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Nookie1986 [14]

2 years ago

5

Finding Ordered Pairs from an Equation For the equation below, x is the input and y is the output. Choose the ordered pair that

makes a true statement for the equation.

1 answer:

Burka [1]2 years ago

4 0

Answer: The answer is dddddd

Step-by-step explanation:

fbfgzgsdgrhzdfgreyty

You might be interested in

Evaluate the question

mylen [45]

Answer:

A) 5/6

B) -67/30 (or) -2 7/30

A) -9/2 (or) -4 1/2

Step-by-step explanation:

A) 2/3(5)-5/2=3 1/3-5/2 = 10/3-5/2 = 20/6-15/6= 5/6

B) 2/3(2/5)-5/2= 4/15-5/2 = 8/30-75/30 = -67/30

A) 2/3(-3)-5/2=-2-5/2= -4 1/2

7 0

2 years ago

A 2.2 kg ball strikes a wall with a velocity of 7.4 m/s to the left. The ball bounces off with a velocity of 6.2 m/s to the righ

Naya [18.7K]

Answer:

The constant force exerted on the ball by the wall is 119.68 N.

Step-by-step explanation:

Consider the provided information.

It is given that the mass of the ball is m = 2.2 kg

The initial velocity of the ball towards left is 7.4 m/s

So the momentum of the ball when it strikes is = $2.2\times 7.4=16.28$

The final velocity of the ball is -6.2 m/s

So the momentum of the ball when it strikes back is = $2.2\times -6.2=-13.64$

Thus change in moment is: $16.28-(-13.64)=29.92$

The duration of force exerted on the ball t = 0.25 s

Therefore, the constant force exerted on the ball by the wall is:

$\frac{29.92}{0.25}=119.68$

Hence, the constant force exerted on the ball by the wall is 119.68 N.

6 0

3 years ago

Q3: Identify the graph of the equation and write and equation of the translated or rotated graph in general form. (Picture Provi

natta225 [31]

Answer:

b. circle; $2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

Step-by-step explanation:

The given conic has equation;

$x^2-5x+y^2=3$

We complete the square to obtain;

$(x-\frac{5}{2})^2+(y-0)^2=\frac{37}{4}$

This is a circle with center;

$(\frac{5}{2},0)$

This implies that;

$x=\frac{5}{2},y=0$

When the circle is rotated through an angle of $\theta=\frac{\pi}{3}$ ,

The new center is obtained using;

$x'=x\cos(\theta)+y\sin(\theta)$ and $y'=-x\sin(\theta)+y\cos(\theta)$

We plug in the given angle with x and y values to get;

$x'=(\frac{5}{2})\cos(\frac{\pi}{3})+(0)\sin(\frac{\pi}{3})$ and $y'=--(\frac{5}{2})\sin(\frac{\pi}{3})+(0)\cos(\frac{\pi}{3})$

This gives us;

$x'=\frac{5}{4} ,y'=\frac{5\sqrt{3} }{4}$

The equation of the rotated circle is;

$(x'-\frac{5}{4})^2+(y'-\frac{5\sqrt{3} }{4})^2=\frac{37}{4}$

Expand;

$(x')^2+(y')^2-\frac{5\sqrt{3} }{2}y'-\frac{5}{2}x'+\frac{25}{4} =\frac{37}{4}$

Multiply through by 4; to get

$4(x')^2+4(y')^2-10\sqrt{3}y'-10x'+25 =37$

Write in general form;

$4(x')^2+4(y')^2-10x'-10\sqrt{3}y'-12 =0$

Divide through by 2.

$2(x')^2+2(y')^2-5x'-5\sqrt{3}y'-6 =0$

8 0

3 years ago

The box plots show attendance at a school dance competition and school badminton games.

Helen [10]

Answer:

The IQR is a better measure of spread for dance competitions than for badminton games.

Step-by-step explanation:

I understand from "Dance Competition box plot has interquartile ranges approximately between 60 and 130" that quartile 1 is at 60 and quartile 3 is at 130, then, the interquartile range (IQR) is 130 - 60 = 70.

Similarly, from Badminton box plot, quartile 1 is at 90 and quartile 3 is at 150, then, the interquartile range is 150 - 90 = 60.

IQR is a measure of statistical dispersion, that is, as IQR increase, spread increase.

4 0

2 years ago

Read 2 more answers

Help. plz. i really need it

Cloud [144]

I think is b not sure

5 0

2 years ago