OMG HELP PLS IM PANICKING OMG OMG I GOT A F IN MATH AND I ONLY HAVE 1 DAY TO CHANGE MY GRADE BECAUSE TOMORROW IS THE FINAL REPORT CARD RESULTS AND I DONT WANNA FAIL PLS HELP-

(a) The top of the ladder is sliding down the wall at a rate of 3/2 m/sec.

(b) The angle between the ladder and the ground is changing at a rate of 1/2 rad/sec.

To solve this problem, we can use related rates, considering the ladder as a right triangle formed by the ladder itself, the wall, and the ground.

Let's denote:

x: the distance from the base of the ladder to the house (in meters)

y: the height of the ladder on the wall (in meters)

θ: the angle between the ladder and the ground (in radians)

Given:

dx/dt = -2 m/sec (the rate at which the base of the ladder is moving away from the house)

Using the Pythagorean theorem, we have:

x^2 + y^2 = 5^2 (since the ladder has a length of 5 meters)

Taking the derivative of both sides with respect to time (t), we get:

2x(dx/dt) + 2y(dy/dt) = 0

(a) To find dy/dt:

We can solve the equation above for dy/dt:

2x(dx/dt) + 2y(dy/dt) = 0

2(3)(-2) + 2y(dy/dt) = 0 (substituting x = 3 and dx/dt = -2)

-12 + 2y(dy/dt) = 0

2y(dy/dt) = 12

dy/dt = 12/(2y)

dy/dt = 6/y

Now, we need to find y. Using the Pythagorean theorem again:

x^2 + y^2 = 5^2

3^2 + y^2 = 5^2

9 + y^2 = 25

y^2 = 25 - 9

y^2 = 16

y = 4 (taking the positive value as y represents a length)

Substituting y = 4 into dy/dt = 6/y:

dy/dt = 6/4

dy/dt = 3/2 m/sec

Therefore, the top of the ladder is sliding down the wall at a rate of 3/2 m/sec.

(b) To find dθ/dt:

We can use trigonometry to relate θ, x, and y:

tan(θ) = y/x

Differentiating both sides with respect to time (t), we get:

sec^2(θ)dθ/dt = (x(dy/dt) - y(dx/dt))/x^2

Substituting the given values:

sec^2(θ)dθ/dt = (3(3/2) - 4(-2))/3^2

sec^2(θ)dθ/dt = (9/2 + 8)/9

sec^2(θ)dθ/dt = (25/2)/9

sec^2(θ)dθ/dt = 25/18

Since sec^2(θ) is equal to 1 + tan^2(θ) and tan(θ) = y/x:

sec^2(θ) = 1 + (y/x)^2

sec^2(θ) = 1 + (4/3)^2

sec^2(θ) = 1 + 16/9

sec^2(θ) = (9 + 16)/9

sec^2(θ) = 25/9

Substituting sec^2(θ) = 25/9 into the equation:

(25/9)dθ/dt = 25/18

Simplifying and solving for dθ/dt:

dθ/dt = (25/18) * (9/25)

dθ/dt = 1/2 rad/sec

Therefore, (b) the angle between the ladder and the ground is changing at a rate of 1/2 rad/sec.

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The question is incomplete. The compete question is :

Saul decides to use the IQR to measure the spread of data. Saul calculates the IQR of the data set to be 27. Saul asks Jasmine to check his work. Saul copies the data in numerical order from least to greatest for Jasmine. 25, 30, 50, 50, 50, 50, 56, n, 250. What is the value of n that will make the IQR of the data set equal to 27?

Solution :

The inter quartile range is being measured as : 3rd quartile - 1st quartile

The 1st quartile = 25% mark

The 3rd quartile = 75% mark

The data set as given in the question is :

25, 30, 50, 50, 50, 50, 56, n, 250

Therefore, the total number of the sample = 9

And the median is = 50

The median separates the given data sets into two equal halves-- that is the upper half and the lower half.

Let the middle of a lower half is the 1st quartile be [tex]$Q_1$[/tex]

And the middle of a 2nd half is the 3rd quartile be [tex]$Q_3$[/tex]

∴ [tex]$Q_1=\frac{30+50}{2}$[/tex]

        [tex]$=40$[/tex]

  [tex]$Q_3=\frac{56+n}{2}$[/tex]

Now for the IQR of the data set to be equal to [tex]$27$[/tex],

[tex]$\frac{56+n}{2}-40=27$[/tex]

[tex]$\frac{56+n}{2}=27+40$[/tex]

[tex]$\frac{56+n}{2}=67$[/tex]

[tex]$56 +n=134$[/tex]

[tex]$n=134-56$[/tex]

[tex]$n=78$[/tex]

Answer: the answer is 3/10

Step-by-step explanation: because its 3/10

Answer:

es

3/ 10

Step-by-st

Answer:

No such a triangle can not exist.  

We prove with the help of Strong Induction.

Let P(n) be the statement that Pn+1/Pn E N.

