Find the value of y.
​

Answer: The 1-kg ball has 1/10 as much kinetic energy as the 10-kg ball.

Step-by-step explanation:

Option (a) - "less momentum" - accurately describes the relationship between the momentum of the 1 kg ball compared to the 10 kg ball.

The momentum of an object is directly proportional to its mass. When comparing a 10 kg ball to a 1 kg ball traveling at the same speed, the 10 kg ball has more momentum due to its greater mass. Since momentum is determined by both mass and velocity, and the velocity is the same for both balls, the difference in momentum is solely due to the difference in mass. The 10 kg ball has ten times the mass of the 1 kg ball. Therefore, the 1 kg ball has less momentum compared to the 10 kg ball. In fact, it has one-tenth of the momentum of the 10 kg ball.

Complete question: A 10 kg ball is traveling at the same speed as a 1 kg ball. Compared with the 10 kg ball, the 1 kg ball has

a-less momentum.

b-twice momentum.

c-five times the momentum.

d-the same momentum.

Answer:

6/11

Step-by-step explanation:

3 white circles, 3 black circles, so six circles out of the total 11 peices

Answer:

Probably of getting both = 3/10 *100 = 30%

Step-by-step explanation:

According to the given statement we have four possible combinations:

Blue pants/white shirt, tan pants/white shirt, blue pants/blue shirt, blue pants/white shirt

.probability of getting tan pants = 2/4

probability of getting white shirt = 3/5

probably of getting both = 2/4 x 3/5 = 6/20 = 3/10

probably of getting both = 3/10 *100 = 30% ...

Answer:

4) Reflect over the y-axis, reflect over the x-axis, rotate 180°

Step-by-step explanation:

Since, the rule of 180° rotation,

[tex](x,y)\rightarrow (-x,-y)[/tex]

Rule of reflection over x-axis,

[tex](x,y)\rightarrow (x,-y)[/tex]

Rule of reflection over y-axis,

[tex](x,y)\rightarrow (-x,y)[/tex]

Rule of reflection over line y = x,

[tex](x,y)\rightarrow (y,x)[/tex]

∵ In the set of reflection,

Rotate 180°, reflect over the x-axis, reflect over the line y=x,

[tex](x,y)\rightarrow (-x,-y)\rightarrow (-x,y)\rightarrow (y,-x)[/tex]

Reflect over the x-axis, rotate 180°, reflect over the x-axis,

[tex](x,y)\rightarrow (x,-y)\rightarrow (-x,y)\rightarrow (-x,-y)[/tex]

Rotate 180°, reflect over the y-axis, reflect over the line y=x,

[tex](x,y)\rightarrow (-x,-y)\rightarrow (x,-y)\rightarrow (-y,x)[/tex]

Reflect over the y-axis, reflect over the x-axis, rotate 180°,

[tex](x,y)\rightarrow (-x,y)\rightarrow (-x,-y)\rightarrow (x,y)[/tex]

Hence, the set of reflections and rotations that would carry rectangle ABCD onto itself is,

Reflect over the y-axis, reflect over the x-axis, rotate 180°.

Option '4' is correct.

Answer:

Reflect over the y-axis, reflect over the x‒axis, rotate 180°

Step-by-step explanation:

[tex]\tan^3\theta-\dfrac1{\tan\theta-1}-\sec^2\theta+1=0[/tex]

Recall that [tex]\tan^2\theta+1=\sec^2\theta[/tex], so we can write everything in terms of [tex]\tan\theta[/tex]:

[tex]\tan^3\theta-\dfrac1{\tan\theta-1}-\tan^2\theta=0[/tex]

Let [tex]x=\tan\theta[/tex], so that

[tex]x^3-\dfrac1{x-1}-x^2=0[/tex]

With some rewriting we get

[tex]x^3-x^2-\dfrac1{x-1}=0[/tex]

[tex]x^2(x-1)-\dfrac1{x-1}=0[/tex]

[tex]\dfrac{x^2(x-1)^2-1}{x-1}=0[/tex]

Clearly we cannot have [tex]x=1[/tex], or [tex]\tan\theta=1[/tex].

The numerator determines when the expression on the left reduces to 0:

[tex]x^2(x-1)^2-1=0[/tex]

[tex]x^2(x-1)^2=1[/tex]

[tex]\sqrt{x^2(x-1)^2}=\sqrt1[/tex]

[tex]|x(x-1)|=1[/tex]

[tex]x(x-1)=1\text{ or }x(x-1)=-1[/tex]

Completing the square gives

[tex]x(x-1)=x^2-x=x^2-x+\dfrac14-\dfrac14=\left(x-\dfrac12\right)^2-\dfrac14[/tex]

so that

[tex]\left(x-\dfrac12\right)^2=\dfrac54\text{ or }\left(x-\dfrac12\right)^2=-\dfrac34[/tex]

The second equation gives no real-valued solutions because squaring any real number gives a positive real number. (I'm assuming we don't care about complex solutions.) So we're left with only

[tex]\left(x-\dfrac12\right)^2=\dfrac54[/tex]

[tex]\sqrt{\left(x-\dfrac12\right)^2}=\sqrt{\dfrac54}[/tex]

[tex]\left|x-\dfrac12\right|=\dfrac{\sqrt5}2[/tex]

which again gives two cases,

[tex]x-\dfrac12=\dfrac{\sqrt5}2\text{ or }x-\dfrac12=-\dfrac{\sqrt5}2[/tex]

[tex]x=\dfrac{1+\sqrt5}2\text{ or }x=\dfrac{1-\sqrt5}2[/tex]

Then when [tex]x=\tan\theta[/tex], we can find [tex]\cot\theta[/tex] by taking the reciprocal, so we get

[tex]\boxed{\cot\theta=\dfrac2{1+\sqrt5}\text{ or }\cot\theta=\dfrac2{1-\sqrt5}}[/tex]

Answer: A. -1

Step-by-step explanation:

Edg 2020

Answer:

(x - 3)(x - 12)

Step-by-step explanation:

Given

x² - 15x + 36

Consider the factors of the constant term (+ 36) which sum to give the coefficient of the x- term (- 15)

The factors are - 3 and - 12, since

- 3 × - 12 = + 36 and - 3 - 12 = - 15, hence

x² - 15x + 36 = (x - 3)(x - 12) ← in factored form

The factored form of the polynomial x² − 15x + 36 is (x − 3)(x − 12). This can be obtained by using the steps of splitting the middle term method.

Given the polynomial, x² − 15x + 36

(-3x)×(-12x)=36x² and (-3x)+(-12x)= - 15x

-3x and -12x are the required factors of 36x²

x² - 3x - 12x+36

x² - 3x - 12x+36 = x(x - 3) - 12(x - 3)

                          =(x - 3)(x - 12)

Hence the factored form of the polynomial x² − 15x + 36 is (x − 3)(x − 12).

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Answer: [tex]\bold{x^4+6x^3-12x-72}[/tex]

Step-by-step explanation:

[tex]f(x) = \sqrt{x^2+12x+36} \implies f(x)=\sqrt{(x+6)^2}\implies f(x) = x+6\\\\\\f(x)\cdot g(x)=(x+6)(x^3-12)\\.\qquad \qquad =x(x^3-12)+6(x^3-12)\\.\qquad \qquad =x^4-12x +6x^3-72\\.\qquad \qquad =\large\boxed{x^4+6x^3-12x-72}[/tex]

The expression equal to f(x)·g(x) is  [tex]x^{4} + 6x^{3} - 12x - 72[/tex] .

The expressions given are f(x) = [tex]\sqrt{x^{2}+12x+36}[/tex]  and g(x) = [tex]x^{3} - 12[/tex] .

Thus we have

⇒ f(X) = [tex]\sqrt{x^{2}+12x+36}[/tex]

⇒ f(x) = [tex]\sqrt{(x+6)^{2} }[/tex]

⇒ f(x) = x + 6 .

Therefore the expression of multiplication of f(x)·g(x) is =

= [tex](x^{3} - 12)*(x + 6)[/tex]

= [tex]x^{4} + 6x^{3} - 12x - 72[/tex]

Thus, the expression equal to f(x)·g(x) is  [tex]x^{4} + 6x^{3} - 12x - 72[/tex] .

