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3 years ago

What is the solution to the system of equations?

Mathematics

2 answers:

mixer [17]3 years ago

8 0

The answer is (1,11). You can solve for x first and plugged it in to find y.

Sveta_85 [38]3 years ago

7 0

Answer:

(1, 11)

Step-by-step explanation:

If you plug both of these values in for the x and y in both equations you will get a true statement.

10(1) + 3(11) = 43

10 + 33 = 43

-9(1) - (11) = -20

-9 - 11 = -20

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The diagram shows the width and height of a table. Mr. Howell wants to build a table with the same ratio of width and height as

just olya [345]

Based on the diagram and the ratio of width and height in it, the width that Mr. Howell should make his table is 48 inches .

<h3>How wide should Mr. Howell's table be?</h3>

The width of the table in the diagram is 12 cm and the height is 6cm.

This means that the ratio of width to height is:

12 : 6

2 : 1

As Mr. Howell wants his table to be 24 inches high, the width of the table would be:

2 : 1

x : 24

Cross-multiply to get:

x = 48 inches

Find out more on ratios at brainly.com/question/20594266

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2 years ago

Ling drove at speed of 45 Miles per hour for 225 miles. How many hours did she drive?

oksian1 [2.3K]

225 miles / 45 miles/hr = 5hr

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3 years ago

Give an answer to the picture above or below

Ostrovityanka [42]

Answer:

Graph

Step-by-step explanation:

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3 years ago

True or false? a triangle can be circumscribed about a given circle

Sphinxa [80]

Answer:\

The answer to you is:

True

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3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago