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SVEN [57.7K]
3 years ago
10

Can someone plz help me!!! I don't know what to do

Mathematics
1 answer:
pashok25 [27]3 years ago
5 0
Length = l
width = 1/3 of length = l/3

perimeter = 2(l + l/3)
Answer is option B

-----------------------------------
2(l+ l/3) = 56
⇒ 8l/3 = 56
⇒ l = 3*56/8
⇒ l = 21 inch
width = 21/3 = 7inch

Area = 21*7 inch² = <span>147 inch²</span>
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If the area of △PQR is A, give the expressions that complete the equation for the measure of ∠R.
Alik [6]

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I think that you have to find the perimeter of p and q

Step-by-step explanation:

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3 years ago
Solve for x please thanks​
N76 [4]

Answer:

32

Step-by-step explanation:

So, lets go over what we know:

The equation to find x is the same on both sides.

Now, lets find this equation:

So, the total area of the two lines on the left side of the triangle are 24.

Lets first subtract the first line from the total of the two lines to find the value of the second line:

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6 0
3 years ago
Help pls im on a timed test<br><br> * For people who search it up (3x)^3=3^1
wel

Answer:

  • Option D. x = -2

Step-by-step explanation:

<u>Given equation:</u>

  • (3^x)^3 = 3^1.
  • (3^3^+^(^-^2^)) = 3^1
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6 0
2 years ago
Find the point on the parabola y^2 = 4x that is closest to the point (2, 8).
guapka [62]

Answer:

(4, 4)

Step-by-step explanation:

There are a couple of ways to go at this:

  1. Write an expression for the distance from a point on the parabola to the given point, then differentiate that and set the derivative to zero.
  2. Find the equation of a normal line to the parabola that goes through the given point.

1. The distance formula tells us for some point (x, y) on the parabola, the distance d satisfies ...

... d² = (x -2)² +(y -8)² . . . . . . . the y in this equation is a function of x

Differentiating with respect to x and setting dd/dx=0, we have ...

... 2d(dd/dx) = 0 = 2(x -2) +2(y -8)(dy/dx)

We can factor 2 from this to get

... 0 = x -2 +(y -8)(dy/dx)

Differentiating the parabola's equation, we find ...

... 2y(dy/dx) = 4

... dy/dx = 2/y

Substituting for x (=y²/4) and dy/dx into our derivative equation above, we get

... 0 = y²/4 -2 +(y -8)(2/y) = y²/4 -16/y

... 64 = y³ . . . . . . multiply by 4y, add 64

... 4 = y . . . . . . . . cube root

... y²/4 = 16/4 = x = 4

_____

2. The derivative above tells us the slope at point (x, y) on the parabola is ...

... dy/dx = 2/y

Then the slope of the normal line at that point is ...

... -1/(dy/dx) = -y/2

The normal line through the point (2, 8) will have equation (in point-slope form) ...

... y - 8 = (-y/2)(x -2)

Substituting for x using the equation of the parabola, we get

... y - 8 = (-y/2)(y²/4 -2)

Multiplying by 8 gives ...

... 8y -64 = -y³ +8y

... y³ = 64 . . . . subtract 8y, multiply by -1

... y = 4 . . . . . . cube root

... x = y²/4 = 4

The point on the parabola that is closest to the point (2, 8) is (4, 4).

4 0
3 years ago
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