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Answer:

5

Step-by-step explanation:

[tex]A_{s} = l^2[/tex], where l - is a side

[tex]l = \sqrt{A_s}[/tex]

[tex]l = \sqrt{25} = 5[/tex]

Answer:

[tex]\large\boxed{y=3x-11}[/tex]

Step-by-step explanation:

[tex]\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\===================================\\\\y=3x-4\to m_1=3\\\\\text{Therefore}\ m_2=3.\\\\\text{We have the equation:}\ y=3x+b.\\\\\text{Put the coorsdinates of the point (2, -5) to the equation of the line}\\\\-5=3(2)+b\\-5=6+b\qquad\text{subtract 6 from both sides}\\-11=b\to b=-11[/tex]

Answer:

[tex]\large\boxed{y=(x-3)^2-3}[/tex]

Step-by-step explanation:

The vertex form of a quadratic equation y = ax² + bx + c:

y = a(x - h)² + k

(h, k) - coordinates of a vertex

We have the equation y = x² - 6x + 6.

Convert to the vertex form:

[tex]y=x^2-2(x)(3)+6\\\\y=\underbrace{x^2-2(x)(3)+3^2}_{(*)}-3^2+6\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2\qquad(*)\\\\y=(x-3)^2-9+6\\\\y=(x-3)^2-3\to h=3,\ k=-3[/tex]

The given system of equation can be solved if x = 2/3 and y = -4/3.

So we need to find another system which can be solved by these values.

From the given 4 options lets find out which can be solved by x = 3/2 and y = -4/3

1.

y = -2y^2 -2 , x = y - 2

lets put the value of y here

y = -2(16/9)-2 = -32/9-2= -50/ 9

So not an equivalent equation.

2.

y = -2y^2 + 2 , x = y + 2

y = -2(16/9)+2=-32/9+2=-14/9

So not an equivalent system.

3.

y = -2y^2 - 8y - 8 , x = y + 2

y = -2(16/9)-8(-4/3)-8 = -32/9+32/3-8=64/9-8=-8/9

So not an equivalent system.

4.

y = -2y^2 + 8y - 8 , x = y -2

y = -2(16/9)+8(-4/3)-8=-32/9-32/3-8=-128/9-8=-22.22

So this is not an equivalent equation too

Answer:

C. [tex]\left \{ {{y=-2y^{2}-8y-8 } \atop {x=y+2}} \right.[/tex]

Step-by-step explanation:

Edge 2020

Answer:

2 and 5

Step-by-step explanation:

The slope-intercept form of a line is y=mx-b where m is slope and b is y-intercept.

The point-slope form of a line is y-y1=m(x-x1) where m is the slope and (x1,y1) is a point on the line.

The standard form a line is ax+by=c.

So anyways parallel lines have the same slope.

So if we are looking for a line parallel to 3x-4y=7 then we need to know the slope of this line so we can find the slope of our parallel line.

3x-4y=7

Goal: Put into slope-intercept form

3x-4y=7

Subtract 3x on both sides:

  -4y=-3x+7

Divide both sides by -4:

   [tex]y=\frac{-3}{-4}x+\frac{7}{-4}[/tex]

Simplify:

   [tex]y=\frac{3}{4}x+\frac{-7}{4}[/tex]

So the slope of this line is 3/4.  So our line that is parallel to this one will have this same slope.

So we know our line should be in the form of [tex]y=\frac{3}{4}x+b[/tex].

To find b we will use the point that is suppose to be on our new line here which is (x,y)=(-4,-2).

So plugging this in to solve for b now:

[tex]-2=\frac{3}{4}(-4)+b[/tex]

[tex]-2=-3+b[/tex]

[tex]3-2=b[/tex]

[tex]b=1[/tex]

so the equation of our line in slope-intercept form is [tex]y=\frac{3}{4}x+1[/tex]

So that isn't option 1 because the slope is different.  That was the only option that was in slope-intercept form.

The standard form of a line is ax+by=c and we have 2 options that look like that.

So let's rearrange the line that we just found into that form.

