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3 years ago

What is the equation of a line, in general form, that passes though points (-1,2) and (5,2)?

Mathematics

1 answer:

Tom [10]3 years ago

3 0

Answer:

Step-by-step explanation:

To write the equation of the line. first calculate the slope using the slope formula.

$m = \frac{y_2-y_1}{x_2-x_1} = \frac{2-2}{5--1}= \frac{0}{6} = 0$

Since the slope of this line is 0, it is horizontal and has the form y=b where b is the y-coordinate. So y = 2 is the equation. In general form, it would be y-2 = 0

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Evaluate each expression if a=4, b=6 and c=3<br><br> 5b-6c

iragen [17]

Answer:

it is 12

Step-by-step explanation:

4 0

3 years ago

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Which graph represents y=1/2x+2 <br><br> please help!!

wel

Answer:

Step-by-step explanation:

Let’s first look at the y-intercept. Since the equation is in y = mx + b form, we know that m is the slope and b is the y-intercept, so the slope is 1/2 and the y-intercept is 2, or (0,2). We can narrow down our search by noticing that C and D don’t intersect (0,2). That leaves A and B.

The slope is calculated by (y_1 - y_2)/(x_1 - x_2) given points (x_1, y_1) and (x_2, y_2). The slope for A is (4-0)/(2-(-2)) = 4/4 = 1 and the slope for B is (2-0)/(0-(-4)) = 2/4 = 1/2. The slope of B matches the slope that we are looking for. So, the answer is B.

8 0

2 years ago

F(n+1)=f(n)-8 if f(1)=100 what is f(6)

Svet_ta [14]

$f(n+1)=f(n)-8\\\\ f(2)=100-8=92\\ f(3)=92-8=84\\ f(4)=84-8=76\\ f(5)=76-8=68\\ f(6)=68-8=60$

8 0

2 years ago

What is the product of -6x3y5 and -4y2? A. -24x3y7 B. -24x3y10 C. 24x3y7 D. 24x3y10

Leto [7]

Answer:

Step-by-step explanation:

So we have the product of:

$(-6x^3y^5)\cdot(-4y^2)$

Multiply. Combine like terms:

$=(-6\cdot -4)(x^3)(y^5\cdot y^2)$

The first term is 24.

The second term stays as it is.

The third term will equate to y^7. Recall how to multiply exponents. The base stays constant and you add the exponents. Thus:

$=24x^3y^7$

Our answer is C.

And we're done!

5 0

2 years ago

(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

2 years ago