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3 years ago

Solving systems by Substitution Homework y=x+4 y=x-3

1 answer:

8 0

There is no solution for this equation since they have the same slope, x/1. The only difference between these two is that their y intercepts are different meaning that they will be parallel lines that will never intersect among one other. For example, think of it as two separate lines that are have the same slope and never gain more distance/units from one another.

Solve:

To solve, you have to get one of this equations into a Ax+By=C equation form, standard equation. Let’s change y=x+4 into a standard equation.

We have to get x and y together and 4 as C.
So let’s subtract x from both sides;

y=x+4
-x -x
————————
-x+y=4

This is a standard equation.

Now let’s substitute.

take the standard equation and plug in y which is x+4 since there is a equation meaning it’s y=x+4

-x+(x+4)=4

Let’s simplify this mess.

-x+x equals 0. So we are left with 4=4.

Subtract 4 from both sides and we get 0=0

This means there is no solution. Hoped this helped.

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

3 years ago

Urgent please help..

Alexxx [7]

Answer:

x ≈ 13.5

Step-by-step explanation:

Using the sine rule in the triangle, then

$\frac{sin34}{x}$ = $\frac{sin27}{11}$ ( cross- multiply )

x × sin27° = 11 × sin34° ( divide both sides by sin27° )

x = $\frac{11sin34}{sin27}$ ≈ 13.5 ( to the nearest tenth )

5 0

3 years ago

Solving systems by Substitution Homework y=x+4 y=x-3​

Solving systems by Substitution Homework y=x+4 y=x-3