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Triss [41]
3 years ago
9

HELP AGAIN PLEASE 20 POINTS What methods can be used to solve a system of equations and when would you use each? Be sure to incl

ude systems with nonlinear equations.​
Mathematics
1 answer:
Yuliya22 [10]3 years ago
7 0

Answer:

The three ways to solve a system of equations are substitution, elimination, and graphing. You can use these methods to solve any system of equations, it just depends on which method you use.

Step-by-step explanation:

In the substitution method, you get one variable isolated on one side of an equation, then use that equation to substitute the variable with what it's equal to into the second equation to solve the equation.

In the elimination method, you get all the variables in your equations together one side, and the constant on the other. Then, multiply or divide one or two equations to get one variable with the same coefficient for both equations (be sure to multiply or divide both sides to each equation), and add or subtract the equations together to get rid of that variable (make sure to add the left sides of the equations together and the right sides of the equations together, no the opposite!) and solve the equation.

In the graphing method, you solve for y in each equation, graph both equations on a coordinate plane, and find the point of intersection for both lines to solve the equation (For example, if the point of intersection is (2, 7), the answer to your equation would be X = 2 and Y = 7).

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How can you prove that csc^2(θ)tan^2(θ)-1=tan^2(θ)
Oxana [17]

Answer:

Make use of the fact that as long as \sin(\theta) \ne 0 and \cos(\theta) \ne 0:

\displaystyle \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.

\displaystyle \csc(\theta) = \frac{1}{\sin(\theta)}.

\sin^{2}(\theta) + \cos^{2}(\theta) = 1.

Step-by-step explanation:

Assume that \sin(\theta) \ne 0 and \cos(\theta) \ne 0.

Make use of the fact that \tan(\theta) = (\sin(\theta)) / (\cos(\theta)) and \csc(\theta) = (1) / (\sin(\theta)) to rewrite the given expression as a combination of \sin(\theta) and \cos(\theta).

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \left(\frac{1}{\sin(\theta)}\right)^{2} \, \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} - 1 \\ =\; & \frac{\sin^{2}(\theta)}{\sin^{2}(\theta)\, \cos^{2}(\theta)} - 1\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1\end{aligned}.

Since \cos(\theta) \ne 0:

\displaystyle 1 = \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)}.

Substitute this equality into the expression:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots\\ =\; & \frac{1}{\cos^{2}(\theta)} - 1 \\ =\; & \frac{1}{\cos^{2}(\theta)} - \frac{\cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}.

By the Pythagorean identity, \sin^{2}(\theta) + \cos^{2}(\theta) = 1. Rearrange this identity to obtain:

\sin^{2}(\theta) = 1 - \cos^{2}(\theta).

Substitute this equality into the expression:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{1 - \cos^{2}(\theta)}{\cos^{2}(\theta)} \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\end{aligned}.

Again, make use of the fact that \tan(\theta) = (\sin(\theta)) / (\cos(\theta)) to obtain the desired result:

\begin{aligned}& \csc^{2}(\theta) \, \tan^{2}(\theta) - 1\\ =\; & \cdots \\ =\; & \frac{\sin^{2}(\theta)}{\cos^{2}(\theta)}\\ =\; & \left(\frac{\sin(\theta)}{\cos(\theta)}\right)^{2} \\ =\; & \tan^{2}(\theta)\end{aligned}.

5 0
2 years ago
To which sets of numbers does 1 1/2 belong to
baherus [9]

Answer:

A rational number because it's negative it can't be natural and cuz it's fraction it can't be whole nor integers.

<em>p.s. thanks to salma1148 for the answer</em>

7 0
3 years ago
Solve 8/t+5 = t-3/t+5 + 1/3
Aleks04 [339]

\frac{8}{t}  + 5 = t -  \frac{3}{t}  + 5 +  \frac{1}{3}  \\  \\ 1. \:  \frac{8}{t} =  -  \frac{3}{t} +  \frac{1}{3}   \\  \\ 2. \: 24 = 3t^{2}  - 9 + t \\  \\ 3. \: 24 - 3t^{2} + 9 - t = 0 \\  \\ 4. \: 33 - 3t ^{2}- t = 0 \\  \\ 5. \: t =  \frac{1 +  \sqrt{397} }{ - 6}  \:  \frac{1 -  \sqrt{397} }{ - 6}  \\  \\ 6. \: t =  -  \frac{1 +  \sqrt{397} }{6}  \:  -  \frac{1 -  \sqrt{397} }{6}

7 0
4 years ago
Read 2 more answers
future value of 10% savings from earnings of 36000 earns 6.25% annual interest compounded quarterly for 15 years​
hoa [83]

without further ado

10% of 36000 is simply 3600, we chopped a "0" off of it hmmm, ok so

~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$3600\\ r=rate\to 6.25\%\to \frac{6.25}{100}\dotfill &0.0625\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{quarterly, thus four} \end{array}\dotfill &4\\ t=years\dotfill &15 \end{cases} \\\\\\ A=3600\left(1+\frac{0.0625}{4}\right)^{4\cdot 15}\implies A=3600(1.015625)^{60}\implies A\approx 9126.53

7 0
3 years ago
Betsy, a recent​ retiree, requires ​$5000 per year in extra income. She has ​$60000 to invest and can invest in​ B-rated bonds p
solniwko [45]

Betsy needs to  invest $32,000 in the B-rated bonds paying 13% interest and $28,000 in the certificate of deposit paying 3% interest.

Interest is calculated by multiplying the rate with the principal.

Let x = the amount Betsy will invest in the B-rated bonds paying 13% per annum. The remaining amount, $60,000-x , she will invest in the certificate of deposit paying 3% per annum.

We can express the earnings on these two investments, after converting the percentages to their decimal equivalents, as: 0.13x+0.03(60000-x) and this is to earn a total of $5,000

Now we can write the necessary equation to solve for x.

or, 0.13x+0.03(60000-x) = 5000

or, 0.13x+1800-0.03x = 5000

or, 0.1x+1800 = 5000

or, 0.1x = 3200

or, x = 32000

So Betsy needs to invest $32,000 in the B-rated bonds paying 13% interest and the remainder ($60,000-$32,000 = $28,000) in the certificate of deposit paying 3% interest to obtain $5,000 in interest per annum.

To learn more about interest:

brainly.com/question/13324776

#SPJ9

3 0
1 year ago
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