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Yakvenalex [24]
3 years ago
13

Please yall im crying this is hard

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
4 0

Answer:

I think ur and is 4140. but I'm not sure, sorry if its incorrect.

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Questions are in the picture
gtnhenbr [62]

The closest point is (3.5, 1.9) and the distance is 1.96 units

<h3>How to determine the point and the distance?</h3>

The coordinate is given as:

(4, 0)

The equation of the function is

y = √x

The distance between two points is calculated using

d = √(x2 - x1)^2 + (y2 - y1)^2

So, we have the following points

(x1, y1) = (4, 0) and (x2, y2) = (x, 0)

This gives

d = √(x - 4)^2 + (√x - 0)^2

Evaluate the difference

d = √(x - 4)^2 + (√x)^2

Evaluate the exponent

d = √x^2 - 8x + 16 + x

Evaluate the like terms

d = √x^2 - 7x + 16

Next, we differentiate using a graphing calculator

d' = (2x - 7)/[2√(x^2 - 7x + 16)]

Set to 0

(2x - 7)/[2√(x^2 - 7x + 16)] = 0

Cross multiply

2x - 7 = 0

Add 7 to both sides

2x = 7

Divide by 2

x = 3.5

So, we have:

Substitute x = 3.5 in y = √x

y = √3.5

Evaluate

y = 1.9

So, the point is (3.5, 1.9)

The distance is then calculated as:

d = √(x2 - x1)^2 + (y2 - y1)^2

This gives

d = √(3.5 - 4)^2 + (1.9 - 0)^2

Evaluate

d = 1.96

Hence, the closest point is (3.5, 1.9) and the distance is 1.96 units

Read more about distance at:

brainly.com/question/23848540

#SPJ1

3 0
2 years ago
Which statements best describe displacement? Check all that apply.
natima [27]
I think it’s how far an object travels from starting point to ending point.
7 0
3 years ago
Read 2 more answers
If A= (7,9) and B= (3,12) what is the length of AB?
MissTica
The length can be found using the Pythagorean Theorem...

c^2=a^2+b^2 and in this case:

d^2=(dx^2)+(dy^2)

d^2=(3-7)^2+(12-9)^2

d^2=-4^2+3^2

d^2=16+9

d^2=25

d=5 

So the length of AB=5 units.
8 0
3 years ago
Read 2 more answers
What is the value of this expression <br>5+(5-9)^2+3(16÷8)​
kaheart [24]

Answer:

27/4 or 6 3/4

Step-by-step explanation:

I can't put it in the box because its gonna take to long but it is correct

3 0
3 years ago
Read 2 more answers
Line segment 19 units long running from (x,0) ti (0, y) show the area of the triangle enclosed by the segment is largest when x=
Debora [2.8K]
The area of the triangle is

A = (xy)/2

Also,

sqrt(x^2 + y^2) = 19

We solve this for y.

x^2 + y^2 = 361

y^2 = 361 - x^2

y = sqrt(361 - x^2)

Now we substitute this expression for y in the area equation.

A = (1/2)(x)(sqrt(361 - x^2))

A = (1/2)(x)(361 - x^2)^(1/2)

We take the derivative of A with respect to x.

dA/dx = (1/2)[(x) * d/dx(361 - x^2)^(1/2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(x) * (1/2)(361 - x^2)^(-1/2)(-2x) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(361 - x^2)^(-1/2)(-x^2) + (361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2)/(361 - x^2)^(1/2) + (361 - x^2)/(361 - x^2)^(1/2)]

dA/dx = (1/2)[(-x^2 - x^2 + 361)/(361 - x^2)^(1/2)]

dA/dx = (-2x^2 + 361)/[2(361 - x^2)^(1/2)]

Now we set the derivative equal to zero.

(-2x^2 + 361)/[2(361 - x^2)^(1/2)] = 0

-2x^2 + 361 = 0

-2x^2 = -361

2x^2 = 361

x^2 = 361/2

x = 19/sqrt(2)

x^2 + y^2 = 361

(19/sqrt(2))^2 + y^2 = 361

361/2 + y^2 = 361

y^2 = 361/2

y = 19/sqrt(2)

We have maximum area at x = 19/sqrt(2) and y = 19/sqrt(2), or when x = y.
3 0
3 years ago
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