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3 years ago

7

Trey found three pieces of wire in his tool drawer. Here are their lengths (in centimeters). 9 , 5.59 , 16.8 What is the total l

ength of the three pieces?

1 answer:

lara [203]3 years ago

8 0

To find the total length we justadd the llength of the vBulletin peices together 9 +5.59 + 16 . 8 = 31.39 The total length is 31.39

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Avery’s water bottle holds 300 milliliters. Dashawn’s container holds 3/4 as much water how much do both containers hold?

GarryVolchara [31]

Answer: Avery's bottle holds 300 milliliters. If Dashawn's container holds 3/4 as much, then his container can hold 225 milliliters. Added together, both containers can hold 525.

Step-by-step explanation: 3/4 of 300 is 225. 300 + 225 = 525.

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3 years ago

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lukranit [14]

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3 years ago

A triangle ABC has its vertices at A(-2, -3), B(2, 1), and C(5,-1).

maksim [4K]

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill ~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-3})\qquad B(\stackrel{x_2}{2}~,~\stackrel{y_2}{1}) ~\hfill AB=\sqrt{( 2- (-2))^2 + ( 1- (-3))^2} \\\\\\ AB=\sqrt{(2+2)^2+(1+3)^2}\implies AB=\sqrt{32}\implies \boxed{AB=4\sqrt{2}} \\\\[-0.35em] ~\dotfill$

$B(\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad C(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{( 5- 2)^2 + ( -1- 1)^2} \\\\\\ BC=\sqrt{3^2+(-2)^2}\implies BC=\sqrt{9+4}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ C(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad A(\stackrel{x_2}{-2}~,~\stackrel{y_2}{-3}) ~\hfill CA=\sqrt{( -2- 5)^2 + ( -3- (-1))^2} \\\\\\ CA=\sqrt{(-7)^2+(-3+1)^2}\implies CA=\sqrt{49+(-2)^2}\implies \boxed{CA=\sqrt{53}}$

$\stackrel{\textit{\large perimeter of ABC}}{4\sqrt{2}+\sqrt{13}+\sqrt{53}~~\approx~~ 16.54}$

3 0

2 years ago

Find the indefinite integral using the substitution provided.

Nady [450]

Answer: $7\text{Ln}\left(e^{2x}+10\right)+C$

This is the same as writing 7*Ln( e^(2x) + 10) + C

=======================================================

Explanation:

Start with the equation $u = e^{2x}+10$

Apply the derivative and multiply both sides by 7 like so

$u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\$

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

$\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\$

Integrating leads to

$\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\$

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of $\frac{14e^{2x}}{e^{2x}+10}$

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2 years ago

Marcos is flying a kite 35 feet above an oak tree. The angle of elevation to the top of the tree is 16 degrees and the angle of

Elza [17]

9514 1404 393

Answer:

195 ft

Step-by-step explanation:

The relationship between the horizontal distance, angle, and vertical height is given by ...

Tan = Opposite/Adjacent

Referring to the attached diagram, we have ...

tan(16°) = BT/MB

tan(25°) = BK/MB

Solving for BT and BK, we can describe their difference as ...

BT = MB·tan(16°)

BK = MB·tan(25°)

BK -BT = 35 = MB(tan(25°) -tan(16°))

Dividing by the coefficient of MB gives ...

MB = 35/(tan(25°) -tan(16°)) = 35/0.179562

MB = 195 . . . feet

Marcos is standing 195 feet from the base of the tree.

3 0

3 years ago