If you jog 8 times around a track that is 600 meters long, how many kilometers have you covered?

Answer:‍♀️

Step-by-step explanation:

Answer:

54, 63

Step-by-step explanation:

6 x 9 = 54

7 x 9 = 63

Answer:

Step-by-step explanation:

Answer:

it's less than

Step-by-step explanation:

12/8 simplified is 3/2

11/6 cannot be further simplified

3/2<11/6

Answer:

<8 corresponds with <3

Answer:

1.1.1) -61

1.1.2) Term 42

Step-by-step explanation:

Tn= dn (difference) + t0 (term 0)

Tn= -4n+ 11

You can now take this general rule to find out the other terms.

Tn= -4(18)+11

Tn= -61

And the term -157

Tn= -4n+11

-157= -4n + 11

-157 - 11= - 4n

-168= -4n

-168/-4= n

42= n

Answer:

angle 7

Step-by-step explanation:

Answer:

angle 7 angle forms a pair of alternate

exterior angles with angle 2.

Answer:

5.5×5.5=30.5

we rounded 30.25 in nearest tenth it becomes 30

Answer:

C.] 5.5

Step-by-step explanation:

It is mentioned that the number was multiplied by the number itself.

Let's take 'x' as the number. It was multiplied by itself and the product is 30. So the equation will be:

x × x = 30

[tex]x^{2}[/tex] = 30

x = [tex]\sqrt{30}[/tex]

x = 5.47 = 5.5 (to the nearest tenth)

Answer:

A,B

Step-by-step explanation:

I think bc I had the same question well something like this in A in B was it

Answer:

ok but I know this is a lie I'm just willing to help

Answer:

c=5/72

Step-by-step explanation:

5/9=14c-5/12

Switch sides

14c-5/12=5/9

Add 5/12 tp both sides

14c-5/12+5/12=5/9+5/12

Simplify;

14c-5/12+5/12

=14c

Simplify;

5/9+5/12

LCM of 9, 12;

36

The least common multiplier of a, b is the smallest positive number that is divisible by both a and b.

Prime factorization of 9;

3*3

Prime factorization of 12;

2*6

=2*2*3

=3*3*2*2

=36

5/9=5*4/9*4=20/36

5/12=5*3/12*3=15/36

=20/36+15/36

20+15/36

=35/36

14c=35/36

14c/14=35/36/14

14c/14=c

35/36*14

=35/504

=5/72

Answer c=5/72

Answer to the following question is as follows;

Value of C = 5 / 72

Step-by-step explanation:

Given:

[tex]\frac{5}{9} = 14c - \frac{5}{12}\\\\[/tex]

Find:

Value of C

Computation:

[tex]\frac{5}{9} = 14c - \frac{5}{12}\\\\[/tex]

By taking 5/12 to LHS Side

[tex]\frac{5}{9} + \frac{5}{12} = 14c\\\\[/tex]

By taking LCM;

[tex]\frac{(4)(5) + (3)(5)}{36} = 14c\\\\[/tex]

[tex]\frac{20 + 15}{36} = 14c\\\\[/tex]

[tex]\frac{35}{36} = 14c\\\\[/tex]

By transfer 14 to LHS side

[tex]\frac{35}{36 \times 14 } = c\\\\[/tex]

[tex]\frac{5}{36 \times 2 } = c\\\\[/tex]

[tex]\frac{5}{72} = c\\\\[/tex]

Value of C = 5 / 72

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Answer:

(a) [tex]A = 1500(1 - e^{-0.01t})[/tex]

(b) [tex]t = 0[/tex]

Step-by-step explanation:

Given

[tex]V= 300gal[/tex] --- Volume of tank

[tex]B = 5lb/gal[/tex] --- Brine solution

[tex]R_1 = 3gal/min[/tex]  --- Rate in, in gallon/min

[tex]R_2 = 6gal/min[/tex] --- Rate out, in gallon/min

Required

Determine A(t)

