I’ll give brainlest I don’t know the answer:/ someone help bruh

If Adam and Simon start on bicycle trips from the same point at the same time and Adam travels east at 8mph and Simon travels south at 6mph then after 4.28 hours the distance between them is 60 miles.

Speed is the distance an object covers in a particular time period. It can be calculated as under:

Speed=Distance/ Time.

We have been given the speed of Adam and Simon as 8 mph and 6 mph and we are required to find the time taken by both to be 60 miles apart.

Time=Distance/speed

We cannot find individually but we can find the average time.

Time=60/(6+8)

=60/14

=2.28 hours

Hence if If Adam and Simon start on bicycle trips from the same point at the same time and Adam travels east at 8mph and Simon travels south at 6mph then after 4.28 hours the distance between them is 60 miles.

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Answer:

3.89 minutes

Step-by-step explanation:

Firstly, we have this in a table using the range

There are 6 ranges

We put the frequency alongside

Afterwards we calculate the midpoint of the range

For example, midpoint of range 1-4 is (1+4)/2 = 5/2 = 2.5

We multiply the midpoint by the frequency which gives another column

Then we calculate the mean, where the mean is the summation of all frequencies and their midpoints

We subtract each midpoint from the mean, square the results, then multiply each by their frequency

Lastly, we apply the standard deviation formula

Please check attachment for the tabular representation and formulas

Answer:

[tex]\mathbf{P(X \le 4 ) \simeq 0.0006722}[/tex]

Step-by-step explanation:

From the information given:

p = x/n

p = 7/10

p = 0.7

sample size n = 15

Suppose X be the number of accidents involved by a single-vehicle.

Then;

[tex]X \sim Binom (15,0.7)[/tex]

Thus, the required probability that at most  4 involve in a single-vehicle is

[tex]P(X\le 4) \\ \\P(X \le 4) = P(X = 0) + P(X =1 ) + ... + P(X = 4)[/tex]

[tex]P(X \le 4 ) = (^{15}_0) *0.7^0 *0.3^{15-0} + (^{15}_1) *0.7^1 *0.3^{15-1} + (^{15}_2) *0.7^2 *0.3^{15-2} + (^{15}_3) *0.7^3 *0.3^{15-3} + (^{15}_4) *0.7^4 *0.3^{15-4}[/tex]

[tex]P(X \le 4 ) = (\dfrac{15!}{0!(15-0)!}) *0.7^0 *0.3^{15-0} + (\dfrac{15!}{1!(15-1)!}) *0.7^1 *0.3^{15-1} + (\dfrac{15!}{2!(15-2)!}) *0.7^2 *0.3^{15-2} + (\dfrac{15!}{3!(15-3)!}) *0.7^3 *0.3^{15-3} + (\dfrac{15!}{4!(15-4)!}) *0.7^4 *0.3^{15-4}[/tex]

[tex]P(X \le 4 ) =1.4348907 \times 10^{-8} +5.02211745 \times 10^{-7} + 8.20279183 \times 10^{-6} + 8.29393397 \times 10^{-5} + 5.80575378 \times 10^{-4}[/tex]

[tex]P(X \le 4 ) =6.7223407 \times 10^{-4}[/tex]

[tex]\mathbf{P(X \le 4 ) \simeq 0.0006722}[/tex]

Step-by-step explanation:

it is the formula to calculate the time

in the formula

t = I / pr

t stands for time.

I stands for interest.

p stands for principal.

r stands for rate.

Hope it helps. :)

Answer:

3, 2, 0

Step-by-step explanation:

There are three modes, all numbers have at least two appearances.

Please give thanks by clicking the heart! I solved your other problem too :)

Answer:

$10 per hour

Step-by-step explanation:

As you can see, at (2, 20) the employee earns 20 dollars for 2 hours and at   (4, 40) we can see the employee earn $40 per 4 hours.  So we can find the slope of these two points to get the answer.