[tex]y=\frac{3}{4}x+1[/tex]

Clear the fractions because we only want integer coefficients by multiplying both sides by 4.

This gives us:

[tex]4y=3x+4[/tex]

Subtract 3x on both sides:

[tex]-3x+4y=4[/tex]

I don't see this option either.

Multiply both sides by -1:

[tex]3x-4y=-4[/tex]

I do see this as a option. So far the only option that works is 2.

Let's look at point slope form now.

We had the point that our line went through was (x1,y1)=(-4,-2) and the slope,m, was 3/4 (we found this earlier).

y-y1=m(x-x1)

Plug in like so:

y-(-2)=3/4(x-(-4))

y+2=3/4 (x+4)

So option 5 looks good too.

Answer:

OPTION 2.

OPTION 5.

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

[tex]y=mx+b[/tex]

Where "m" is the slope and "b" is the y-intercept.

Given the line [tex]3x - 4y = 7[/tex], solve for "y":

[tex]3x - 4y = 7\\\\-4y=-3x+7\\\\y=\frac{3}{4}x-\frac{7}{4}[/tex]

The slope of this line is:

[tex]m=\frac{3}{4}[/tex]

Since the slopes of parallel lines are equal, the slope of the other line is:

[tex]m=\frac{3}{4}[/tex]

Substitute the slope and the given point into [tex]y=mx+b[/tex] and solve for "b":

[tex]-2=\frac{3}{4}(-4)+b\\\\-2+3=b\\\\b=1[/tex]

Then, the equation of this line in Slope-Intercept form is:

[tex]y=\frac{3}{4}x+1[/tex]

The equation of the line in Standard form is:

[tex]Ax+By=C[/tex]

Then, manipulating the equation [tex]y=\frac{3}{4}x+1[/tex] algebraically, we get:

[tex]y-1=\frac{3}{4}x\\\\4(y-1)=3x\\\\4y-4=3x\\\\-4=3x-4y\\\\3x-4y=-4[/tex]

Answer:

First angle = 50, second angle = 26 and the third angle = 104 degrees.

Largest angle is 104 degrees.

Step-by-step explanation:

Let the second angle be x degrees, the the first will be x + 24 and the third is

4x degrees.

x + x+ 24 + 4x = 180

6x = 180 - 24

6x = 156

x = 26

so the other 2 angles are 26 + 24 = 50 and 4*26 = 104 degrees.  

Triangle is the polygon with three edges, three vertices and three angle. The total number of degrees in a triangle is always 180 degrees. The value of the largest angle in the given triangle is 104 degrees.

Let The measure of the second angle is x degrees.

The measure of the first angle is 24 degrees more than the measure of the second angle. Thus first angle is (24+x) degrees.

The third angle is four times the measure of the second angle. Thus the measure of the third angle is 4x.

Triangle is the polygon with three edges, three vertices and three angle. The total number of degrees in a triangle is always 180 degrees.

As we know that the sum of the all the three angle in the triangle is 180 degrees. Therefore,

[tex](24+x)+x+4x=180[/tex]

[tex]6x=124-48[/tex]

[tex]6x=156[/tex]

[tex]x=\dfrac{156}{6}[/tex]

[tex]x=26[/tex]

Hence the the measure of the second angle is 26 degrees. Now The measure of the first angle is 24 degrees more than the measure of the second angle. Thus,

[tex]=24+26[/tex]

[tex]=50[/tex]

Hence measure of the first angle is 50 degrees. Now the third angle is four times the measure of the second angle. Thus,

[tex]=4\times 26=104[/tex]

Hence, the measure of the third angle is 104 degrees.

Thus the value of the largest angle in the given triangle is 104 degrees.