First, calculate the rate in (R in) and (R out) in lb/min

[tex]R_{in} = B * R_1[/tex]

[tex]R_{in} = 5lb/gal * 3gal/min[/tex]

[tex]R_{in} = 15lb/min[/tex]

R(out) is calculated by multiply the rate at which brine leaves the tank (lb/gal) * R2

So, we have:

[tex]R_{out} = \frac{A(t)}{300} lb/gal * 3gal/min[/tex]

[tex]R_{out} = \frac{3*A(t)}{300} lb/min[/tex]

[tex]R_{out} = \frac{A(t)}{100} lb/min[/tex]

The change in the amount of brin e in the tank at time t is given as:

[tex]A'(t) = R_{in} - R_{out}[/tex]

[tex]A'(t) = 15 - \frac{A(t)}{100}[/tex]

Rewrite:

[tex]\frac{dA}{dt} = 15 - \frac{A}{100}[/tex]

Multiply through by dt

[tex]dA = [15 - \frac{A}{100}]* dt[/tex]

Make dt stand alone

[tex]\frac{dA}{15 - \frac{A}{100}} = dt[/tex]

[tex]ln|15 - \frac{A}{100}| = -0.01t + ln\ c[/tex]

Express as exponents

[tex]15 - \frac{A}{100} = ce^{-\frac{t}{100}}[/tex]

Multiply through by 100

[tex]1500 - A = 100ce^{-\frac{t}{100}}[/tex]

[tex]A = 1500 - 100ce^{-\frac{t}{100}}[/tex]

[tex]A = 1500 - 100ce^{-0.01t}[/tex]

Apply initial conditions

[tex]A(0) = 0[/tex]

Substitute 0 for t in: [tex]A = 1500 - 100ce^{-0.01t}[/tex]

[tex]0 = 1500 - 100ce^{-0.01*0}[/tex]

[tex]0 = 1500 - 100ce^{0}[/tex]

[tex]0 = 1500 - 100c[/tex]

[tex]100c = 1500[/tex]

[tex]c = 15[/tex]

Substitute 15 for c in [tex]A = 1500 - 100ce^{-0.01t}[/tex]

[tex]A = 1500 - 100*15e^{-0.01t}[/tex]

[tex]A = 1500 - 1500e^{-0.01t}[/tex]

Factorize:

[tex]A = 1500(1 - e^{-0.01t})[/tex]

How long to empty the tank

Set A to 0

[tex]1500(1 - e^{-0.01t}) = 0[/tex]

Divide through by 1500

[tex]1 - e^{-0.01t} = 0[/tex]

Collect Like Terms

[tex]e^{-0.01t} = 1[/tex]

Take ln of both sides

[tex]ln\ (e^{-0.01t}) = ln\ 1[/tex]

[tex]-0.0t = 0[/tex]

[tex]t = 0[/tex]

Answer:

5u-15=10 |+15

5u=25 |:5

u=5

The solution is 25

I hope it helps

have anice day

Answer:

SAS

Step-by-step explanation:

ITS CORRECT CUZ SAME ANGLE, AND TWO OF THE SAME SIDES

Answer A is correct. C is correct.

Step-by-step explanation:

Answer:

B

Step-by-step explanation:

2/5 is equal to 40 percent

Step-by-step explanation:

step 1. the standard form of a quadratic equation is f(x) = ax^2 + bx + c

step 2. since they are all in standard form you need to provide more information.

Answer:

0.9692 = 96.92% probability of getting someone in the 22-29 age bracket or someone who agreed to respond.

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

Find the probability of getting someone in the 22-29 age bracket or someone who agreed to respond.

357 people.

281 in the 22-29 age bracket + 65 in the 18-21 that agree to respond. So

[tex]p = \frac{281+65}{357} = 0.9692[/tex]

0.9692 = 96.92% probability of getting someone in the 22-29 age bracket or someone who agreed to respond.