Slope formula y1-y2/x1-x2

so plug it in    

40-20/4-2 and we get 20/2 =10

Answer:

secondary

Step-by-step explanation:

Answer:

you search on doutnut ok dear

[tex]f(x) = 3 {x}^{2} [/tex]

function g has to be the transformation of function f. Therefore,

[tex]g(x) = 3 {(x - 6)}^{2} + 5[/tex]

is the answer.

Answer:

Step-by-step explanation:

Surface area comprises of:

Correct option is C

Answer:

its 31.25 i took the test and got it right

Step-by-step explanation:

Answer:-16x+17

Step-by-step explanation:

Answer:

Step-by-step explanation:

Sally owes 7 dollars to Matthew. She opens her piggy bank and takes out 7 dollars. Now, how much money does she owe to Matthew?

-7 + 7 = 0

-7 + 7 is the same as  7 - 7

You can clearly see that 7-7 is 0.

Answer:

You are in debt of $7. You pay the bank $7. How many money do you owe now?

Answer:

Step-by-step explanation:

4: 4 8 12 16 20 24 28 32 36 40 44 48 52 56

7: 7 14 21 28 35 42 49

lcm is 28

hope this helps <3

Answer:

using divisional law of indices logab-logan=log(b/n)therefore logbM-logbN=logb(M/N)

Answer:

B. $2080

Step-by-step explanation:

A.P.E.X

Answer:

94/39 is the correct answer

Answer:

The answer above is correct! A. 94/39

Step-by-step explanation:

E2020!

Answer:

well for me

Step-by-step explanation:

I think it's 3/8

Answer:

8πcm²/sec

Step-by-step explanation:

The radius of the circle increases at a rate of dr/dt = 0.25cm/sec

The radius of the circle r = 16cm

The area of the circle increases at a rate of dA/dt = ?

Area of a circle = πr²

We take the derivative with respect to t

dA/dt = 2πrdr/dt

dA/dt = 2 π (16)(0.25)

dA/dt = 8πcm²/sec

Thank you!!!

Answer:

8pi cm^2/sec

Step-by-step explanation:

took the unit test, also give other person brainliest, their answer was REALLY good.

Answer:

Usha is correct

Step-by-step explanation:

See attachment

Given

Dimension of the development

[tex]Height = \frac{1}{2}\ mile[/tex]

[tex]Base = \frac{1}{2}\ mile[/tex]

Dimension of a football field

[tex]Length = 100m[/tex]

[tex]Width = 50m[/tex]

First, we calculate the area of the development.

Because it is triangular in shape, the area is:

[tex]Area = \frac{1}{2} * Height * Base[/tex]

[tex]Area = \frac{1}{2} * \frac{1}{2}mile * \frac{1}{2}mile[/tex]

Convert mile to meter

[tex]Area = \frac{1}{2} * \frac{1}{2}* 1600m * \frac{1}{2} * 1600m[/tex]

[tex]Area = \frac{1}{2} * 800m * 800m[/tex]

[tex]Area = 400m * 800m[/tex]

[tex]Area = 320000m^2[/tex]

The area of 50 football pitch is then calculated as:

[tex]Area = 50 * Length * Width[/tex]

[tex]Area = 50 * 100m * 50m[/tex]

[tex]Area = 250000m^2[/tex]

By comparison:

[tex]320000 > 250000[/tex]

Hence, Usha's thought is correct

Answer:

4 quarters

4 nickels

Step-by-step explanation:

Step-by-step explanation:

When the inequality uses < or >, it is a broken line.

For the point (0, 0) it means that x = 0 and y = 0.

Since (0) < (0)² + 4(0) + 5 is true, the point is part of the solution.

Yes, the point (0,0) is part of the solution of y < - x² + 4x + 5.

Inequality is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are not equal.

A graph can be defined as a pictorial representation or a diagram that represents data or values.

To determine the border line of the graph of y < - x² + 4x + 5

A broken line results from the inequality using < or >.

x and y are both equal to zero at the coordinates (0, 0).

Substitute the values of x = 0 nad y = 0 in the inequality  y < - x² + 4x + 5

So, (0) < (0)² + 4(0) + 5

Since (0) < (0)² + 4(0) + 5 is valid,

Thus, the coordinates (0, 0) are part of the solution.