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Answer:

[tex]\large\boxed{y=\dfrac{1}{4}x+4}[/tex]

Step-by-step explanation:

[tex]Let:\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\============================\\\\\text{We have the equation of the line:}\ y=-4x+9\to m_1=-4.\\\\\text{Therefore}\ m_2=-\dfrac{1}{-4}=\dfrac{1}{4}.\\\\\text{Put the value of the slope and the coodrdinates of the given point (4, 5)}\\\text{to the equation of a line}\ y=mx+b:\\\\5=\dfrac{1}{4}(4)+b\\\\5=1+b\qquad\text{subtract 1 from both sides}\\\\4=b\to b=4\\\\\text{Finally:}\\\\y=\dfrac{1}{4}x+4[/tex]

Answer:

Step-by-step explanation:

(f - g)(x) = f(x) - g(x)

f(x) = 3x + 2, g(x) = 2x - 2.

Substitute:

(f - g)(x) = (3x + 2) - (2x - 2) = 3x + 2 - 2x - (-2) = 3x + 2 - 2x + 2

combine like terms

(f - g)(x) = (3x - 2x) + (2 + 2) = x + 4

$9.38

0.0625×150=$9.38

Have a good day

Answer:

[tex]\large\boxed{y=2}[/tex]

Step-by-step explanation:

In this question, we're going to solve the equation.

What we would do is plug in the given information in the right variable. In this case, we would plugin -8 to x, since x = -8

We are solving for y, so we would nee to find how much y = ?

Your equation should look like this:

[tex]4(-8)+9y=-14[/tex]

Lets solve:

[tex]4(-8)+9y=-14\\\\\text{Multiply}\, 4(-8)\\\\-32+9y=-14\\\\\text{Add 32 to both sides}\\\\9y=18\\\\\text{Divide}\\\\y=2[/tex]

When you're done solving, you should get y = 2

This means that y = 2

Answer:

  [tex]-100x - 400[/tex]

Step-by-step explanation:

The derivative of a sum is the sum of the derivatives by the sum rule, and this also extends to differences by the constant multiple rule

  [tex]\dfrac{d}{dx}(28000 - 50x^2 - 400x) = \dfrac{d}{dx}(28000) - \dfrac{d}{dx}(50x^2) - \dfrac{d}{dx}(400x)[/tex]

By the constant multiple rule, we have

  [tex]= \dfrac{d}{dx}(28000) - 50\dfrac{d}{dx}(x^2) - 400\dfrac{d}{dx}(x)[/tex]

The derivative of any constant is 0.

The power rule says that for any real number [tex]n[/tex], [tex]\frac{d}{dx} x^n = nx^{n-1}[/tex]. And note that [tex]x = x^1[/tex]. Thus we have

  [tex]= 0 - 50(2)x - 400(1)x^0 = -100x - 400[/tex]

since [tex]x^0 = 1[/tex]

Answer:

The popcorn was $5.50

Answer:

$5.50

Step-by-step explanation:

22.50 - 11.50 = 11

divide 11 by 2 and you get 5.5

Hope this helps!

Answer:

prime

not factorable over the reals

Step-by-step explanation:

30x^2+40xy+51y^2 is not factorable (also know as prime)

because the discriminant is negative.

That is when I computed b^2-4ac, in this case 40^2-4(30)(51)=-4520.

If the discriminant is negative, then it isn't factorable over the reals.

Answer:

See below.

Step-by-step explanation:

You fill the 5 gallon up to the top then you pour 3 gallons into the 3 gallon jug, leaving 2 gallons in the  bigger jug.

You throw the contents of the 3 gallon jug away then transfer the 2 gallons from the bigger jug into the 3 gallon jug. So you have  2 gallons in the 3 gallon jug.

You then fill the 5 gallon jug and pour 1 gallon into the 3 gallon jug, so you are left with 4 gallons of water in the 5 gallon jug.

Answer:

Multiplication is basically repeating addition

Step-by-step explanation:

So, on a numberline, start at 0 then add -6 three times (since you are multiplying it by three) and you should get -18

Answer:

3,5,7

Step-by-step explanation:

to find which set it is you have to add the first too like this

3+5 >7 if greater than 7 move on too this

5+7>3 if greater move on to this

3+7>5 if greater that means it is a set you can use

3,5,7 is the set of three numbers that could be the side lengths of a triangle. This can be obtained by using the triangle inequality theorem.

