frac{2}{3}x-4[/tex]Step 2.
Interchange the variables:
[tex]x=-\frac{2}{3}y-4[/tex]Step 3.
Solve for y
Add 4 to both sides:
[tex]\begin{gathered} x+4=-\frac{2}{3}y-4+4 \\ \\ x+4=-\frac{2}{3}y \end{gathered}[/tex]Multply all terms by 3:
[tex]\begin{gathered} 3x+3(4)=-\frac{2}{3}y\ast3 \\ \\ 3x+12=-2y \end{gathered}[/tex]Divide all terms by -2:
[tex]\begin{gathered} -\frac{3}{2}x+\frac{12}{-2}=\frac{-2y}{-2} \\ \\ -\frac{3}{2}x-6=y \\ \\ y=-\frac{3}{2}x-6 \end{gathered}[/tex]Therefore, the inverse of the function is:
[tex]f^{-1}(x)=-\frac{3}{2}x-6[/tex]Let's graph both functions.
To graph each function let's use two points for each.
• Main function:
Find two point usnig the function.
When x = 3:
[tex]\begin{gathered} f(3)=-\frac{2}{3}\ast3-4 \\ \\ f(3)=-2-4 \\ \\ f(3)=-6 \end{gathered}[/tex]When x = 0:
[tex]\begin{gathered} f(0)=-\frac{2}{3}\ast(0)-4 \\ \\ f(-3)=-4 \end{gathered}[/tex]For the main function, we have the points:
(3, -6) and (0, -4)
Inverse function:
When x = 2:
[tex]\begin{gathered} f^{-1}(2)=-\frac{3}{2}\ast(2)-6 \\ \\ f^{-1}(2)=-3-6 \\ \\ f^1(2)=-9 \end{gathered}[/tex]When x = -2:
[tex]\begin{gathered} f^{-1}(-2)=-\frac{3}{2}\ast(-2)-6 \\ \\ f^1(-2)=3-6 \\ \\ f^{-1}(2)=-3 \end{gathered}[/tex]For the inverse function, we have the points:
(2, -9) and (-2, -3)
To graph both functions, we have:
ANSWER:
[tex]\begin{gathered} \text{ Inverse function:} \\ f^{-1}(x)=-\frac{3}{2}x-6 \end{gathered}[/tex]Find the prime factorization of the following number write any repeated factors using exponents
Answers
Notice that 100=10*10, and 10=2*5. 2 and 5 are prime numbers; therefore,
[tex]\begin{gathered} 100=10\cdot10=(2\cdot5)(2\cdot5)=2\cdot2\cdot5\cdot5=2^2\cdot5^2 \\ \Rightarrow100=2^2\cdot5^2 \end{gathered}[/tex]The answer is 100=2^2*5^2
Han and clan are stuffing enveloppes Han can stuff 20 envelopes in one minute and Clare can stuff 10 envelopes in one minute. They start working together on a pile of 1000 envelopes. How long does it take them to finish the pile.
Answers
uff = Given
Han can stuff 20 envelopes in one minute
Clare can stuff 10 envelopes in one minute
Together they start working on a pile off 1000 envelope.
Find
How long does it take them to finish the pile.
Explanation
as we have given
in one minute , Han can stuff = 20 envelope
in one minute , Clare can stuff = 10 envelope
together in one minute , they can stuff =
[tex]\begin{gathered} 20+10=30 \\ \\ \end{gathered}[/tex]we know that the number of time it will take to finish stuffing would be number of envelope / joint rate = 1000/30
so , time taken to finish the pile =
[tex]\begin{gathered} \frac{1000}{30} \\ \\ \frac{100}{3} \\ \\ 33\frac{1}{3} \\ or \\ 33min20sec \end{gathered}[/tex]Final Answer
Hence , the time taken by them to finish the pile is 33 minutes 20 seconds
how do you find the domain in a range of number 2?
Answers
The domain is all the x values included in the function, while the range are all the y values included in the function.