In order to prove this statement, we will utilize strong induction.So we are given that p + 1/p E N. We will show that P(n) is true for all n >= 1.

Let's consider the base case

P(1):P2/P1 = (p + 1/p)^2 - 2 = (p^2 + 2 + 1/p^2) - 2p/p = (p^2 + 1/p^2) - (2p - 2/p)

Since p + 1/p E N, both p and 1/p must be integers.

Hence, p^2 and 1/p^2 are also integers. This implies that (p^2 + 1/p^2) is an integer.

It only remains to show that (2p - 2/p) is an integer. This is equivalent to showing that 2p^2 - 2 E 0 mod p. But this is clearly true, since 2p^2 - 2 = 2(p^2 - 1) and p^2 - 1 is divisible by p.

Let's assume that P(k) is true for all k such that 1 <= k <= n. We need to prove that P(n+1) is true as well.

Now we need to prove that P(n+1) is true. In other words, we need to show that P(n+2)/P(n+1) E N, assuming that P(n+1)/P(n) E N and P(n)/P(n-1) E N.

Using the definition of P(n), we have:P(n+1)/P(n) E N and P(n)/P(n-1) E N imply that P(n+1) = aP(n) and P(n) = bP(n-1) for some integers a and b. Then:P(n+2)/P(n+1) = (P(n+2)/P(n+1)) * (P(n)/P(n)) = (P(n+2)P(n))/(P(n+1)P(n)) = (aP(n+1)P(n))/(bP(n)P(n+1)) = a/bIf we can show that a/b E N, then P(n+2)/P(n+1) E N, and P(n+1) satisfies the inductive hypothesis.

But this follows from the fact that a and b are integers and the product of two integers is always an integer.

Hence, P(n+1) is true for all n >= 1, by strong induction.Therefore, by strong induction, we have proved that if p + 1/p E N, then Pn+1/Pn EN for all nEN.

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Given, an equivalence relation ~ on a set S. Let a and b be two elements in the set S. Assuming that [a] and [b] are two equivalence classes under the equivalence relation ~. Now we need to prove that either [a] = [b] or [a] ∩ [b] = ∅ (disjoint).

Proof:If [a] and [b] are not equal, then there must be some element c in the intersection of the equivalence classes [a] and [b]. i.e, c belongs to [a] and c belongs to [b].Thus, [a] ∩ [b] is not empty.

Let x be an element in [a], then x~a, and a~c (since c belongs to [a]) and hence x~c. So, x belongs to [c] which implies that [a] is a subset of [c].Now, let y be an element in [b], then y~b, and b~c (since c belongs to [b]) and hence y~c. So, y belongs to [c] which implies that [b] is a subset of [c].Thus, both [a] and [b] are subsets of [c].

Therefore, if [a] and [b] are not equal, then [a] and [b] are both subsets of [c] and hence the intersection of [a] and [b] is not empty. Thus, [a] and [b] are not disjoint. Hence, the proof by contradiction.

Conversely, if [a] and [b] are disjoint, then [a] ∩ [b] = ∅. And thus, [a] is not equal to [b].Therefore, two equivalence classes under the equivalence relation ~ are either equal or disjoint.

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Answer:

3.14 * r * r = area of circle, 2 * 3.14 * r = circumference

Step-by-step explanation:

Remember these formulas, very important.

The probability of getting four consecutive heads in four 2-sided coin tosses is 1/16 or 0.0625 or 6.25%.

The Probability is defined as a measure of the likelihood or chance that an event will occur, expressed as a value between 0 and 1, where 0 represents impossibility and 1 represents certainty.

To calculate the probability of getting four consecutive heads in a row, we multiply the probabilities of each individual toss.

The Probability of heads (H) is = 1/2,

So, The Probability of four consecutive heads (HHHH) is :

= (1/2) × (1/2) × (1/2) × (1/2) = (1/2)⁴ = 1/16,

Therefore, the required probability is 1/16 or 6.25%.

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Answer:

7,-7 are the solutions

Step-by-step explanation:

Not a perfect square so its

+-7

I would go with -7 since its the lowest amount

Answer:

-7

Step-by-step explanation:

Answer:

answer is 65 in²

hope this helps

Answer:

uhuhuhuhuhuhuhuhuhuhuh oh yeah

Step-by-step explanation:

Answer:

D) 156/100 = n/70

Step-by-step explanation:

n = 156% of 70

% indicates a fraction over 100.

"of" indicates multiplication.

n = 156/100 • 70

Let's divide both sides by 70 to have one fraction one each side.

n/70 = 156/100

This is equal to D) [tex](\frac{156}{100}) = (\frac{n}{70})[/tex]

a. The area of the shaded region = (3x)(5x + 2y).

b. The area of shaded region = 2x(2xπ - 3x).

Given that,

We have to write an expression in factored form to represent the area of each shaded region.

We know that,

a. In the picture we can see that the shaded region.

The area of the shaded region is area of rectangle with vertical + area of rectangle Horizontal

The area of rectangle is length × width.

