1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lubasha [3.4K]
3 years ago
9

Which matrix can be used to solve the system?

Mathematics
1 answer:
Brut [27]3 years ago
4 0
It's the last choice.  In the top row you have the 1's that are the coefficients on the x and the y, and the solution for that equation is 21; in the second row you have the 5 from the x and the 4 from the y and the solution for that equation which is 20.  I'm more used to writing them down instead of across, but there you have it anyways
You might be interested in
Equation 1: m = 8 + 2n
slava [35]

Answer:

1:×4

m×4=(8+2n)×4

4m=32+8n

now,

1:×4 - 2:

4m-4m = 32+8n -4+4n

0=(32-4) +(8n-4n)

0=28+4n

28=-4n

n= -7

n=-7. 1:,

m=8+2n

m=8+2×(-7)

m=8-14

m=-6

7 0
2 years ago
Help please:Suppose that x and y inversely, and x=10 when y=8 write the function that models the inverse variation
jarptica [38.1K]

SOLUTION

From the question, we are told that y varies inversely as x

This is mathematically written as

\begin{gathered} y\propto\frac{1}{x} \\ \propto\text{ is a sign of proportionality } \end{gathered}

Now, we will remove the proportionality sign and replace it with equal to sign =

If we do this, we will intoduce a constant k

\begin{gathered} y\propto\frac{1}{x} \\ y=k\times\frac{1}{x} \\ y=\frac{k}{x} \end{gathered}

So we have the formula

y=\frac{k}{x}

We will substitute the values of x for 10 and y for 8 into the formula to get k, we have

\begin{gathered} y=\frac{k}{x} \\ 8=\frac{k}{10} \\ k=8\times10 \\ k=80 \end{gathered}

Now, we will substitute k for 80 back into the formula to get the inverse function, we have

\begin{gathered} y=\frac{k}{x} \\ y=\frac{80}{x} \end{gathered}

Hence the answer is option C

7 0
1 year ago
I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be
Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: g'(-1)\,(1) since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: g'(-1)= 7 as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1

6 0
3 years ago
the pitch, or frequency, of a vibrating string varies directly with the square root of the tension. if a string vibrates at a fr
Naily [24]
\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------\\\\
\begin{array}{llll}
\stackrel{frequency}{f}\textit{ of a vibrating string varies directly}\\
\qquad \qquad \textit{with the square root of the tension }\stackrel{tension}{t}
\end{array}

\bf f=k\sqrt{t}\quad \textit{we also know that }
\begin{cases}
f=300\\
t=8
\end{cases}\implies 300=k\sqrt{8}
\\\\\\
\cfrac{300}{\sqrt{8}}=k\implies \cfrac{300}{\sqrt{2^2\cdot 2}}=k\implies \cfrac{300}{2\sqrt{2}}=k\implies \cfrac{150}{\sqrt{2}}=k
\\\\\\
\textit{and we can \underline{rationalize} it to }\cfrac{150\sqrt{2}}{2}\implies 75\sqrt{2}=k

\bf thus\qquad f=\stackrel{k}{75\sqrt{2}}\sqrt{t}\implies \boxed{f=75\sqrt{2t}}\\\\
-------------------------------\\\\
\textit{now, when t = 72, what is \underline{f}?}\qquad f=75\sqrt{2(72)}
5 0
3 years ago
Read 2 more answers
What is the difference?<br><br> 52 – 13 =<br><br> 29<br> 39<br> 41<br> 65
mylen [45]

Answer:

52-13 = 39

Step-by-step explanation:

This is simple, next time use a calculator and calculate yourself

6 0
3 years ago
Read 2 more answers
Other questions:
  • How do i find points that divide a line segment into three parts
    10·1 answer
  • I need someone to help me solve this ASAP!!!
    6·1 answer
  • 1. Five more than twice a number is 7.
    7·2 answers
  • -3 M squared + x 4 M cubed N squared over 2 m squared n old squared x 3 m n where m equals 4 + n equals -3
    15·1 answer
  • What is the answer to 7x+3 (4x-8)=109
    14·1 answer
  • Simplify. Multiply and remove all perfect squares from inside the square roots. Assume z is positive. z* 30z2* 35z3
    11·1 answer
  • Mrs. Barlow's math class took a test yesterday. Out of 80 students, 70% of them passed. How many of Mrs. Barlow's students passe
    6·1 answer
  • Will give 100 points and Brainliest! <br> Take a factor out of the sqaure root: √3y^2, where y&lt;0
    6·1 answer
  • 1. Solve the following system of equations by elimination. What is the value of x?
    12·2 answers
  • Expand the brackets and simplify: 2(5x-y)-3(3x-y)​
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!