The net of a triangular prism is shown. Use the ruler provided to measure the dimension of the net to the nearest half centimeter.

Which measurement is closest to the total surface area of the triangular prism in square centimeters?

A.20cm2
B.14cm2
C.8cm2
D.6cm2

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:

...

Step-by-step explanation:

Answer:

68

Step-by-step explanation:

Divide the circumference by pi, approximately 3.14, to calculate the diameter of the circle. For example, if the circumference equals 56.52 inches, divide 56.52 by 3.14 to get a diameter of 18 inches. Multiply the radius by 2 to find the diameter.

Answer:

the answer is attached to the picture

Answer:

y=2/5 or 0.4

Step-by-step explanation:

Isolate the variable by dividing each side by factors that don't contain the variable.

3y + 6 = 18y -->

subtract 3y from both sides to get rid of it.

6 = 15y -->

isolate the variable by dividing it by 15.

6/15 = 15y/15 -->

simplify.

2/5 = y

Answer: just get gud at math lol

Step-by-step explanation: no

Answer:

0.3529411765

Step-by-step explanation:

add red and white and divide it by the total

Answer:

The real solutions are

[tex]x=\sqrt[6]{6}\approx 1.35\\\\\:x=-\sqrt[6]{6}\approx -1.35[/tex]

Step-by-step explanation:

The solution, or root, of an equation is any value or set of values that can be substituted into the equation to make it a true statement.

To find the real solutions of the equation [tex]5x^6=30[/tex]:

[tex]\mathrm{Divide\:both\:sides\:by\:}5\\\\\frac{5x^6}{5}=\frac{30}{5}\\\\\mathrm{Simplify}\\\\x^6=6\\\\\mathrm{For\:}x^n=f\left(a\right)\mathrm{,\:n\:is\:even,\:the\:solutions\:are\:}x=\sqrt[n]{f\left(a\right)},\:-\sqrt[n]{f\left(a\right)}\\\\x=\sqrt[6]{6}\approx 1.35\\\\\:x=-\sqrt[6]{6}\approx -1.35[/tex]

｡☆✼★ ━━━━━━━━━━━━━━  ☾  

You need to find the length of A F and BC first

Lets call BC --> 'x'

Lets form an equation

9 + 9 + x + x = 28

18 + 2x = 28

- 18

2x = 10

/ 2

x = 5

So now we have the length BC

We can subtract this length from the 11cm to find the vertical height of the trapezium

11 - 5 = 6

Now we have all we need to work it out.

area = (a + b) / 2 x h

area = (5 + 9) / 2 x 6

area = 42 cm^2

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Answer:

42 cm²

Step-by-step explanation:

We can see this shape is made of 2 shapes that we are familiar with which are a rectangle and a trapezium.

⇒ The first step in working out the area of the trapezium is working out the length of CB and then working out the height of trapezium. We are given that the perimeter of the rectangle is 28 cm and we know that opposite sides of a rectangle are equal so we will call the length we want to work out x

→ x + x + 9 + 9 = 28

⇒ Simplify

→ 2x + 18 = 28

⇒ Minus 18 from both sides to isolate 2x

→ 2x = 10

⇒ Divide both sides by 5 to isolate x

→ x = 5

5 cm is the length of CB, we will need to minus that from 11 to find the height of the trapezium so,

11 - 5 = 6. The height of the trapezium is 6

Now we have the height of the trapezium (6), we have the base (9) and we have the top length (5). All we do now is substitute these numbers into the trapezium formula which is

→ 0.5 × ( a + b ) × h

Where 'a' and 'b' are the parallel sides and 'h' is the height

Now we begin to substitute in the values,

→ 0.5 × ( a + b ) × h

⇒ Substitute in the values

→ 0.5 × ( 5 + 9 ) × 6

⇒ Simplify

→ 0.5 × ( 14 ) × 6

⇒ Simplify further

→ 42

The area of the trapezium is 42 cm²

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!

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:

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:

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:

None! .7632 is smaller than 9.

Step-by-step explanation:

Answer:

D = 23.4 units

Step-by-step explanation:

In the attached file

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:

C)

Step-by-step explanation:

*Look at the picture*

Answer:

Step 1: Write down the decimal divided by 1, like this: decimal 1.

Step 2: Multiply both top and bottom by 10 for every number after the decimal point. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc.)

Step 3: Simplify (or reduce) the fraction.

Answer:

7 b + 3 r

Step-by-step explanation:

Answer: A) DE = 24 yd and EF = 10 yd

Step-by-step explanation:

Shape EFD is just a dilate of 2, meaning BCA x 2.

