A. Circle
B. Kite
C. Rectangle
D. Trapezoid
ANY ONE PLZZ THIS IS DUE IN TWO HOURS!!!!!!!!

Answer:

nine

Step-by-step explanation:

Answer:

a bc they are parallel

Step-by-step explanation:

Answer:

65

Step-by-step explanation:

Order the numbers in order

Find the middle number

Since there were 2, find the average

Answer:65

Step-by-step explanation:

median is the middle

there is 2 and they are 50 and 80

add it up and divide by 2

Answer:

Step-by-step explanation:

To prove that [tex]w_1,\dots w_n[/tex] form a basis for W, we must check that this set is a set of linearly independent vector and it generates the whole space W. We are given that T is an isomorphism. That is, T is injective and surjective. A linear transformation is injective if and only if it maps the zero of the domain vector space to the codomain's zero and that is the only vector that is mapped to 0. Also, a linear transformation is surjective if for every vector w in W there exists v in V such that T(v) =w

Recall that the set [tex]w_1,\dots w_n[/tex] is linearly independent if and only if  the equation

[tex]\lambda_1w_1+\dots \lambda_n w_n=0[/tex] implies that

[tex]\lambda_1 = \cdots = \lambda_n[/tex].

Recall that [tex]w_i = T(v_i)[/tex] for i=1,...,n. Consider [tex]T^{-1}[/tex] to be the inverse transformation of T. Consider the equation

[tex]\lambda_1w_1+\dots \lambda_n w_n=0[/tex]

If we apply [tex]T^{-1}[/tex] to this equation, then, we get

[tex] T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) =T^{-1}(0) = 0[/tex]

Since T is linear, its inverse is also linear, hence

[tex]T^{-1}(\lambda_1w_1+\dots \lambda_n w_n) = \lambda_1T^{-1}(w_1)+\dots +  \lambda_nT^{-1}(w_n)=0[/tex]

which is equivalent to the equation

[tex]\lambda_1v_1+\dots +  \lambda_nv_n =0[/tex]

Since [tex]v_1,\dots,v_n[/tex] are linearly independt, this implies that [tex]\lambda_1=\dots \lambda_n =0[/tex], so the set [tex]\{w_1, \dots, w_n\}[/tex] is linearly independent.

Now, we will prove that this set generates W. To do so, let w be a vector in W. We must prove that there exist [tex]a_1, \dots a_n[/tex] such that

[tex] w = a_1w_1+\dots+a_nw_n[/tex]

Since T is surjective, there exists a vector v in V such that T(v) = w. Since [tex]v_1,\dots, v_n[/tex] is a basis of v, there exist [tex]a_1,\dots a_n[/tex], such that

[tex]a_1v_1+\dots a_nv_n=v[/tex]

Then, applying T on both sides, we have that

[tex]T(a_1v_1+\dots a_nv_n)=a_1T(v_1)+\dots a_n T(v_n) = a_1w_1+\dots a_n w_n= T(v) =w[/tex]

which proves that [tex]w_1,\dots w_n[/tex] generate the whole space W. Hence, the set [tex]\{w_1, \dots, w_n\}[/tex] is a basis of W.

Consider the linear transformation [tex]T:\mathbb{R}^2\to \mathbb{R}^2[/tex], given by T(x,y) = T(x,0). This transformations fails to be injective, since T(1,2) = T(1,3) = (1,0). Consider the base of [tex]\mathbb{R}^2[/tex] given by (1,0), (0,1). We have that T(1,0) = (1,0), T(0,1) = (0,0). This set is not linearly independent, and hence cannot be a base of [tex]\mathbb{R}^2[/tex]

Answer:

YOU MAY NEED TO CHAN

Step-by-step explanation:

Answer:

The solution is rotation, then translation. The figure has been rotated about the origin by 90° and then translated 6 units to the right.

Answer:

51 rows

Step-by-step explanation:

Given

Length of bookmark = 20cm

Distance between beads = 4mm

Required

Number of rows of beads

First, the distance between the rows of beads must be converted to cm

if 1mm = 0.1cm

then

4mm = 4*0.1cm

4mm = 0.4 cm

This means that each row of beads is placed at 0.4 cm mark.

