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4 years ago

Solve each system of equations by substitution. 2x + 2y = 3 x - 4y = -1

Mathematics

1 answer:

Shtirlitz [24]4 years ago

4 0

Step-by-step explanation:

2x + 2y = 3 ( x2 for everything in the equation)

x - 4y = -1

4x+ 4y = 6

x - 4y = -1

Add the 1st line to the 2nd line to get rid of the y:

5x = 5

x - 4y = -1 ( you can pick any line from this system to be the second equation in this step, you can put 2x + 2y = 3 or 4x+ 4y = 6, anything above this step, but to make it simple chose the one you think is the easiest)

Solve for x, then plug the x value into the 2nd equation

x = 1

1 - 4y = -1

x = 1

-4y = -2 =) y = 1/2

x = 1

y= 1/2

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Answer:

It's B,

Step-by-step explanation:

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2 years ago

Lauren Rob need to find a decimal equivalent to 39/50 Laura said she could write an equivalent fraction with 100 as the denomina

Lina20 [59]

Answer:

Laura and Rob are correct

Step-by-step explanation:

we have

39/50

1) Rob said he could divide the numerator by the denominator

using a calculator

39/50=0.78

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2) Laura said she could write an equivalent fraction with 100 as the denominator to converted into a decimal

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(39/50)*(2/2)=78/100=0.78

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3 years ago

Prove that:

Lorico [155]

A.)

$\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)$

Use the identities $\csc x = 1 / \sin x$ and $\tan x = \sin x / \cos x$ on the left-hand side

$\begin{aligned}
\text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\
&= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\
&= \frac{1}{\cos^2 (x)} - 1
\end{aligned}$

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1 - \cos^2 x}{\cos^2 (x)}
\end{aligned}$

Use Pythagorean identity for the numerator.

If $\sin^2 (x) + \cos^2(x) = 1$ then subtracting both sides by $\cos^2 (x)$ yields $\sin^2(x) = 1 - \cos^2(x)$ . We can substitute that into the numerator

$\begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\
&= \frac{\sin^2 (x)}{\cos^2 (x)} \\
&= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\
&= \text{RHS}
\end{aligned}$

======

b.)

$\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1$

For the left-hand side:
By definition, $\sec(x) = 1/\cos(x)$ and $\tan (x) = 1/\cot (x)$

$\begin{aligned}
\text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} \\
&= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} 
\end{aligned}$

Since $\cot (x) = \cos (x) / \sin (x)$

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}$

Using Pythagorean identity, $\cos^2(x) = 1 - \sin^2(x)$ so

$\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}$

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3 years ago

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ArbitrLikvidat [17]

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4 years ago

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Harman [31]

Use the trig ratio which gives you the most information. Here we have opposite and hypotenuse relative to angle x, so think sin( x ) = 14.1 / 27.9<span>Now we want to solve for x, but x is stuck inside the sine function
So how do we do this?
We can 'undo' the sine function by using the inverse sine. apply the inverse Sin to both sides of the equation.
sin^-1 ( sin(x) ) = sin^-1 ( 14.1 / 27.9)
the left side 'cancels' , leaving you with x
x = sin^-1 ( 14.1 / 27.9)</span><span>Now you need a calculator.
Make sure you are in degree mode.
Calculators can give you answer in radian or degree.
If you get a decimal answer, you are probably in radian mode.
Round up if you want nearest hundredth
30.3563 ≈ 30.36
Round up again if you want nearest tenth
30.36 ≈ 30.4
Hope this helps! :)

</span>

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3 years ago