Based on the graph:
Answer:
• Domain:
[tex](-\infty,\text{ }\infty)[/tex]• Range:
[tex](0,\infty)[/tex]8You are asked to draw a triangle withside lengths of 10 inches, 7 inches, and2 inches. How many triangles like thiscan you draw?A. OneB. ThreeC. TwoD. Zero
Answers
ANSWER
D. Zero
EXPLANATION
The triangle inequality states that the sum of any two sides of a triangle is greater than the third side,
All these inequalities must be true to be able to form a triangle with the given sides,
[tex]\begin{gathered} 7+10>2\Rightarrow17>2\Rightarrow true \\ 2+10>7\Rightarrow12>7\Rightarrow true \\ 7+2>10\Rightarrow9>10\Rightarrow false \end{gathered}[/tex]Hence, no triangle can be formed with these side lengths.
The position of an open-water swimmer is shown in the graph. The shortest route to the shoreline is one that is perpendicular to the sh Ay 10 00 6 water 4 shore |(2, 1) swimmer 19 -2 2 1 3 4 5X N -2 An equation that represents the shortest path is y=
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Answer:
Explanation:
From the graph, we ca
Find the length to the nearest whole number of the diagonal (hypotenuse) of a square with 30 cm on a side. Round answers to the nearest tenth if necessary. Your answer
Answers
Notice that we can draw a triangle in the square , and that the length of the square's diagonal is the same as the length of the triangle's hypotenuse. The triangle is a right triangle therefore it satisfies the Phytagorean Theorem. To calculate for it's hypotenuse , we will use:
[tex]c^2=a^2+b^2[/tex]where c is the hypotenuse, and a, b are the other legs of the triangle.
[tex]\begin{gathered} c^2=30^2+30^2 \\ c^2=1800 \\ c=\sqrt[]{1800} \\ c=42.43 \end{gathered}[/tex]Since the hypotenuse of the triangle is 42.43 cm. Therefore, the square's diagonal is also 42.43 cm
Answer:
The square's diagonal is 42.43 cm
Given that the two triangles are similar find the unknowns length of the side labeled in
Answers
The unknown length of the side labeled n is 10.5 units
Explanation:Given:
Two similar triangles with one unknown
To find:
the unknown length of the side labelled n
For two triangles to be similar, the ratio of their corresponding sides will equal
[tex]\begin{gathered} side\text{ with 36 corresponds to side with 27} \\ side\text{ with 14 corresponds to side with n} \\ The\text{ ratio:} \\ \frac{14}{n}\text{ = }\frac{36}{27} \end{gathered}[/tex][tex]\begin{gathered} crossmultiply: \\ 14(27)\text{ = 36\lparen n\rparen} \\ 36n\text{ = 378} \\ \\ divide\text{ both sides by n:} \\ \frac{36n}{36}\text{ = }\frac{378}{36} \\ n\text{ = 10.5} \end{gathered}[/tex]The unknown length of the side labeled n is 10.5 units
Below is the graph of a parabola with its vertex and another point on the parabola labeled.Write an equation of the parabola.(-2,4).(1, -2)
Answers
The vertex form of a parabola is given by:
[tex]x=a(y-k)^2+h[/tex]Where the vertex is:
[tex]\begin{gathered} V(h,k)=(-2,4) \\ so\colon \\ x=a(y-4)^2-2 \\ x=a(y-4)^2-2 \end{gathered}[/tex]for (1,-2):
[tex]\begin{gathered} 1=a(-2-4)^2-2 \\ 1=a(-6)^2-2 \\ 1=36a-2 \\ solve_{\text{ }}for_{\text{ }}a\colon \\ 36a=1+2 \\ 36a=3 \\ a=\frac{3}{36} \\ a=\frac{1}{12} \\ \end{gathered}[/tex]therefore:
[tex]x=\frac{1}{12}(y-4)^2-2[/tex]if the area of polygon A is 72 and Q is a scaled copy and the area of Q is 5 what scale factor got 72 to 5
Answers
A area= 72
Q area =5
So, if we multiply the A area by the square of the scale factor ( since they are areas) we obtain area Q:
72 x^2 = 5
Solving for x:
x^2 = 5/72
x = √(5/72)
x= 0.26
The first 19 terms of the sequence 9, 2, -5, -12,... find the sum of the arithmetic sequence
Answers
To find the sum of the ari
Zachary is designing a new board game, and is trying to figure out allthe possible outcomes. How many different possible outcomes arethere if he spins a spinner with three equal-sized sections labeledWalk, Run, Stop, spins a spinner with four equal-sized sections labeledRed, Green, Blue, Orange, and spins a spinner with 5 equal-sizedsections labeled Monday, Tuesday, Wednesday, Thursday, Friday?
