pa help po pls pls pls​

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Explanation:

A proportional linear relationship is of the form y = kx

We can solve for k to get k = y/x.

Based on that, we can prove a table is a proportional linear relationship if and only if each column of x,y values divide to the same number.

Choice A shows that y/x = 2/8 = 1/4 for the first column; however, y/x = 8/16 = 1/2 for the second column. The outputs 1/4 and 1/2 not being the same indicates we don't have a constant k value for all of the columns. This is sufficient to show that table A is not a proportional linear relationship.

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In contrast, table B is a proportional linear relationship because we have these four divisions that lead to the same value (1/8)

Showing that k = 1/8 for table B. The equation here is y = (1/8)x. We can rule out choice B.

I'll let you check out tables C and D on your own, but you should find that a similar situation shows up as did with table B. This indicates tables C and D are also proportional linear relationships. We can rule out choices C and D.

Answer:

This isn't the answer but if you can take a pic of this and put it in this app called "Math-way" or "Photo-math" those help me

Answer:

Try to do it yourself, since you have a test tomorrow

Good luck

Step-by-step explanation:

5: 4605

13 (8+5):

[tex] \frac{4605}{5} \times 13 = [/tex]

=Your answer

Answer:

[tex]{ \bf{2x + y = 8 - - - (a)}} \\ { \bf{x + y = 4 - - - (b)}} \\ a - b : \\ x = 4 \\ 4 + y = 4 \\ y = 0 \\ { \tt{they \: intersect \: at \: (4, \: 0)}}[/tex]

2^4×3^2 = 144

___________

Answer:

144 would be the answer.

Step-by-step explanation:

Question:- [tex]2^{4}[/tex] · [tex]3^{2}[/tex]

[tex]2^{4}[/tex] = 2 x 2 x 2 x 2

   = 4 x 2 x 2

   = 8 x 2

   = 16

[tex]3^{2}[/tex] = 3 x 3

   = 9

So, [tex]2^{4}[/tex] · [tex]3^{2}[/tex] = 16 x 19

              = 144

[tex]{\color{red}{\huge{\underbrace{\overbrace{\mathfrak{\:\:\:\:\:\:\:꧁"Answer"꧂\:\:\: }}}}}}[/tex]

Because ∆XZW and WYZ is the same height and

Because ∆XZW and WYZ is the same height and the are of ∆WXZ to the the WYZ

[tex]so \:( \frac{1}{2}xw \times h)( \frac{1}{2} w)=7:2[/tex]

Because xy=21

[tex]so \: WY=2 \frac{1}{2}(7 + 2x2) \\ = 21 \frac{1}{2}p \times 2 \\ =\small\color{blue}{{{\boxed{\tt\red{} \:\:\:\:\:\:\:\:\:\: WY=\frac{14}{3}\:\:\:\: }}}}[/tex]

The length of the line segment WY is [tex]\frac{14}{3}[/tex] unit.

The area of a triangle is defined as the total region that is enclosed by the three sides of any particular triangle.

Area of triangle = (1/2)base × height

According to the given question.

The ratio of the area of triangle WXZ to the area of triangle WYZ is 7:2.

Since, the height WZ for both the triangles WXZ and WYZ is same.

Let, WZ = h

Therefore, the ratio of the area of the triangles is given by  

 [tex]\frac{\frac{1}{2}WX(h) }{\frac{1}{2}(WY)(h) } =\frac{7}{2}[/tex]

⇒ [tex]\frac{WX}{WY} =\frac{7}{2}[/tex]

⇒ [tex]WX = \frac{7}{2} WY..(i)[/tex]

Since, in the given figure

[tex]XY = WY + WX[/tex]

⇒ [tex]21 = WY + \frac{7}{2} WY[/tex]   (from i)

⇒ [tex]21 = \frac{9}{2}WY[/tex]

⇒ [tex]WY = \frac{14}{3}[/tex]

Hence, the length of the line segment WY is [tex]\frac{14}{3}[/tex] unit.