Learn more about inequalities here:

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Answer:

The value of [tex]x[/tex] is approximately -1.531.

Step-by-step explanation:

Let [tex]3.3^{2\cdot x + 1}-103^{x+1} = 0[/tex], we proceed to solve this expression by algebraic means:

1) [tex]3.3^{2\cdot x + 1}-103^{x+1} = 0[/tex]  Given

2) [tex]3.3^{2\cdot x}\cdot 3.3 -103^{x}\cdot 103 = 0[/tex] [tex]a^{b}\cdot a^{c} = a^{b+c}[/tex]

3) [tex](3.3^{x})^{2}\cdot 3.3 -\left[\left( \sqrt{103} \right)^{2}\right]^{x}\cdot 103 = 0[/tex] [tex](a^{b})^{c} = a^{b\cdot c}[/tex]

4) [tex](3.3^{x})^{2}\cdot 3.3 - \left[\left(\sqrt{103}\right)^{x}\right]^{2}\cdot 103 = 0[/tex] [tex](a^{b})^{c} = a^{b\cdot c}[/tex]/Commutative property

5) [tex]\left[\left(\frac{3.3}{\sqrt{103}}\right)^{x}\right] ^{2}-\frac{103}{3.3} = 0[/tex] Existence of multiplicative inverse/Definition of division/Modulative property/[tex]a^{b}\cdot a^{c} = a^{b+c}[/tex]

6) [tex]\left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\frac{103}{3.3}[/tex] Existence of additive inverse/Modulative property/[tex](a^{b})^{c} = a^{b\cdot c}[/tex]

7) [tex]\log \left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\log \frac{103}{3.3}[/tex] Definition of logarithm.

8) [tex]2\cdot x\cdot \log \left(\frac{3.3}{\sqrt{103}} \right)= \log \frac{103}{3.3}[/tex]     [tex]\log_{b} a^{c} = c\cdot \log_{b} a[/tex]

9) [tex]2\cdot x \cdot [\log 3.3-\log \sqrt{103}] = \log 103 - \log 3.3[/tex]      [tex]\log_{b} \frac{a}{d}[/tex]

10) [tex]x\cdot (2\cdot \log 3.3-\log 103) = \log 103 - \log 3.3[/tex]     [tex]\log_{b} a^{c} = c\cdot \log_{b} a[/tex]/Associative property

11) [tex]x = \frac{\log 103-\log 3.3}{2\cdot \log 3.3-\log 103}[/tex]   Existence of multiplicative inverse/Definition of division/Modulative property

12) [tex]x \approx -1.531[/tex]  Result

The value of [tex]x[/tex] is approximately -1.531.

Answer:

13g*4g+7

Step-by-step explanation:

Answer:

4:120

Step-by-step explanation:

for 120 student they would be 4 teacher needed per 30 student

4:120

Answer:

Nimer's family paid $489.25 for the car rental

Step-by-step explanation:

Nimer's family rents an SUV from a company that charges a flat rate of $19.99 plus $0.79 per mile driven.

They travel from St. Louis to Chicago and back. The distance between these cities is 297 miles, so they drive 2*297 = 594 miles.

The total charges include a flat fee of $19.99 and the cost of 594 miles:

$19.99 + $0.79 * 594 = $19.99 + $469.26 = $489.25

Nimer's family paid $489.25 for the car rental

Answer:

y = [tex]\frac{5}{26}[/tex]

Step-by-step explanation:

[tex]5y + \frac{1}{5}y - 25 * \frac{1}{25} = 0[/tex]

Simplify, remember the order of operarions;

1. Parentehsis

2. Exponents

3. Multiplication/division

4. Addition/subtraction

[tex]5y + \frac{1}{5}y - 25 * \frac{1}{25} = 0[/tex]

[tex]5y + \frac{1}{5}y - 1 = 0[/tex]

[tex]5y +0.2y - 1 = 0[/tex]

[tex]5.2y -1 =0[/tex]

Inverse operations;

[tex]5.2y -1 =0[/tex]

        +1     +1

5.2y = 1

/5.2    /5.2

y = [tex]\frac{5}{26}[/tex]