Answers
ANSWER
60 possible outcomes
EXPLANATION
If he spins the 3-section spinner, there are 3 possible outcomes: Walk, Run, Stop.
If he spins the 4-section spinner, there are 4 possible outcomes: red, green, blue, orange.
If he spins the 5-section spinner, there are 5 possible outcomes: Monday, Tuesday, Wednesday, Thursday, Friday.
If he has to spin the three spinners, the total possible outcomes is the product of the possible outcomes of each spinner: 3x4x5 = 60.
Find the tangent of each angle that is not the right angle. Drag and drop the numbers into the boxes to show the tangent of each angle. B 76 tan ZA tan ZB 2.45 0.38 0.93
Answers
From the trignometric ratio of right angle triangle :
The ratio for the tangent of any angle of right angle triangle is the ratio of the side Opposite to that angle to the adjacent side of that angle :
[tex]\tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}}[/tex]In the given triangle :The side opposite to the angle A is BC and the adjacent side AC
So,
[tex]\begin{gathered} \tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}} \\ \tan A=\frac{BC}{AC} \end{gathered}[/tex]In the figure : we have AC = 76, BC = 31 and AB = 82.1
Substitute the value and simplify :
[tex]\begin{gathered} \tan A=\frac{BC}{AC} \\ \tan A=\frac{31}{76} \\ \tan A=0.407 \\ \tan A=0.41 \end{gathered}[/tex]Thus, tan A = 0.41
Now, the side opposite to the angle B is AC and the adjacent side is BC
thus :
[tex]\begin{gathered} \tan \theta=\frac{Opposite\text{ Side}}{Adjacent\text{ Side}} \\ \tan B=\frac{AC}{BC} \end{gathered}[/tex]In the figure : we have AC = 76, BC = 31 and AB = 82.1
Substitute the value and simplify :
[tex]\begin{gathered} \tan B=\frac{AC}{BC} \\ \tan B=\frac{76}{31} \\ \tan B=2.451 \end{gathered}[/tex]tan B = 2.451
Answer :
tanA = 0.41
tanB = 2.45
Determine the value of each limit for the function below.f(x)=x/(x-2)^2(a) lim f(x). (b) lim f(x)x---2^-. x---2^+
Answers
We will have the following:
a)
[tex]\lim _{x\rightarrow2^-}\frac{x}{(x-2)^2}=\infty[/tex]b)
[tex]\lim _{x\rightarrow2^+}\frac{x}{(x-2)^2}=\infty[/tex]An object moves in simple harmonic motion with period 6 seconds and amplitude 4cm. At time =t0 seconds, its displacement d from rest is 0cm, and initially it moves in a negative direction. Give the equation modeling the displacement d as a function of time t.
Answers
The general function for describing the displacement from the mean position in harmonic motion is:
[tex]d(t)=A\cdot\sin (\frac{2\pi}{T}\cdot t+\phi)\text{.}[/tex]Where:
• A is the amplitude,
,• T is the period,
,• φ is initial phase displacement.
From the statement, we know that:
• the amplitude is 4 cm,
,• at time t = 0 its displacement d from the rest is 0 → d(t = 0) = 0,
,• initially, it moves in a negative direction.
s
Analyze the equations in the graphs to find the slope of each equation the y-intercept of each equation in the solution for the system of equations equation 1: y = 50x + 122
Answers
Given:
[tex]y=50x+122\ldots\text{ (1)}[/tex][tex]y=1540-82x\ldots\text{ (2)}[/tex]The general equation is
[tex]y=mx+c[/tex]m is a slope and c is the y-intercept.