The distance between each row follows an arithmetic progression and it can be solved as follows;

[tex]T_n = a + (n-1)d[/tex]

Where [tex]Tn = 20cm[/tex] (The last term)

[tex]a = 0 cm[/tex] (The first term)

[tex]d = 0.4cm[/tex] (The distance between each row of beads)

n = ?? (number of rows)

Solving for n; we have the following;

[tex]T_n = a + (n-1)d[/tex] becomes

[tex]20 = 0 + (n-1)0.4[/tex]

[tex]20 = (n-1)0.4[/tex]

DIvide both sides by 0.4

[tex]\frac{20}{0.4} = \frac{(n-1)0.4}{0.4}[/tex]

[tex]50 = (n-1)[/tex]

[tex]50 = n-1[/tex]

Add 1 to both sides

[tex]50 + 1= n-1 + 1[/tex]

[tex]n = 51[/tex]

Hence, the number of rows of beads is 51

Answer:

Perimeter of the figure is 2r([tex]\pi[/tex] + 6)

Step-by-step explanation:

Perimeter of a shape is total length of the boundaries of the shape. In the given question, we have two semicircles and a rectangle.

The circumference of a circle = 2[tex]\pi[/tex]r, thus the length of the arc of a semicircle = [tex]\pi[/tex]r.

The height of the rectangle is h, radius 'r' of the semicircle = [tex]\frac{h}{2}[/tex]

⇒                             h = 2r

The perimeter of a rectangle = 2(l +b).

Given that: the length is twice its height, so that:

length = 2h

Perimeter of the rectangle = 2 (2h + h)

                                            = 6h

But,        h = 2r

Perimeter of the rectangle = 6 × 2r  

                                            = 12r      

Perimeter of the figure in terms of the semicircle's radius = [tex]\pi[/tex]r + 12r  + [tex]\pi[/tex]r

                                                                                               = 2[tex]\pi[/tex]r + 12r

                                                                                   = 2r([tex]\pi[/tex] + 6)

Answer:

c

Step-by-step explanation:

Answer:

X + 3 = 12

6g = 18

P/3.5 = 2.25

9 = 100 - f

Step-by-step explanation:

Equations have equal signs.

Expressions do not.

Therefore, any expression with an equal sign is an equation, making it one of the correct answers.

Answer:

X + 3 = 12

6g = 18

P/3.5 = 2.25

9 = 100 - f

Step-by-step explanation:

Just did  it!

Step-by-step explanation:

Hope this helps you... Lemme know if you don't understand

Answer:

a.i [tex]\frac{dQ(t)}{dt} = kr_{in} - \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t}\\\frac{dQ(t)}{dt} + \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}[/tex]

ii. time at which vat will be empty

[tex]T = \frac{Q(T)r_{out} }{[V_{0} + (r_{in} - r_{out} )]kr_{in}}[/tex] when dQ/dt = 0

b. i Amount of salt at time t

Q(t) = [[exp(rout/(rin - rout)]krin[V(0)t + [exp(rout/(rin - rout)](rin - rout)t²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)t]

ii. Concentration at t = T

c(T) = [krin[(V(0)T + (rin - rout)T²/2)exp(rout/(rin - rout) + Q(0)V(0)]/([V(0) + exp(rout/(rin - rout)[(rin - rout)T]][V₀ +  (rin - rout)T])

Step-by-step explanation:

a.

i. We determine the differential equation for the solution in the vat at time,t.

Let Q be the quantity of salt in the vat at any time,t.

So, dQ/dt = rate of change of quantity of salt in the vat. = rate of change of quantity of salt into the vat - rate of change of quantity of salt out of the vat.

Now, since a concentration of salt, k enter the vat at a rate of rin, rate of change of quantity of salt into the vat = krin.

Let V(0) = V₀ be the volume of water int the vat at time t = 0. Now, the volume of water in the vat increases at a rate of (rin - rout). The increase in volume after a time t is  (rin - rout)t. So the volume after a time, t is V(t) = V₀ +  (rin - rout)t. The concentration of this liquid is thus Q(t)/V(t) = Q(t)/V₀ +  (rin - rout)t. Now, the rate of change of quantity of salt out of the vat is thus [Q(t)/V₀ +  (rin - rout)t]rin = Q(t)rout/[V₀ +  (rin - rout)t].

So, dQ(t)/dt = krin - Q(t)rout/[V₀ +  (rin - rout)t].

[tex]\frac{dQ(t)}{dt} = kr_{in} - \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t}\\\frac{dQ(t)}{dt} + \frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}[/tex]

ii Time T at which vat will be empty

At time T, when the vat is empty, dQ(T)/dt = 0.