Answer:

a. [tex](x - 3)^2 + 16[/tex]

b. [tex]8(x -7)^2[/tex]

c. [tex](a^2 - 1)(7x - 6)[/tex] or [tex](a+1)(a-1)(7x-6)[/tex]

d. [tex](x^2-4)(x^2+3)[/tex] or [tex](x-2)(x+2)(x^2+3)[/tex]

e. [tex](a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})[/tex]

Step-by-step explanation:

[tex]a.\ (x + 1)^2 - 8(x - 1) + 16[/tex]

Expand

[tex](x + 1)(x + 1) - 8(x - 1) + 16[/tex]

Open brackets

[tex]x^2 + x + x + 1 - 8x + 8 + 16[/tex]

[tex]x^2 + 2x + 1 - 8x + 24[/tex]

Collect Like Terms

[tex]x^2 + 2x - 8x+ 1 + 24[/tex]

[tex]x^2 - 6x+ 25[/tex]

Express 25 as 9 + 16

[tex]x^2 - 6x+ 9 + 16[/tex]

Factorize:

[tex]x^2 - 3x - 3x + 9 + 16[/tex]

[tex]x(x -3)-3(x - 3) + 16[/tex]

[tex](x - 3)(x - 3) + 16[/tex]

[tex](x - 3)^2 + 16[/tex]

[tex]b.\ 8(x - 3)^2 - 64(x-3) + 128[/tex]

Expand

[tex]8(x - 3)(x - 3) - 64(x-3) + 128[/tex]

[tex]8(x^2 - 6x+ 9) - 64(x-3) + 128[/tex]

Open Brackets

[tex]8x^2 - 48x+ 72 - 64x+192 + 128[/tex]

Collect Like Terms

[tex]8x^2 - 48x - 64x+192 + 128+ 72[/tex]

[tex]8x^2 -112x+392[/tex]

Factorize

[tex]8(x^2 -14x+49)[/tex]

Expand the expression in bracket

[tex]8(x^2 -7x-7x+49)[/tex]

Factorize:

[tex]8(x(x -7)-7(x-7))[/tex]

[tex]8((x -7)(x-7))[/tex]

[tex]8(x -7)^2[/tex]

[tex]c.\ 7a^2x - 6a^2 - 7x + 6[/tex]

Factorize

[tex]a^2(7x - 6) -1( 7x - 6)[/tex]

[tex](a^2 - 1)(7x - 6)[/tex]

The answer can be in this form of further expanded as follows:

[tex](a^2 - 1^2)(7x - 6)[/tex]

Apply difference of two squares

[tex](a+1)(a-1)(7x-6)[/tex]

[tex]d.\ x^4 - x^2 - 12[/tex]

Express [tex]x^4[/tex] as [tex]x^2[/tex]

[tex](x^2)^2 - x^2 - 12[/tex]

Expand

[tex](x^2)^2 +3x^2- 4x^2 - 12[/tex]

[tex]x^2(x^2+3) -4(x^2+3)[/tex]

[tex](x^2-4)(x^2+3)[/tex]

The answer can be in this form of further expanded as follows:

[tex](x^2-2^2)(x^2+3)[/tex]

Apply difference of two squares

[tex](x-2)(x+2)(x^2+3)[/tex]

[tex]e.\ a^{4n} -b^{4n}[/tex]

Represent as squares

[tex](a^{2n})^2 -(b^{2n})^2[/tex]

Apply difference of two squares

[tex](a^{2n} -b^{2n})(a^{2n} +b^{2n})[/tex]

Represent as squares

[tex]((a^{n})^2 -(b^{n})^2)(a^{2n} +b^{2n})[/tex]

Apply difference of two squares

[tex](a^n+b^n)(a^n-b^n)(a^{2n} +b^{2n})[/tex]

Answer: D is 0.93

Step-by-step explanation:

(3/4 x 1/8 = 0.93 ) We are looking for 0.93.

4/3 x 1/8 = 0. 16

4/3 × 8 = 10.6

3/4 × 1/8 0.93