Triangle inequality theorem: The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

If ΔABC is a triangle with sides a, b and c, then it should satisfy the following conditions, a+b>c

                                   b+c>a

                                   a+c>b

Use the triangle inequality theorem in each set,

3+5=8<9, does not satisfy the theorem

2+4=6, does not satisfy the theorem

2+4=6<8, does not satisfy the theorem

3+5=8>7

3+7=10>5

5+7=12>3, this set satisfies the theorem

Hence 3,5,7 is the set of three numbers that could be the side lengths of a triangle.

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[tex]|\Omega|=6^2=36\\A=\{(3,5),(5,3),(4,4),(2,6),(6,2)\}\\|A|=5\\\\P(A)=\dfrac{5}{36}\approx13.9\%[/tex]

Answer:

Step-by-step explanation:

Look at the picture.

9. The perimeter of the rectangle l × w:

P = 2l + 2w

Substitute l = 2x + 8 and w = x + 8:

P = 2(2x + 8) + 2(x + 8)     use the distributive property

P = (2)(2x) + (2)(8) + (2)(x) + (2)(8)

P = 4x + 16 + 2x + 16     combine like terms

P = (4x + 2x) + (16 + 16)

P = 6x + 32

10. Solve the equation:

6x + 32 = 158           subtract 32 from both sides

6x = 126     divide both sides by 6

x = 21

Put the value of x to the expression 2x + 8:

2(21) + 8 = 42 + 8 = 50

Answer:

f(4) = 14

Step-by-step explanation:

f(x) = 3x+2

Let x=4

f(4) = 3(4)+2

    = 12+2

    = 14

Answer:

Choice A

Step-by-step explanation:

Use the F. O. L. D method while solving

Answer:

the question is not written correctly 16t2 can't be used in the equation

Step-by-step explanation:

do it it correctly you need to combine the like numbers, subtract the 48 from both sides and then divide the number with the t to both sides.

We get the value of t as:

[tex]t=3+\sqrt{6}\ and\ t=3-\sqrt{6}[/tex]

We are asked to solve the given equation for t.

The quadratic equation in terms of the variable t is given by:

[tex]16t^2-96t+48=0[/tex]

On dividing both side of the equation by 16 we get:

[tex]t^2-6t+3=0[/tex]

Now, we know that any quadratic equation of the type:

[tex]at^2+bt+c=0[/tex] has solution as :

[tex]t=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]

Here,

[tex]a=1,\ b=-6\ and\ c=3[/tex]

i.e.

[tex]t=\dfrac{-(-6)\pm \sqrt{(-6)^2-4\times 1\times 3}}{2\times 1}\\\\t=\dfrac{6\pm \sqrt{36-12}}{2}\\\\t=\dfrac{6\pm \sqrt{24}}{2}\\\\t=\dfrac{6\pm 2\sqrt{6}}{2}\\\\t=3\pm \sqrt{6}[/tex]

i.e.

[tex]t=3+\sqrt{6}\ and\ t=3-\sqrt{6}[/tex]

Answer: 35

Step-by-step explanation:

The measure of ∠Z is 35°

The correct answer is an option (D)

For given question,

We have been given a parallelogram WXYZ.

The measure of angle X is 35°

We know that, the opposite angles of parallelogram are equal.

⇒ ∠X = ∠Z

⇒ ∠Z = 35°

Therefore, the measure of ∠Z is 35°

The correct answer is an option (D)

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Answer:

Option C

Step-by-step explanation:

As per the contract:

"Any rental not received by midnight on the day it is due is subject to a late charge of $1.50 for each day it is late."

this makes option a: "Pay a $1.50 late fee per movie per day past the due date" responsibility of customer

"All rentals are due back by midnight of the due date as printed on the transaction receipt."

this makes option b: "Get his or her rentals back by midnight the day they are due." responsibility of customer

"Any rental not returned by the fifth day after the due date will be transferred to a sale. The Customer will then be required to pay the purchase price of the item in addition to five (5) days of late fees. "

this makes option d: "Pay the purchase price and five days of late fees for any item kept more than five days past the due date." responsibility of customer

Hence option C is correct: Return a rental item kept out thirteen days past the due date!

The option that is not the responsibility of the customer is Return a rental item kept out thirteen days past the due date.

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