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3 years ago

Finding solutions to linear systems

Mathematics

2 answers:

vladimir2022 [97]3 years ago

8 0

Khan academy has very good videos on how to solve them

Inessa05 [86]3 years ago

6 0

To find the solution to systems of linear equations, you can any of the methods below:

1. Solve By Graphing.
2. Solve By Elimination.
3. Solve By Substitution.
4. Solve With Meta Calculator.I
5. interactive System of Linear Equations.

Hope this helps.

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Write the equation of the line that is perpendicular to y=2x-7 and passes through the point (6, 5)

Pani-rosa [81]

Answer:

Step-by-step explanation:

we are given a equation of a line

we want to figure out the equation of the perpendicular line passes through the (6,5) points

in order to do so

recall that,

$\displaystyle m_{ \text{perpendicular}} = - \frac{1}{m}$

we got from our given equation that m=2

because equation of a line is y=mx+b

thus

$\displaystyle m_{ \text{perpendicular}} = - \frac{1}{2}$

remember that, when we want to figure out perpendicular line or parallel line we should the formula given by

$\displaystyle y - y_{1} = m(x - x_{1})$

since we got our perpendicular m is -½, $x_1=6$ and $y_1=5$ , substitute

$\displaystyle y - 5 = - \frac{1}{2} (x - 6)$

to get the perpendicular equation you should simplify the above equation to y=mx+b form

distribute -½:

$\displaystyle y - 5 = - \frac{1}{2} x + 3$

add 5 to both sides:

$\displaystyle y = - \frac{1}{2} x + 8$

hence,

our answer choice is B

3 0

3 years ago

Read 2 more answers

Here's another one :)

jekas [21]

Area Of Triangle:

4/2 = 2
2x6 = 12

12x2 = 24ft^2

Remaining area:

150 - 24 = 126ft^2

Maximum length:

126/6 = 21

x = 21ft

8 0

3 years ago

Perimeter of 316 and 67 m

storchak [24]

Answer:

383 m

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Select the correct answer from each drop down menu.

KATRIN_1 [288]

Answer: x^2, x<2, 5, 2</=x<4

Step-by-step explanation:

7 0

3 years ago

If u(x)=-2x²+3 and v(x)=1/x, what is the range of (u ° v)(x)?

Ludmilka [50]

$u(x)=-2x^2+3\\\\v(x)=\dfrac{1}{x}\\\\(u\ \circ\ v)(x)=-2\left(\dfrac{1}{x}\right)^2+3=-2\left(\dfrac{1}{x^2}\right)+3=-\dfrac{1}{x^2}+3\\\\\text{The range of}\ y=\dfrac{1}{x^2}\ \text{is all positive real numbers.}\\\\\text{The range of}\ y=-\dfrac{1}{x^2}\ \text{is all negative real numbers.}\\\\\text{The range of }\ (u\ \circ\ v)(x)=-\dfrac{1}{x^2}+3\ is\ (-\infty,\ 3)$

6 0

3 years ago

Read 2 more answers