13 x 2 = 26

12 x 2 = 24

5 x 2 = 10

Therefore, DE = 24 yd and EF = 10 yd

I hope this helps!

Answer:

DE = 24, EF = 10

Step-by-step explanation:

We can use ratios to solve

13 ft           12 ft

---------- = -----------

26 ft          x ft

Using cross products

13x = 12*26

Divide each side by 13

13x/13 = 12*26/13

x = 24

13 ft           5 ft

---------- = -----------

26 ft          y ft

Using cross products

13y = 5*26

Divide each side by 13

13y/13 = 5*26/13

y = 10

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!

Step-by-step explanation:

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

Answer:

  none of the above

Step-by-step explanation:

The attached graph shows that none of f(-2), f(2), or f(4) matches any of g(-2) or g(4).

None of the offered statements is true.

___

Please report this problem to your teacher.

Answer:

The answer is not C

Step-by-step explanation:

I think the answer is D sooooo Pray it will be

UPDATE IT IS D

HOPE THIS HELP STAY SAFE

Answer:

Okay well the 3.4 yard pieces are the constant in which Michael will be cuttings that number stays the same an will not change. The (X) represents the unknown number of how many times the ribbon will be cut. As the whole inequality states the unknown number of pieces that are 3.4 yards long are less than 100 yards.

Answer:

The variable, x, is the number of pieces of ribbon Michael cuts. Since each piece is 3.4 yards long, 3.4x is the total length of all the pieces together. This total length cannot be more than the 100 yards of ribbon he has, so 3.4x must be less than or equal to 100.

Answer:

its C

Step-by-step explanation:

i just did it on edge and plus its really not cool to post answers when you dont actually know the answer bc people that actually want to help cant bc there only two answers allowed per question. so please dont answer if your not trying to help.  

The geometric series [tex]\sum_{n=1}^{\infty} 5(-1)^{n-1}[/tex] converges to zero.

A mathematical term denoting an endless series of the form

a + ar + ar² + ar³ + ⋯, where r is referred to as the common ratio. The geometric series for a = 1 and r = 1/2, or 1 + 1/2 + 1/4 + 1/8 +, is a straightforward example.

A series is considered to be convergent if the partial sums gravitate to a certain value, also known as a limit.

In contrast, a divergent series is one whose partial sums do not reach a limit. The Divergent series usually approaches infinity.

From the given options let us try with [tex]\sum_{n=1}^{\infty} 5(-1)^{n-1}[/tex].

[tex]= 5\sum_{n=1}^{\infty} (-1)^{n-1}.[/tex]

[tex]= 5\sum_{n=1}^{\infty} (-1)^{n}.\frac{1}{-1}[/tex].

[tex]= -5\sum_{n=1}^{\infty} (-1)^{n}.[/tex]

= - 5[- 1 + 1 - 1 + 1 ...+ 1].

= - 5[0].

= 0.

learn more about geometric sequences here :

https://brainly.com/question/19235539

#SPJ7

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:

You can simplify the integrand:

[tex]\dfrac{3x^2+4x+1}{2x}=\dfrac{3x}2+2+\dfrac1{2x}[/tex]

Then integrate term-by-term:

[tex]\displaystyle\int\frac{3x^2+4x+1}{2x}\,\mathrm dx=\frac32\int x\,\mathrm dx+2\int\mathrm dx+\frac12\int\frac{\mathrm dx}x[/tex]

[tex]=\dfrac{3x^2}4+2x+\dfrac12\ln|x|+C[/tex]

Answer:

[tex]8 - 3x > -25 : \left[\begin{array}{ccc}solution: x < 11\\Interval Notation:(-\infty,11)\end{array}\right][/tex]

Step-by-step explanation:

[tex]8-3x>-25[/tex]

Subtract [tex]8[/tex] from both sides:

[tex]8-3x-8>-25-8[/tex]

Simplify:

[tex]-3x>-33[/tex]

Multiply both sides by [tex]-1[/tex] (reverse the inequality):

[tex]\left(-3x\right)\left(-1\right)<\left(-33\right)\left(-1\right)[/tex]

Simplify:

[tex]3x<33[/tex]

Divide both sides by [tex]3[/tex] :

[tex]\frac{3x}{3}<\frac{33}{3}[/tex]

Simplify:

[tex]x < 11[/tex]

Hope I helped. If so, may I get brainliest and a thanks?

Thank you, have a good day! =)