So

[tex]\\\frac{dQ(T)}{dt} + \frac{Q(T)r_{out} }{V_{0} + (r_{in} - r_{out} )T} = kr_{in}\\0 + \frac{Q(T)r_{out} }{V_{0} + (r_{in} - r_{out} )T} = kr_{in}\\T = \frac{Q(T)r_{out} }{[V_{0} + (r_{in} - r_{out} )]kr_{in}}[/tex]

b. i

i. We solve the differential equation to find the amount of salt at time,t

The integrating factor is ex

[tex]exp(\int\limits {\frac{r_{out} }{V_{0} + (r_{in} - r_{out})t} } \, dt ) = exp(\frac{r_{out}}{(r_{in} - r_{out})} \int\limits {\frac{ (r_{in} - r_{out})}{V_{0} + (r_{in} - r_{out})t} } \, dt )\\= exp(\frac{r_{out}}{(r_{in} - r_{out})} ln [V_{0} + (r_{in} - r_{out})t} ]) \\\= [V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})[/tex]

Multiplying both side of the equation by the integrating factor, we have

[tex][V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})\frac{dQ(t)}{dt} + [V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})\frac{Q(t)r_{out} }{V_{0} + (r_{in} - r_{out} )t} = kr_{in}[V_{0} + (r_{in} - r_{out})t} ]exp(\frac{r_{out}}{(r_{in} - r_{out})})[/tex]dQ(t)[V(0) + (rin - rout)texp(rout/(rin - rout))]/dt = krin[V(0) + (rin - rout)texp(rout/(rin - rout))]

Integrating both sides we have

∫(dQ(t)[V(0) + (rin - rout)texp(rout/(rin - rout))]/dt)dt = ∫(krin[V(0) + (rin - rout)texp(rout/(rin - rout))])dt

Let exp(rout/(rin - rout) = A

Q(t)[V(0) + A(rin - rout)t] = Akrin[V(0)t + A(rin - rout)t²/2 + C

At t = 0, Q(t) = Q(0),

So,

Q(0)[V(0) + A(rin - rout)×0] = AkrinV(0)t + A(rin - rout)×0²/2 + C

C = Q(0)V(0)

Q(t)[V(0) + A(rin - rout)t] = Akrin[V(0)t + A(rin - rout)t²/2 + Q(0)V(0)

Q(t) = [[exp(rout/(rin - rout)]krin[V(0)t + [exp(rout/(rin - rout)](rin - rout)t²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)t]

The above is the amount of salt at time ,t.

ii. The concentration of the last drop of salt at time, t = T

To find the concentration of the last drop of salt at time t = T, we insert T into Q(t) to find its quantity and insert t = T into V(t) =  V₀ +  (rin - rout)t.

So

Q(T) = [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/[V(0) + A(rin - rout)T]

Q(T) = [[exp(rout/(rin - rout)]krin[V(0)T + [exp(rout/(rin - rout)](rin - rout)T²/2 + Q(0)V(0)]/[V(0) + [exp(rout/(rin - rout)](rin - rout)T]

the volume at t = T is

V(T) =  V₀ +  (rin - rout)T.

The concentration at t = T is c(T) = Q(T)/V(T) =  [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/[V(0) + A(rin - rout)T]/V₀ +  (rin - rout)T.

=  [Akrin[V(0)T + A(rin - rout)T²/2 + Q(0)V(0)]/([V(0) + A(rin - rout)T][V₀ +  (rin - rout)T])

= [krin[(V(0)T + (rin - rout)T²/2)exp(rout/(rin - rout) + Q(0)V(0)]/([V(0) + exp(rout/(rin - rout)[(rin - rout)T]][V₀ +  (rin - rout)T])

Answer:

3) 0.30

The probability a randomly selected student plays a sport given they work part time = 0.30

Step-by-step explanation:

Step(i):-

Given  'A' plays a sport

          B work part time

Given P(A) = 0.48

         P(B)  = 0.40

        P(A∩B) =0.12

       P(A∪B)¹ =0.24

Step(ii):-

By using conditional probability

[tex]P(B/A) = \frac{P(BnA}{P(A)}[/tex]

and similarly [tex]P(A/B) = \frac{P(AnB}{P(B)}[/tex]

The probability a randomly selected student plays a sport given they work part time

Now [tex]P(A/B) = \frac{P(AnB}{P(B)}[/tex]

       [tex]P(A/B) = \frac{0.12}{0.40}= 0.30[/tex]

Final answer:-

The probability a randomly selected student plays a sport given they work part time = 0.30

Answer:

C

Step-by-step explanation:

Since the output is just the input multiplied by 3, the output will be 15.4*3=46.2, or answer choice C. Hope this helps!

