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2 years ago

(-2,-3), y=2x-1 write the equation in slope intercecpt form

Mathematics

1 answer:

Damm [24]2 years ago

7 0

The slope intercept form y=2x^2-x-13. Hope this helps :)

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Write out the sample space for the given experiment. Use the following letters to indicate each choice: M for mushrooms, A for a

MrMuchimi

Answer: {(MBV), (MBH) , (MEV) , (MEH) , (OBV), (OBH) , (OEV) , (OEH)}

Step-by-step explanation:

Sample space is a set that contains all the possible outcomes for an event.

For salad:

Choices for extra toppings (mushrooms, olives) =2

Choices for meats (bacon, eggs) =2

Choices for dressings (vinaigrette, honey mustard.)= 2

By Fundamental counting principal .

The number of different salads can be made= (Choices for extra toppings) x (Choices for meats ) x (Choices for dressings)

= 2 x 2 x 2 = 8

All the possible outcomes = (MBV), (MBH) , (MEV) , (MEH) , (OBV), (OBH) , (OEV) , (OEH)

Required Sample space would be : {(MBV), (MBH) , (MEV) , (MEH) , (OBV), (OBH) , (OEV) , (OEH)}

5 0

1 year ago

In J, K is the midpoint. The coordinates of ) are (2, 2), and the coordinates of Kare (10, 11). What are the coordinates of L?

DiKsa [7]

Answer:

L(18, 20)

Step-by-step explanation:

In JL, K is the midpoint. The coordinates of J are (2, 2), and the

coordinates of K are (10, 11). What are the coordinates of L?

Solution:

If O(x, y) is the midpoint between two points A( $x_1,y_1$ ) and B( $x_2,y_2$ ). The equation to determine the location of O is given by:

$x=\frac{x_1+x_2}{2} \\\\y=\frac{y_1+y_2}{2}$

Since JL is a line segment and K is the midpoint. Given the location of J as (2, 2) and K as (10, 11). Let ( $x_2,y_2$ ) be the coordinate of L. Therefore:

$10=\frac{2+x_2}{2} \\\\20=2+x_2\\\\x_2=18$

$11=\frac{2+y_2}{2} \\\\22=2+y_2\\\\y_2=20$

Therefore L = (18, 20)

4 0

2 years ago

Find the other endpoint of the line segment with the given endpoint and midpoint. Endpoint: (-2, -6) Midpoint: (-5, 1)

FromTheMoon [43]

Answer:

(-8,8)

Step-by-step explanation:

The midpoint $M(x_M,y_M)$ of the segment AB with endpoints $A(x_A,y_A)$ and $B(x_B,y_B)$ has coordinates

$x_M=\dfrac{x_A+x_B}{2}\\ \\y_M=\dfrac{y_A+y_B}{2}$

In your case, $A(-2,-6), \ M(-5,1),$ then

$-5=\dfrac{-2+x_B}{2}\Rightarrow -2+x_B=-10,\ x_B=-10+2=-8\\ \\1=\dfrac{-6+y_B}{2}\Rightarrow -6+y_B=2,\ y_B=2+6=8$

3 0

2 years ago

Find the missing denominator or numerator of 1/2 = /4

Jet001 [13]

Simply cross-multiply about,

$\bf \cfrac{1}{2}=\cfrac{x}{4}\implies 4\cdot 1=2\cdot x\implies 4=2x\implies \cfrac{4}{2}=x\implies 2=x$

6 0

2 years ago

Read 2 more answers

1) Determine the discriminant of the 2nd degree equation below:

Aleksandr-060686 [28]

$\LARGE{ \boxed{ \mathbb{ \color{purple}{SOLUTION:}}}}$

We have, Discriminant formula for finding roots:

$\large{ \boxed{ \rm{x = \frac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} }}}$

Here,

x is the root of the equation.
a is the coefficient of x^2
b is the coefficient of x
c is the constant term

1) Given,

3x^2 - 2x - 1

Finding the discriminant,

➝ D = b^2 - 4ac

➝ D = (-2)^2 - 4 × 3 × (-1)

➝ D = 4 - (-12)

➝ D = 4 + 12

➝ D = 16

2) Solving by using Bhaskar formula,

❒ p(x) = x^2 + 5x + 6 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5\pm \sqrt{( - 5) {}^{2} - 4 \times 1 \times 6 }} {2 \times 1}}}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm \sqrt{25 - 24} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 5 \pm 1}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 2 \: or - 3}}}$

❒ p(x) = x^2 + 2x + 1 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{ {2}^{2} - 4 \times 1 \times 1} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm \sqrt{4 - 4} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - 2 \pm 0}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = - 1 \: or \: - 1}}}$

❒ p(x) = x^2 - x - 20 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 1) \pm \sqrt{( - 1) {}^{2} - 4 \times 1 \times ( - 20) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{ 1 \pm \sqrt{1 + 80} }{2} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{1 \pm 9}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 5 \: or \: - 4}}}$

❒ p(x) = x^2 - 3x - 4 = 0

$\large{ \rm{ \longrightarrow \: x = \dfrac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) } }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm \sqrt{9 + 16} }{2 \times 1} }}$

$\large{ \rm{ \longrightarrow \: x = \dfrac{3 \pm 5}{2} }}$

So here,

$\large{\boxed{ \rm{ \longrightarrow \: x = 4 \: or \: - 1}}}$

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5 0

2 years ago

Read 2 more answers