From equation (1),
[tex]\text{Slope = 50 and y intercept is 122}[/tex]From equation (2)
[tex]\text{Slope = -82 and yintercept is }1540[/tex]From equation (1) and (2)
Substitute equation (2) in (1)
[tex]1540-82x=50x+122[/tex][tex]50x+82x=1540-122[/tex][tex]132x=1418[/tex][tex]x=\frac{1418}{132}[/tex][tex]x=44[/tex]Substitute in (2)
[tex]undefined[/tex]What are the domain and range of y = cot x? Select onechoice for domain and one for range.
Answers
ANSWER:
A. Domain: x ≠ n
D. Range: All real numbers
STEP-BY-STEP EXPLANATION:
We have the following function:
[tex]y=\cot\left(x\right)[/tex]The domain of a function is the interval of input values, that is, the interval of x while the range is the interval of output values, that is, the interval of y.
In the cotangent function, x cannot take the value of radians (nor its multiples), since it is not defined, while the range is continuous on all real numbers.
That means the correct options are:
A. Domain: x ≠ n
D. Range: All real numbers
Cameron can run 3.6 miles for every 4 miles Juliette runs. If Juliette ran 7.6 miles, how far will Cameron run? 6.84 miles68.4 miles6 miles68 miles
Answers
lets set up a proportion here
cameroon runs 3.6 miles for every 4 miles Juliette runs
3.6 miles(C).................................................... 4 miles (J)
? miles(C)...........................................................7.6 miles (J)
cameron will run= (7.6*3.6)/4=6.84 miles
Cameron will run 6.84 miles
x+y+z=12x+4y+2z = -6-x+9y-3z=-49 Can someone please help me solve this system of equation?
Answers
Let's begin by listing out the information given to us:
[tex]\begin{gathered} x+y+z=1 \\ 2x+4y+2z=-6 \\ -x+9y-3z=-49 \end{gathered}[/tex]To solve this 3 variable equation, let's eliminate one of the variables
add equation 1 & 3, we have:
[tex]\begin{gathered} x-x+y+9y+z-3z=1-49 \\ 10y-2z=-48 \\ Make\text{ z the }subject,we\text{ have:} \\ -2z=-10y-48 \\ divide\text{ through by -2} \\ z=5y+24 \end{gathered}[/tex]Substitute z into equation 1, 2 & 3
[tex]\begin{gathered} x+y+5y+24=1 \\ x+6y=1-24 \\ x+6y=-23 \end{gathered}[/tex][tex]\begin{gathered} 2x+4y+2\left(5y+24\right)=-6 \\ 2x+4y+10y+48=-6 \\ 2x+14y=-6-48 \\ 2x+14y=-54 \end{gathered}[/tex][tex]\begin{gathered} -x+9y-3\left(5y+24\right)=-49 \\ -x+9y-15y-72=-49 \\ -x-6y=-49+72 \\ -x-6y=23 \end{gathered}[/tex]Solve as a simultaneous equation, we have:
[tex]\begin{gathered} x+6y=-23 \\ 2x+14y=-54 \\ \text{Multiply the top equation by 2 \& subtract it from the bottom equation} \\ 2\cdot(x+6y=-23)\Rightarrow2x+12y=-46 \\ 2x+14y=-54-(2x+12y=-46) \\ 2x-2x+14y-12y=-54-(-46) \\ 2y=-8 \\ y=-4 \end{gathered}[/tex]Substitute y = -4 into x + 6y = -23, we have:
[tex]\begin{gathered} x+6\left(-4\right)=-23 \\ x-24=-23 \\ x=-23+24 \\ x=1 \end{gathered}[/tex]Substitute y = -4 into z = 5y + 24, we have:
[tex]\begin{gathered} z=5\left(-4\right)+24 \\ z=-20+24 \\ z=4 \end{gathered}[/tex]Please find the square root. Round your answer to the nearest tenth. [tex] \sqrt{58 } = [/tex]
Answers
Determine the square root of 58.
[tex]\begin{gathered} \sqrt[]{58}=7.615 \\ \approx7.6 \end{gathered}[/tex]So answer is 7.6.