Answer:

C. 46. 2

Step-by-step explanation:

All you need to do is multiply 15.4 by 3, like so:

15.4 · 3 = 46.2

Answer: just get gud at math lol

Step-by-step explanation: no

Answer:

It would be 2/6.

Step-by-step explanation:

2/6 simplified is 1/3 and she is going against two of her friends in which makes three.

The probability that Taylor will win the game is 3/6 or 50% option (B) is correct.

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words, the probability is the number that shows the happening of the event.

We have:

Taylor is playing a board game with two friends. Using a single dice, one friend rolled a one, and the other friend rolled a three.

To win, Taylor needs to roll a 4, a 5, or a 6.  

Total number in a standard roll = 6

Favorable outcomes = 3

Probability = 3/6 or

= (3/6)×100

= 50 %

Thus, the probability that Taylor will win the game is 3/6 or 50% option (B) is correct.

Learn more about the probability here:

brainly.com/question/11234923

#SPJ5

Answer:

Kim's business earns $10,000 per month.

Kim's non-employee expenses are $3,000 per month.

If Kim wants $2,000 in profit per month, then the maximum amount Kim can spend for employee:

10000 - 3000 - 2000 =5000

If each employee costs $1,000 per month, Kim can recruit 5000/1000 = 5 as the maximum number of employees.

Hope this helps

:)

Answer:

5

Step-by-step explanation:

Revenue - profit = expense

10000 - 2000 = 8000

8000 = 3000 + 1000n

1000n = 5000

n = 5

Answer:

...

Step-by-step explanation:

Answer:

ham

tam

mat

hat

at

ha

am

ma

Step-by-step explanation:

Answer:

Step-by-step explanation:

3(2x + 1) = 4x + 7

6x + 3 = 4x + 7

2x = 4

x = 2

y = 2(2) + 1

y = 4 + 1

y = 5

(2,5)

Answer:

2,5

Step-by-step explanation:

Answer:

Mean, Median

Step-by-step explanation:

Outliers affect the mean because it is not resistant to outliers. They will affect the sum and therefore the mean.

The solution is, Mean, Median.

i.e. The outlier of a data set causes the Mean (mean/median) to be greater than the Median (mean/median) when the outlier is in the upper part of the data set.

In statistics and probability theory, the median is the value separating the higher half from the lower half of a data sample, a population, or a probability distribution. For a data set, it may be thought of as "the middle" value.

here, we have,

we know that,

Outliers affect the mean because it is not resistant to outliers.

They will affect the sum and therefore the mean.

so, we get,

The outlier of a data set causes the Mean (mean/median) to be greater than the Median (mean/median) when the outlier is in the upper part of the data set.

hence, The solution is, Mean, and, Median.

To learn more on median click:

brainly.com/question/20425811

#SPJ3

Answer:

The area of base is 47.1 m² or 15π m².

Step-by-step explanation:

Given that the formula of volume of cylinder is V = πr²h where πr² is the base area of cylinder so you have to substitute the following values into the formula :

[tex]v = \pi \times {r}^{2} \times h[/tex]

Let v = 315π,

Let h = 21,

[tex]315\pi = \pi {r}^{2} \times 21[/tex]

[tex]\pi {r}^{2} = 315\pi \div 21[/tex]

[tex]\pi {r}^{2} = 15\pi[/tex]

[tex]base \: area = 15\pi \: {m}^{2} [/tex]

[tex]base \: area = 47.1 \: {m}^{2} (3s.f)[/tex]

Answer:

a

Step-by-step explanation:

Answer:

Answers are below

Step-by-step explanation:

1.) No solution

2.) Unique solution

3.) Unique solution

4.) No solution

I graphed the equation that had a unique solution on the graph below.

If these answers are correct, please make me Brainliest!

Answer:

B. Life insurance

Step-by-step explanation:

A. Federal income tax   required by law

B. Life insurance  usually employer paid for a small amount and then optional  

C. Medicare  required by law

D. Social Security required by law

Answer:

460

Step-by-step explanation: multiply all numbers to get your answer

Hey!!

Here's your answer:

prime numbers which are factors of 30 are:

1,2,3 and 5

Hope it helps

Answer:

1, 2, 3, 5

sorry for the confusion i didn't read it correctly

Answer:

5

Step-by-step explanation:

4 people you didnt kill and that were already there and you

can i have brainliest if i got it right pls

Answer:

31

Step-by-step explanation:

there were 34 people and you killed 30. the other people would have gotten scared and ran away leaving you with 30 other bodies in the room