Expand 4(-2y + 3). 2 y + 7 -8 y + 12 2 y + 12 -8 y + 3

Find the distance of point U(2, 3) from point V(5, 7).

aalyn [17]

Answer:

5

Step-by-step explanation:

You can draw a point at (5,3) and call it point A.

Connect point A with V and U to get VA = 4 and UA = 3.

You get a right triangle and use pythagoream theroem. The answer is the root of 4^2 + 3^2 which equals to 5.

The answer is 5

5 0

3 years ago

t hits the square dartboard shown below at a random point. Find the probability that the dart lands in the shaded circular regio

masha68 [24]

Given:

Required:

To find the probability that the dart land will be in the shaded region.

Explanation:

Area of the circle is given by the formula:

$A=\pi r^2$

Where r = radius

Thus the area of the circular region

$\begin{gathered} =(3.14)\times(2)^2 \\ =3.14\times4 \\ =12.56\text{ square in.} \end{gathered}$

The area of the square is given by the formula:

$=(side)^2$

Thus the area of the given square

$\begin{gathered} =(6)^2 \\ =36\text{ square in.} \end{gathered}$

The probability of an event is given by the formula:

$P=\frac{number\text{ of possible outcomes}}{Total\text{ number of outcomes}}$

The probability that the dart land will be in the shaded region

$=\frac{Area\text{ of the shaded region}}{Area\text{ of the square}}$

Thus probability

$\begin{gathered} P=\frac{12.56}{36} \\ P=0.3488 \\ P=0.349 \end{gathered}$

Final answer:

Thus the probability that the dart land will be in the shaded region is 0.349.

6 0

1 year ago

If discriminant (b^2 -4ac>0) how many real solutions

MaRussiya [10]

Answer:

If Discriminant, $b^{2} -4ac >0$

Then it has Two Real Solutions.

Step-by-step explanation:

To Find:

If discriminant (b^2 -4ac>0) how many real solutions

Solution:

Consider a Quadratic Equation in General Form as

$ax^{2} +bx+c=0$

then,

$b^{2} -4ac$ is called as Discriminant.

So,

If Discriminant, $b^{2} -4ac >0$

Then it has Two Real Solutions.

If Discriminant, $b^{2} -4ac < 0$

Then it has Two Imaginary Solutions.

If Discriminant, $b^{2} -4ac=0$

Then it has Two Equal and Real Solutions.

4 0

3 years ago

Anbody know the answer??

LiRa [457]

I'm going to have to go with choice "c". 10x - 1 with the remainder of 5

8 0

3 years ago

1. A sine function has the following key features:

andrew11 [14]

Problem 1

See the attached image (figure 1)

16pi seems like a typo. I'm going to assume that it's a fraction and it is 1/(6pi)
f = 1/(6pi) = frequency
T = 1/f = 1/(1/(6pi)) = 6pi
Amplitude = 2
a = 2
b = 2pi/T = 2pi/(6pi) = 1/3
Midline: y = 3
d = 3

The function is
y = a*sin(bx-c)+d
y = 2*sin(1/3*x-0)+3
y = 2*sin(x/3)+3

===============================================

Problem 2

See the attached image (figure 2)

T = 12 is the period
a = 4 is the amplitude
b = 2pi/T = 2pi/12 = pi/6
y = 1 is the midline so d = 1
The y intercept is (0,1) which is the midline, which indicates no phase shifts have occurred so c = 0

The function is
y = a*sin(bx-c)+d
y = 4*sin((pi/6)x-0)+1
y = 4*sin((pi/6)x)+1

===============================================

Problem 3

See the attached image (figure 3)

Period = 4pi
T = 4pi
b = 2pi/T = 2pi/(4pi) = 1/2 = 0.5
Amplitude = 2
a = 2
Midline: y = 3
d = 3
y-intercept: (0,3)
The function is a reflection of its parent function over the x-axis, so 'a' is negative meaning a = -2 instead of a = 2

The function is
y = a*sin(bx-c)+d
y = -2*sin(0.5x-0)+3
y = -2*sin(0.5x)+3

===============================================

Problem 4

See the attached image (figure 4)

a = 10 which is half of the distance between the highest and lowest points
T = 8 is the period
b = 2pi/T = 2pi/8 = pi/4
c = -pi/2 is the phase shift since its really a cosine graph
d = 0 is the midline

The function is
y = a*sin(bx-c)+d
y = 10*sin((pi/4)*x+(-pi/2))+0
y = 10*sin((pi/4)*x+pi/2)

===============================================

Problem 5

See the attached image (figure 5)

a = 2 is the amplitude since it bobs up and down this distance from the midline
T = 8 seconds is the period (double that of the time it takes for it to go from the highest to the lowest point)
b = 2pi/T = 2pi/8 = pi/4
c = 0 is the phase shift as the buoy starts at normal depth of 20 meters
d = 20 is the midline

The function is
y = a*sin(bx-c)+d
y = 2*sin((pi/4)x-0)+20
y = 2*sin((pi/4)x)+20

===============================================

8 0

3 years ago

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