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3 years ago

Find the slope of the line that passes through the points (3,0) and (3,-2).

2 answers:

8 0

When given the points (x1,y1) and (x2,y2)

slope=(y2-y1)/(x2-x1)

we have
(3,0) and (3,-2)
(x,y)
x1=3
y1=0
x2=3
y2=-2

slope=(-2-0)/(3-3)=-2/0=undefined since you caon't divide by zero

answer is D

valkas [14]3 years ago

4 0

$\frac{-2-0}{3-3}= \frac{-2}{0}=$ undefined

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!PLEASE HELP!

charle [14.2K]

$\bold{ANSWER:}$
C) 5.4

$\bold{SOLUTION:}$

8 0

2 years ago

Subtract the following polynomials, then place the answer in the proper location on the grid.

melamori03 [73]

When we add or subtract polynomials, the addition or subtraction is applied to the coefficients of the terms of equal powers.

For the given polynomials, we are required to do subtraction as follows:
(-4.1 x^2 + 0.9 x - 9.8) - (1.7 x^2 - 2.4 x - 1.6)
= -4.1 x^2 + 0.9 x - 9.8 - 1.7 x^2 + 2.4 x + 1.6
= (-4.1-1.7) x^2 + (0.9+2.4) x + (-9.8+1.6)
= -5.8 x^2 + 3.3 x - 8.2

The final expression (solution) is:
-5.8 x^2 + 3.3 x - 8.2

8 0

3 years ago

Help me with this please!!

valina [46]

<h2><u>Given:</u><u>-</u></h2>

Points C = (-7,2) → $\sf{(X_1,Y_1)}$
D = (3,12) → $\sf{(X_2,Y_2)}$

The Midpoint of CD.

<h2><u>Required</u><u> </u><u>Response</u><u>:</u><u>-</u></h2>

Let,

Midpoint of CD be (x,y).

WKT,

$\boxed{\sf{(x,y) = \frac{X_1+X_2}{2},\frac{Y_1+Y_2}{2}}}$

$→\;{\sf{\frac{-7+3}{2},\frac{2+12}{2}}}$

$→\;{\sf{\frac{-4}{2},\frac{14}{2}}}$

$→\;{\sf{-2,7}}$

The Midpoint of CD ◕➜ $\Large{\red{\mathfrak{(-2,7)}}}$

Let,

The centre be O

Radius = CO & OD

Here, C = (-7,2) → $\sf{(X_1,Y_1)}$

O = (-2,7) → $\sf{(X_2,Y_2)}$

$\boxed{\sf{Distance = \sqrt{(X_2-X_1)²+(Y_2-Y_1)²}}}$

$→\;{\sf{\sqrt{(-2+7)²+(7-2)²}}}$

$→\;{\sf{\sqrt{5²+5²}}}$

$→\;{\sf{\sqrt{25+25}}}$

$→\;{\sf{\sqrt{50}}}$

$→\;{\sf{5√2 (or) 7.07}}$

Radius of Circle ◕➜ $\Large{\red{\mathfrak{7.07}}}$

<h2>Option D.</h2>

Hope It Helps You ✌️

3 0

3 years ago

EXAMPLE 5 Evaluate the iterated integral 5 0 5 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are

Dimas [21]

$\displaystyle\int_0^5\int_x^5\sin(y^2)\,\mathrm dy\,\mathrm dx$

The integration region is the triangle in the $x,y$ plane bounded by the lines $y=5$ , $y=x$ , and $x=0$ . Reversing the order of integration, this is equal to

$\displaystyle\int_0^5\int_0^y\sin(y^2)\,\mathrm dx\,\mathrm dy=\int_0^5y\sin(y^2)\,\mathrm dy$

For the remaining integral, let $u=y^2\implies\mathrm du=2y\,\mathrm dy$ :

$\displaystyle\int_0^5y\sin(y^2)\,\mathrm dy=\frac12\int_0^{25}\sin u\,\mathrm du=-\frac12\cos u\bigg|_0^{25}=\frac{1-\cos(25)}2$

6 0

3 years ago

Write an equation of the parabola in vertex form with the coordinates (-2,6) and (-1,3)

Sergio [31]

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4 0

3 years ago