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3 years ago

HELP ME!!! Create a matrix for this linear system: What is the solution of the system?

Mathematics

2 answers:

Aleksandr-060686 [28]3 years ago

5 0

The answer should be A

nydimaria [60]3 years ago

3 0

Answer:

Step-by-step explanation:

I attached the solution

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Answer the following questions.

schepotkina [342]

<h3>Answer the following questions :-</h3>

Raja is junk seller. He sells many things. Here is given the list of junk.

Check the Raja's rate list and solve the following questions.

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \small\boxed{\begin{array}{c|c} \bf{Kind \: of\:junk} & \bf{Price \: of \: 1 \: kg } \\ \dfrac{\qquad\qquad\qquad}{ \sf News \: Paper} &\dfrac{\qquad \qquad\qquad}{ \sf Rs.5} & \\ \dfrac{\qquad\qquad\qquad}{ \sf Iron} &\dfrac{\qquad\qquad\qquad}{ \sf Rs.10} & \\ \dfrac{\qquad\qquad\qquad}{ \sf Brass} &\dfrac{\qquad\qquad\qquad}{ \sf Rs.50} & \\ \dfrac{\qquad\qquad\qquad}{ \sf Plastic} &\dfrac{\qquad\qquad\qquad}{ \sf Rs.8}&\end{array}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

<h2>Solutions :-</h2>

1. How much money will you pay to Raju for 20 kg of newspapers?

$\quad{\longrightarrow{\sf{Cost \: of \: 1 \: kg \: newspaper = Rs.5}}}$

$\quad{\longrightarrow{\sf{Cost \: of \: 20 \: kg \: newspaper = Rs.20 \times 5}}}$

$\quad{\longrightarrow{\sf{Cost \: of \: 20 \: kg \: newspaper = Rs.100}}}$

$\quad{\longrightarrow{\underline{\boxed{\sf{\purple{Cost \: of \: 20 \: kg \: newspaper = Rs.100}}}}}}$

∴ We need to pay Rs.100 for buy 20 kg of newspapers. 

$\rule{200}{1.5}$

2. What is the cost of 5 kg Iron?

$\quad{\longrightarrow{\sf{Cost\: of \: 1 \: kg \: iron = Rs.10}}}$

$\quad{\longrightarrow{\sf{Cost\: of \: 5 \: kg \: iron = Rs.10 \times 5}}}$

$\quad{\longrightarrow{\sf{Cost \: of \: 5 \: kg \: iron = Rs.50}}}$

$\quad{\longrightarrow\underline{\boxed{\sf{\purple{Cost \: of \: 5 \: kg \: iron = Rs.50}}}}}$

∴ The cost of 5 kg iron is Rs.50.

$\rule{200}{1.5}$

3. What is the cost of 9 kg Plastic?

$\quad{\longrightarrow{\sf{Cost \: of \: 1 \: kg \: plastic = Rs.8}}}$

$\quad{\longrightarrow{\sf{Cost \: of \: 9 \: kg \: plastic = Rs.8 \times 9}}}$

$\quad{\longrightarrow{\sf{Cost \: of \: 9 \: kg \: plastic = Rs.72}}}$

$\quad{\longrightarrow\underline{\boxed{\sf{\purple{Cost\: of \: 9 \: kg \: plastic = Rs.72}}}}}$

∴ The cost of 9 kg plastic is Rs.72. 

$\rule{200}{1.5}$

4. What is the cost of 7 kg Brass?

$\quad{\longrightarrow{\sf{Cost \: of \: 1 \: kg \: brass = Rs.50}}}$

$\quad{\longrightarrow{\sf{Cost\: of \: 7 \: kg \: brass = Rs.50 \times 7}}}$

$\quad{\longrightarrow{\sf{Cost \: of \: 7 \: kg \: brass = Rs.350}}}$

$\quad{\longrightarrow\underline{\boxed{\sf{\purple{Cost \: of \: 5 \: kg \: brass = Rs.350}}}}}$

∴ The cost of 7kg brass is Rs.350.

$\underline{\rule{200pt}{2pt}}$

7 0

2 years ago

Determine if the two triangles are congruent. If they are, state how you know.

Studentka2010 [4]

The answers here would be:
16) c
17) a

3 0

2 years ago

The tangents to the curve with equation y = x^3-3x at the points A and B with x coordinates −1 and 4 respectively meet at the po

sladkih [1.3K]

Answer:

$\displaystyle C=\left(\frac{26}{9}, 2\right)$

Step-by-step explanation:

We need to find the equation of the tangent lines of Points A and B.

Differentiate the equation:

$\displaystyle \frac{dy}{dx}=3x^2-3$

Point A has an x-coordinate of -1. Hence, the slope of its tangent line is:

$\displaystyle \frac{dy}{dx}\Big|_{x=-1}=3(-1)^2-3=0$

Find the y-coordinate of Point A using the original equation:

$y(-1)=(-1)^3-3(-1)=2$

Hence, Point A is (-1, 2).

Thus, the tangent line at Point A is:

$y-2=0(x-(-1))$

Simplify:

$y=2$

Point B has an x-coordinate of 4. Hence, the slope of its tangent line is:

$\displaystyle \frac{dy}{dx}\Big|_{x=4}=3(4)^2-3=45$

Find the y-coordinate of Point B:

$y(4)=(4)^3-3(4)=52$

Thus, Point B is at (4, 52).

So, the tangent line at Point B is:

$y-52=45(x-4)$

Simplify:

$y=45x-128$

Point C occurs at the intersections of the tangent lines of Points A and B. Set the two equations equal to each other and solve for x:

$2=45x-128$

We acquire:

$\displaystyle x=\frac{26}{9}$

Since one of our equations is y = 2, the y-coordinate is 2.

Hence, Point C is:

$\displaystyle C=\left(\frac{26}{9}, 2\right)$

5 0

3 years ago

Read 2 more answers

Write the following decimal number in its equivalent fraction form. Show all work for full credit.

Zigmanuir [339]

Let x=0.8888...

x=0.8888 Start with the given equation

10x=8.8888 Multiply both sides by 10. The goal is to move the decimal point to the next repeating portion (which is just one spot to the right)

Now subtract the equation x=0.8888 from 10x=8.8888 to get
10x-x=8.8888-0.8888

9x=8 Combine like terms. Notice how the decimal portions cancel out.

x=8/9 Divide both sides by 9 to isolate x

So the answer is x=8/9 which means that 8/9=0.8888 (where the 8's repeat)

8 0

3 years ago

Which lists the ratios in order from least to greatest 2:9,1:6,7:36

mezya [45]

Answer:

1:6, 7:36, 2:9

Step-by-step explanation:

2 : 9 → 8 : 36

1 : 6 → 6 : 36

7 : 36

Least → Greatest

1:6, 7:36, 2:9

4 0

3 years ago