HW HELP ASAP PLZZZZZ

Answer:

A (-1,-10) ; A (-1,14)

Step-by-step explanation:

[tex]\sqrt{(-1-4)^2 + (y-2)^2} = 13 \\ 25 + y^2 + 4 -4y = 169[/tex]

y^2 -4y - 140 = 0

Δ/4 = 4 + 140 = 144

y1 = 2 + 12 = 14

y2 = 2 -12= -10

Answer:

[tex]{ \bf{r(t) = 7e {}^{0.6t} }} \\ { \tt{r(2) = 7 {e}^{0.6 \times 2} }} \\ = { \tt{7 {e}^{1.2} }} \\ = 23.241 \: thiusand bacteria \: per \: hour[/tex]

Answer:

7

Step-by-step explanation:

|-7|  means find the distance from 0

We take the non negative value

|-7| = 7

Answer:

x ≥ 12

Step-by-step explanation:

-3/4x +2 ≤ -7

Subtract 2 from each side

-3/4x +2-2 ≤ -7-2

-3/4x  ≤ -9

Multiply each side by -4/3, remembering to flip the inequality

-3/4x * -4/3 ≥ - 9 *(-4/3)

x ≥ 12

Answer:

x>=12

Step-by-step explanation:

-3/4x + 2<=-7

-3/4x <= -7 -2

-3/4x<=-9

cross multiply

-3x<=-36

dividing throughout by -3

x>=12

Answer:

x² + 3x - 10 = 0

x² - 25 = 0

Answer:

D

Step-by-step explanation:

Plug in the numbers for the x-value and solve the equation.

| 3(80/3)/4 + 1 | = 16

| 3(-40/3)/4 + 1 | = 16

Answer:

30 hour

Step-by-step explanation:

girls time

12 20 hour

8 x(let)

now,

12/8=x/20

12×20=8×x

240=8x

x=240/8

x=30,,

the answer is b. insurance

Answer:

It will take him 5.85 seconds.

Step-by-step explanation:

12.5 (0.2t - 1) + 21t = 150

Use Distributive Property:

2.5t - 12.5 + 21t = 150

Combine like terms:

23.5t - 12.5 = 150

Subtract 12.5 from both sides:

23.5t = 137.5

Divide both sides by 23.5 to isolate variable t:

5.851063.....

Round to two decimal places (hundredths place):

5.85

Given:

Line A goes through (0,y) and (-2,0).

Line B goes through (1,2) and (3,10).

To find:

The value of y for which the system of given linear equation (equation of line A and line B) has no solutions.

Solution:

Two linear equation has no solutions if they are parallel line.

Slope formula:

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

We know that the slopes of two parallel lines are the same.

So, the given system of given linear equation has no solutions if

Slope of line A = Slope of line B

[tex]\dfrac{0-y}{-2-0}=\dfrac{10-2}{3-1}[/tex]

[tex]\dfrac{-y}{-2}=\dfrac{8}{2}[/tex]

[tex]\dfrac{y}{2}=4[/tex]

Multiply both sides by 2.

[tex]\dfrac{y}{2}\times 2=4\times 2[/tex]

[tex]y=8[/tex]

Therefore, the required value of y is 8.

Answer:

90 kilometers

Step-by-step explanation:

https://www.bing.com/search?q=kilometers+are+there+in+9000000cm

Answer:

$30.21

Step-by-step explanation:

100% -25%= 75%

Discounted price of the book

= 75% ×$38

= $28.50

Since the customer must pay an additional 6% of the discounted price,

percentage of discounted price paid

= 100% +6%

= 106%

Total amount paid

= 106% × $28.50

= $30.21

_________________________________

Alternative working:

Original selling price= $38

Since the book is discounted 25%,

100% ----- $38

1% ----- $0.38

75% ----- 75 ×$0.38= $28.50

Since the sales tax is based on the discounted price, we let the discounted price be 100%.

100% ----- $28.50

1% ----- $0.285

106% ----- 106 ×$0.285= $30.21

∴ The total amount the customer pays for the discounted book is $30.21.

Answer:

Because we know that here we have an oscillation, we can model this with a sine or cosine function.

P = A*cos(k*t) + M

where:

k is the frequency

A is the amplitude

M is the midline

We know that at t = 0, we have the lowest population.