The area of the shaded region = (3x)(3x + y) + (2x + y)(3x)

The area of the shaded region = (3x)(3x + y + 2x + y)

The area of the shaded region = (3x)(5x + 2y)

Therefore, The area of the shaded region = (3x)(5x + 2y)

b. In the picture we can see that circle with shaded region,

The area of shaded region is area of the circle - area of the rectangle

The area of the circle is πr² and area of rectangle is length × width.

The area of shaded region = πr² - (l×w)

The area of shaded region = π(2x)² - (2x × 3x)

The area of shaded region = 4x²π - 6x²

The area of shaded region = 2x(2xπ - 3x)

Therefore, The area of shaded region = 2x(2xπ - 3x).

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Answer:

rap music

Step-by-step explanation:

^

Answer:

i’m not sure if these are right or not but

Step-by-step explanation:

the first one is no

i think the second one is no

i think the third one is yes

i think the fourth one is yes

i am not sure and i’m REALLY sorry if its wrong

Answer:

Rate of change is 1

Step-by-step explanation:

Well rate of change in a function is basically slope

theres 1 rise and.1 run

1/1=1

So rate of change is 1

The derivatives of the vector function are r'(t) = (18 · t, 1, 0), r''(t) = (18, 0, 0) and r'''(t) = (0, 0, 0).

In this question we find the definition of a vector function in terms of time, whose first, second and third derivatives must be found. This can be done by using derivative rules several times. First, define the vector function:

r(t) = (9 · t² + 6, t + 5, 6)

Second, find the first derivative:

r'(t) = (18 · t, 1, 0)

Third, find the second derivative:

r''(t) = (18, 0, 0)

Fourth, find the third derivative:

r'''(t) = (0, 0, 0)

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Answer:

$97,894.33.

Step-by-step explanation

Answer:

The numbers are 3 and 27.

Step-by-step explanation:

The explanation is attached below.

Answer:

3x-13+2x+4=116( the sum of two interior angle is epual to the sum of exterior angle )

5x =116-9

5x=107

x=21.4 degree

then

angle a= 3*21.4-13

a=51.2degree

angle b=2*21.4+2

b =44.8degree

Step-by-step explanation:

plz make me brainliest

Answer:

r = dx - de

Step-by-step explanation:

x = e + r/d

x - e = r/d

d (x - e) = r

dx- de = r

r = dx - de

Answer:

2

Step-by-step explanation:

it's literally 6-4 = 2

Answer: 2

Step-by-step explanation: +6 is simply 6 and 6-4 is 2.

:-):-):-):-):-)

Step-by-step explanation:

1) 40÷10

2) 40÷2

Answer:

72 degrees

Step-by-step explanation:

To determine the value of DBC, we first have to determine the value of q

Angle on a straight line = 180 degrees

7q - 46 + 3q + 6 = 180

10q - 40 = 180

collect like terms

10q = 180 + 40

10q = 220

q = 22

Substitute for q in angle dbc

3(22) + 6 = 72 degrees

Answer:

Two solutions

Step-by-step explanation:

Given the functions Y = -x + 1 y = -x² + 4x − 2

Since they are both y values, we will equate them to have:

-x+1 = -x² + 4x − 2

Equate to zero

x²-x-4x+1+2=0

x²-5x+3 = 0

Since the resulting equation is quadratic (highest degree being 2), hence there will be two solutions to the equation

The fund balance at the end of 2019 will be $8,000.

The given problem has four different parts, where we are supposed to calculate the accumulation of funds at the end of 2019 in different scenarios.

Scenario 1In the first scenario, the first deposit is made on December 31, 2016, and interest is compounded annually.

Using the FVA of $1 table; Payment: $2,000n = 4i = 12%

Fund balance 12/31/2019: $9,559

Hence, the fund balance at the end of 2019 will be $9,559.Scenario 2In the second scenario, the first deposit is made on December 31, 2015, and interest is compounded annually.

Using the FVAD of $1 table;Payment: $2,000n = 4i = 12%

Fund balance 12/31/2019: $10,706 Therefore, the fund balance at the end of 2019 will be $10,706.Scenario 3In the third scenario, the first deposit is made on December 31, 2015, and interest is compounded quarterly. Using the FV of the $1 chart, we get the following calculation:

Deposit Date i = n = Deposit Fund Balance 12/31/2015 3% 16 $2,000 $3,20912/31/2016 3% 12 $2,000 $2,85212/31/2017 3% 8 $2,000 $2,53412/31/2018 3% 4 $2,000 $2,251

The interest rate is 3%, and the payment is $2,000. Hence, the fund balance at the end of 2019 will be $10,846.Scenario 4In the fourth scenario, the first deposit is made on December 31, 2015, interest is compounded annually, and interest earned is withdrawn at the end of each year.

Deposit Amount No. of Payments Interest left in Fund Fund Balance 12/31/2019$2,000 $8,000 Hence, the fund balance at the end of 2019 will be $8,000.

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