Find out more information about area of triangle here:

https://brainly.com/question/19305981

#SPJ3

[tex]\displaystyle\ a(n)=-3(2)^ {n-1}\Longrightarrow a(10)=-3(2)^{10-1}=2^9\cdot -3=\boxed{-1536}[/tex]

average =65%

average = (60+70)/(100+100)*100

= 130/200*100

= 65%

Answer:

6in^2

Step-by-step explanation:

formula of a triangle: l x h : 2

4 x 3 = 12

12 : 2 = 6^2

Step-by-step explanation:

x=4x+11

11=4x-x

11=3x

x=11/3

x=37

Answer: x = -2

Step-by-step explanation:

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

Answer:

B

Step-by-step explanation:

mgh=PE

(5)(9.8)(2.5)=pe

122.5j=pe

Answer:

g-21=129

Step-by-step explanation:

Answer:

56.25metres

Step-by-step explanation:

Given the height of the ball h meters, above the ground after t seconds modeled by the function h(t) = -5t^2 + 15t + 45, at maximum height, the velocity is zero, hence;

dh/dt = 0

Since dh/dt = -10t + 15

-10t+ 15 = 0

-10t = -15

t = 15/10

t = 3/2

t = 1.5secs

Substitute t = 1.5 into the expression given

h(1.5) =  -5(1.5)^2 + 15(1.5) + 45

h(1.5) = -5(2.25)+22.5+45

h(1.5) = -11.25+67.5

h(1.5) = 56.25

Hence the maximum height is 56.25metres

Answer:

[tex]10\sqrt{3}[/tex]

Step-by-step explanation:

One is given the following expression:

[tex]\sqrt{30}*\sqrt{10}[/tex]

One property of multiplying square roots is that one can multiply the numbers under the radical by each other, and then put the radical on top of the result. The following is an example of this general format of this property:

[tex]\sqrt{a}*\sqrt{b}=\sqrt{ab}[/tex]

Apply this to the real problem:

[tex]\sqrt{30}*\sqrt{10}=\sqrt{30*10}[/tex]

Simplify,

[tex]\sqrt{30*10}=\sqrt{300}[/tex]

Rewrite as (300) as the product of two numbers, such that one number is a perfect square:

[tex]\sqrt{300}=\sqrt{3*10^2}[/tex]

Simplify, take a term out of the radical:

[tex]\sqrt{3*10^2}=10\sqrt{3}[/tex]

Answer:

it may be 0.75 i am not sure

Answer:

answer will be 0.75

Answer:

the answer is gonna be

6x^2+1

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\huge\textsf{Just do PEMDAS}\\\\\\\huge\textsf{Parentheses}\\\\\huge\textsf{Exponents}\\\\\huge\textsf{Multiplication}\\\\\huge\textsf{Division}\\\\\huge\textsf{Addition}\\\\\huge\textsf{Subtraction}[/tex]

[tex]\large\textsf{18 }\div\large\textsf{ 3 }\times\large\textsf{ 2 + 1}\\\\\\\large\textsf{18 }\div\large\textsf{ 3 = \bf 6}\\\\\\\large\textsf{6 }\times\large\textsf{ 2 + 1}\\\\\\\large\textsf{6 }\times\large\textsf{ 2 = \bf 12}\\\\\\\large\textsf{12 + 1}\\\\\\\large\textsf{= \bf 13}\\\\\\\boxed{\boxed{\huge\textsf{Answer: \bf 13}}}\huge\checkmark[/tex]

[tex]\huge\textsf{Or you could if your equation is: }\\\\\large\textsf{18}\div\large\textsf{3x}\mathsf{^2}\large\textsf{ + 1}[/tex]

[tex]\large\textsf{18}\div\large\textsf{3x}\mathsf{^2}\large\textsf{ + 1}\\\\\large\textsf{18}\div\large\textsf{3x}\mathsf{^2}\large\textsf{ = \bf 6x}\mathsf{\bf ^2}\\\\\large\textsf{= \bf 6x}\mathsf{\bf ^2 \ + }\large\textsf{ \bf 1}\\\\\\\\\\\boxed{\boxed{\large\textsf{Answer: \bf 6x}\mathsf{\bf ^2}\large\textsf{ \bf + 1}}}\huge\checkmark[/tex]

[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

Step-by-step explanation:

To solve this, we can use polynomial long division.

Seeing the picture, we first divide x (the variable to the largest degree in x+1) from 2x³. We divide from 2x³ because that is the variable with the largest degree in the polynomial. That is equal to 2x², so we put that on top and subtract (x+1) * (2x²) from the polynomial. Then, we repeat the process, but with x² instead of 2x³, and again with -6x as the variable with the largest degree.