We know that the mean is 129, so this is the midline.

We know that the population oscillates 25 above and below this midline,

And we know that for t = 0 we have the lowest population, so:

P = A*cos(k*0) + 129 = 129 - 25

P = A + 129 = 129 - 25

A = -25

So, for now, our equation is

P = -25*cos(k*t) + 129

Because this is a yearly period, we should expect to see the same thing for t = 12 (because there are 12 months in one year).

And remember that the period of a cosine function is 2*pi

Then:

k*12 = 2*pi

k = (2*pi)/12 = pi/6

Finally, the equation is:

P = -25*cos(t*pi/6) + 129

Now we want to find the lowest population was in April instead:

if January is t = 0, then:

February is t = 2

March is t = 3

April is t = 4

Then we would have that the minimum is at t = 4

If we want to still use a cosine equation, we need to use a phase p, such that now our equation is:

P = -25*cos(k*t + p) + 129

Such that:

cos(k*4 + p) = 1

Then:

k*4 + p = 0

p =  -k*4

So our equation now is:

P = -25*cos(k*t  - 4*k) + 129

And for the periodicity, after 12 months, in t = 4 + 12 = 16, we should have the same population.

Then, also remembering that the period of the cosine function is 2*pi:

k*12 - 4*k = 2*pi

k*8 = 2*pi

k = 2*pi/8 = pi/4

And remember that we got:

p = -4*k = -4*(pi/4) = -pi

Then the equation for the population in this case is:

P = -25*cos( t*pi/4 - pi) + 129

Answer:

I'd use A = πr^2

The area is 28.3 if we're using 3.14 as pi (rounded to the nearest tenth)

Answer:

See Explanation

Step-by-step explanation:

The question is incomplete, as the right trianglea are not given. The general explanation is as follows.

Using Pythagoras Theorem, we have:

a² = b² + c²

Where:

a => hypotenuse

Assume that the opposite and the adjacent sides are given as 3 and 4, respectively.

The hypotension becomes

a² = 3² + 4²

a² = 9 + 16.

a² = 25

Take square roots.

a = 5

If any of the other side lengths is missing; you make that side the subject and then solve.

Answer:

Since acceleration is the derivative of velocity, velocity is the antiderivative of acceleration. If you know the acceleration for all time, and if you know the starting velocity, you can figure out the velocity for all time.

Step-by-step explanation:

Answer:

Since acceleration is the derivative of velocity, velocity is the antiderivative of acceleration. If you know the acceleration for all time, and if you know the starting velocity, you can figure out the velocity for all time.

Answe

SI Si olla amigo lel just spammin here

Step-by-step explanation:

Answer:

216

Step-by-step explanation:

=>log   36 = -2/3

        m

=>36=(m)^-2/3

=>(root(-36))^3=m

=>m=(root(36))^3

=>m=6^3

=>m=216

Answer:

Buses - 43 people

Vans - 12 people

Answer:

1

Step-by-step explanation:

Computing Tv(X, X + a ) for any a∈(0,1)

Given that :  X∼Ber(0.5)

∴ The probability mass function

    P(X = 1 ) = 0.5

    P(X = 0) = ( 1 - 0.5 )

and expectation  E[X] = 0.5

hence ; TV ( X, X + a ) = 1

Answer:

A) Independent

B) Dependent

Step-by-step explanation:

A) If we take a marble out and put the marble back, it means we have restored the sample to what it was initially and thus it doesn't affect probability of making another selection.

Thus, this is an independent event.

B) A card is taken from a deck of cards without replacement and set aside. Then after that another card is taken from the first sample, this means that the first sample size has now reduced and thus the first card taken affects the probability of the second card to be picked. Thus, this is a dependent event.

Answer:

[tex]cos \theta = \frac{11}{12}[/tex]

Step-by-step explanation:

[tex]sin \theta = \frac{\sqrt{23}}{12} \ , \ tan \theta = \frac{\sqrt{23}}{11}\\\\tan \theta = \frac{sin \theta }{cos \theta }\\\\ \frac{\sqrt{23}}{11} = \frac{\frac{\sqrt{23}}{12} }{cos \theta}\\\\cos \theta = \frac{\frac{\sqrt{23}}{12} }{\frac{\sqrt{23}}{11} }\\\\cos \theta = \frac{\sqrt{23}}{12 } \times \frac{11}{\sqrt{23}}\\\\cos \theta = \frac{11}{12}[/tex]

Answer:

[tex]16 {x}^{2} + 24xy + 9 {y}^{2} [/tex]

Step-by-step explanation:

U can reduce this into

[tex](4x + 3y) {}^{2} [/tex]

Which is a square.

