Having trbl doing equations.

Contains the point (-1, 2) and is parallel to<br> x – 2y = -3

ivanzaharov [21]

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange x - 2y = - 3 into this form

Subtract x from both sides

- 2y = - x - 3 ( divide all terms by - 2 )

y = $\frac{1}{2}$ x + $\frac{3}{2}$ ← in slope- intercept form

with m = $\frac{1}{2}$

• Parallel lines have equal slopes, thus

y = $\frac{1}{2}$ x + c ← is the partial equation

To find c substitute (- 1, 2) into the partial equation

2 = - $\frac{1}{2}$ + c ⇒ c = 2 + $\frac{1}{2}$ = $\frac{5}{2}$

y = $\frac{1}{2}$ x + $\frac{5}{2}$ ← in slope- intercept form

Multiply through by 2

2y = x + 5 ( subtract 2y from both sides )

0 = x - 2y + 5 ( subtract 5 from both sides )

- 5 = x - 2y, thus

x - 2y = - 5 ← in standard form

4 0

3 years ago

Create a list of steps, in order, that will solve the following equation. (x-5)^2=25(x−5) 2 =25left parenthesis, x, minus, 5, ri

Gekata [30.6K]

The value of x in the equation (x - 5)² = 25 is 10.

<h3>How to solve an equation?</h3>

(x - 5)² = 25

Take the square root of both sides

Therefore,

√(x - 5)² = √25

x - 5 = 5

add 5 to both sides

x - 5 + 5 = 5 + 5

x = 10

learn more on equation here: brainly.com/question/10506642

#SPJ1

4 0

1 year ago

What is the radius of the following circle?

lara [203]

Answer:

The radius is: $2\sqrt{3}$

Step-by-step explanation:

The equation of a circle in center-radius form is:

$(x - h)^2 + (y - k)^2 = r^2$

Where the center is at the point (h, k) and the radius is "r".

So, given the equation of the circle:

$x^2+y^2=12$

You can identify that:

$r^2=12$

Then, solving for "r", you get that the radius of this circle is:

$r=\sqrt{12}\\\\r=2\sqrt{3}$

7 0

2 years ago

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Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

5 0

3 years ago

What is 3/4 + 2/3 ???

erica [24]

To add fractions, first convert them to have equal denominators.

3/4= 9/12 (multiply both sides by 3)
2/3= 8/12 (multiply both sides by 4)

From here, add the numerators while keeping the denominator.
9/12+ 8/12= 17/12

Final answer: 17/12 or 1 5/12

3 0

2 years ago

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