Answer:

22

Step-by-step explanation:

You add up the value of each X and you get a final answer of 22. Have a great day!

Answer:

c = 1/4

-1-1 = -2

-3-5 = -8  

Step-by-step explanation:

Answer:

10

Step-by-step explanation:

Apply difference of squares.

[tex](x + y)(x - y) = {x}^{2} - {y}^{2} [/tex]

[tex](2 + 6i)(2 - 6i) = {2}^{2} - {6i}^{2} [/tex]

[tex] = 4 + 6[/tex]

[tex] = 10[/tex]

Answer:

Step-by-step explanation:

Acres of corn harvested : time taken = 8 3/4 : 5/6

How much corn can she collect in a hour. ?

Let

x = Acres of corn harvested

Acres of corn harvested : time taken = x : 1

8 3/4 : 5/6 = x : 1

35/4 : 5/6 = x : 1

35/4 ÷ 5/6 = x/1

35/4 × 6/5 = x/1

35*6/4*5 = x/1

210/20 = x/1

210 * 1 = 20 * x

210 = 20x

x = 210/20

= 10 1/2 acres

x = Acres of corn harvested = 10 1/2 acres

Answer: the correct answer is A. Between 0 and 90

If an angle is between 0 and 90 it is acute

If an angle is EXACTLY 90 it is a right angle

If an angle is between 90 and 180 it is obtuse angle

If it's 360 then it is a full circle

Answer:

The function is:

[tex]f(t) = -16t^2 + 50t + 3[/tex]

Step-by-step explanation:

Given

[tex]height = 3ft[/tex]

[tex]velocity = 50ft/s[/tex]

[tex]acceleration = -16ft/s^2[/tex]

Required

The function

Let t represents time;

So, we have:

[tex]f(t) = acceleration * t^2 + velocity * t + height[/tex]

So, we have:

[tex]f(t) = -16* t^2 + 50* t + 3[/tex]

[tex]f(t) = -16t^2 + 50t + 3[/tex]

Answer:

b on edge

Step-by-step explanation:

Answer:

$ 100 + $ 60 y

Step-by-step explanation:

The dental clinic Dentes Felizes calculates the value of its services based on a fixed fee of R$ 100.00 and R$ 60.00 for each filling.

Let the total number of fillings is y.

The total cost of fillings = $ 60 x y =  $ 60 y

The total cost of the treatment = $ 100 + $ 60 y

Answer:

E = 7.30 J

PE = 7.30 J

KE = 0

Step-by-step explanation:

Given;

maximum height, h = 1.2 m

mass of the object, m = 0.62 kg

velocity, v = 0 m/s

acceleration due to gravity, g = 9.81 m/s²

To determine;

The potential energy, PE = mgh

                                    PE = 0.62 x 9.81 x 1.2

                                    PE = 7.30 J

The kinetic energy, KE = ¹/₂mv²

                                KE = ¹/₂ x 0.62 x 0

                               KE = 0

The total mechanical energy, E = PE + KE

                                                 E = 7.30 J + 0

                                                E = 7.30 J

Answer: hello some part of your question is missing

Let v=〈−2,5〉 in R^2,and let y=〈0,3,−2〉 in R^3.

Find a unit vector u in R^2 such that u is perpendicular to v. How many such vectors are there?

answer:

One(1) unit vector ( < 5/√29,  2 /√29 >  ) perpendicular to 〈−2,5〉

Step-by-step explanation:

let  

u = < x , y > ∈/R^2  be perpendicular to  v = < -2, 5 > ------ ( 1 )

hence :

-2x + 5y = 0

-2x = -5y

x = 5/2 y

back to equation 1

u = < 5/2y, y >

∴ || u || = y/2 √29

∧

u   = < 5 /2 y * 2 / y√29 ,  y*2 / y√29 >

    = < 5/√29,  2 /√29 >  ( unit vector perpendicular to < -2, 5 > )

Answer:

64

Step-by-step explanation:

-1.5 = (X - 82)12

-18 = X-82

X = 64

Hello,

[tex]\left\{\begin{array}{ccc}2x+5y&=&-5\\x+3y&=&3\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}x&=&3-3y\\6-6y+5y&=&-5\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}x&=&3-3y\\y&=&11\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}x&=&-30\\y&=&11\\\end{array}\right.\\\\[/tex]

Answer A