Now,

16x^2 + 24xy + 9y^2

16x^2 + 12xy + 12xy + 9y^2

4x ( 4x + 3y ) + 3y ( 4x + 3y )

( 4x + 3y ) ( 4x + 3y )

( 4x + 3y )^2

I hope you understand...

Mark me as brainliest...

Answer:

x = 5 or x = - 2

Step-by-step explanation:

Using the rules of logarithms

log x + log y = log (xy)

log x = log y ⇒ x = y

Given

ln(x- 3) + ln(x + 1) = ln(x + 7) , then

ln (x - 3)(x + 1) = ln (x + 7) , so

(x - 3)(x + 1) = x + 7 ← expand left side using FOIL

x² - 2x - 3 = x + 7 ( subtract x + 7 from both sides )

x² - 3x - 10 = 0 ← in standard form

(x - 5)(x + 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 5 = 0 ⇒ x = 5

x + 2 = 0 ⇒ x = - 2

Answer:

Your options are not clear

Step-by-step explanation:

[tex]\sqrt[3]{-1000 \times p^{12} \times q^3} \\\\(-1 \times 10^3 \times p^{12} \times q^3)^{\frac{1}{3} }\\\\(-1^3)^{\frac{1}{3} }\times 10^{3 \times \frac{1}{3} } \times p^{12 \times \frac{1}{3}} \times q^{3 \times \frac{1}{3}} \ \ \ \ \ \ \ \ \ \ \ \ \ \ [ \ (-1)^3 = - 1 \ ] \\\\- 1 \times 10 \times p^4 \times q\\\\-10p^4q[/tex]

Answer:

y = -3x + 13

Step-by-step explanation:

First, find the slope:

[tex]m=\frac{y_1-y_2}{x_1-x_2}\\\\m=\frac{4-10}{3-1}\\\\m=\frac{-6}{2}\\\\m=-3[/tex]

Finally, find the equation:

[tex]y-y_1=m(x-x_1)\\\\y-4=-3(x-3)\\\\y-4=-3x+9\\\\y=-3x+13[/tex]

Answer:

a) the work done in propelling a five-ton satellite to a height of 100 miles above Earth is 487.8 mile-tons  

b) the work done in propelling a five-ton satellite to a height of 300 miles above Earth is 1395.3 mile-tons

Step-by-step explanation:

Given the data in the question;

We know that the weight of a body varies inversely as the square of its distance from the center of the earth.

⇒F(x) = c / x²

given that; F(x) = five-ton = 5 tons

we know that the radius of earth is approximately 4000 miles

so we substitute

5 = c / (4000)²

c = 5 × ( 4000 )²

c = 8 × 10⁷

∴ Increment of work is;

Δw =  [ ( 8 × 10⁷ ) / x² ] Δx

a) For 100 miles above Earth;

W = ₄₀₀₀∫⁴¹⁰⁰ [ ( 8 × 10⁷ ) / x² ] Δx

= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{x}[/tex] [tex]]^{4100}_{4000[/tex]

= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{4100}[/tex] [tex]+\frac{1}{4000}[/tex] [tex]][/tex]

= (8 × 10⁷ ) [ 6.09756 × 10⁻⁶ ]

= 487.8 mile-tons  

Therefore, the work done in propelling a five-ton satellite to a height of 100 miles above Earth is 487.8 mile-tons  

b) For 300 miles above Earth.

W = ₄₀₀₀∫⁴³⁰⁰ [ ( 8 × 10⁷ ) / x² ] Δx

= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{x}[/tex] [tex]]^{4300}_{4000[/tex]

= (8 × 10⁷) [tex][[/tex] [tex]-\frac{1}{4300}[/tex] [tex]+\frac{1}{4000}[/tex] [tex]][/tex]

= (8 × 10⁷ ) [ 1.744186 × 10⁻⁵ ]

= 1395.3 mile-tons

Therefore, the work done in propelling a five-ton satellite to a height of 300 miles above Earth is 1395.3 mile-tons