Answer:

6 / 8 or 3 / 4

Step-by-step explanation:

Subtraction lol

Answer:

1988 years ago

Step-by-step explanation:

2021 - 33 =

1988

Hope that helps!

Answer:

Total possible routes participants can choose = 4×3 = 12 ways

The correct option is - b. 4×3

Step-by-step explanation:

Given - A behavioral economist studies transportation habits in an

            urban center. She identifies 4 bus routes that connect to 3

            different light rail routes available to participants in the study.

To find - Before she interviews participants, she must determine the

              number of possible routes participants can choose to use.

              Which expression represents the number of available options.

Proof -

Given that ,

Total bus routes = 4

So, number of possible ways she use the bus route = 4

Also,

The 4 bus route connects to 3 different light rail

So, possible ways that she use light rail = 3

∴ we get

Total possible routes participants can choose = 4×3 = 12 ways

The correct option is - b. 4×3

Answer: 1.7004 × 10^2

Step-by-step explanation:

Answer:36

Step-by-step explanation: 6 sided dice = 6 possible outcomes

6×6=36

The required number of outcomes is 36 after rolling all six dice. Option B is correct.

Jane has to roll a set of six standard dice and record the results of each
trial. She wonders how many different outcomes are possible after rolling all six dice. the total number of possible outcomes for each trial to be determined.

Probability can be defined as the ratio of favorable outcomes to the total number of events for a system of the certain events.

For 1 dice, sample space = { 1  2  3  4  5  6}
total possible outcome = 6
Now 6 dice roll at once and the total outcome = 6 * 6
                                                                                = 36
Meanwhile, Jane has a total of 36 outcomes.

Thus, the required number of outcomes is 36 after rolling all six dice. Option B is correct.

Learn more about probability here:

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Answer:

i know this isn't the answer but this should help...

Step-by-step explanation:

Answer:

x

∈

R

y

≤

1

Step-by-step explanation:

Answer:

[tex]y\geq -1[/tex]

Step-by-step explanation:

see graph below for reference

Answer: so here is what you have to do

Step-by-step explanation:Determine Whether a Number

Is a Solution to an Equation

2 Solve Linear Equations

3 Determine Whether an Equation Is a Conditional Equation, Identity, or Contradiction

Preparing for Linear Equations

Before getting started, take this readiness quiz. If you get a problem wrong, go back to the section cited and review the material.

P1. Determine the additive inverse of 5.

P2. Determine the multiplicative inverse of - 3.

P3. Use the Reduction Property to simplify 15 # 5x.

P4. Find the Least Common Denominator of 3 and 5 .

8 12

P5. Use the Distributive Property to remove the parentheses: 61z - 22

P6. What is the coefficient of - 4x?

P7. Simplify by combining like terms: 41y - 22 - y + 5 P8. Evaluate the expression -51x + 32 - 8 when x = -2

P9. Isx = 3inthedomainof 2 ?Isx = -3inthe domain? x + 3

[Section R.3, p. 21] [Section R.3, pp. 23–24]

[Section R.3, pp. 25–26] [Section R.3, pp. 27–28]

[Section R.3, pp. 29–30] [Section R.5, p. 41] [Section R.5, pp. 41–43] [Section R.5, pp. 40–41]

[Section R.5, pp. 43–44]

 In Words

In the equation 2

2 + 5 is the left side of the equation and 0 is the right side. In this equation only the left side contains the variable, y. In the equation 4 +5=-2x+10,theexpression

4 + 5 is the left side of the equation and - 2 + 10 is the right side. In this equation both sides have expressions that contain the variable, x.

1 Determine Whether a Number Is a Solution to an Equation

An equation in one variable is a statement made up of two expressions that are equal, and in the statement, at least one of the expressions contains the variable. The expres- sions are called the sides of the equation. Examples of equations in one variable are

2y + 5 = 0 4x + 5 = -2x + 10 3 = 9 z+2

In this section, we will concentrate on solving linear equations in one variable.