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3 years ago

Pls pls help me ASAP

Mathematics

1 answer:

Maru [420]3 years ago

6 0

Answer:

1. 27.5 in.^2

2. 360 in.^2

Step-by-step explanation:

A = bh/2

A = (4 + 7)(5)/2

A = 27.5 in.^2

A = bh/2

A = (35 + 13)(15)/2

A = 360 in.^2

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Find the indefinite integral. (Use C for the constant of integration.) <br> e2x 25 e4x dx.

Sladkaya [172]

Answer:

The solution is $\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

Step-by-step explanation:

From the question

The function given is $f(x) = \frac{e^{2x}}{ 25 + e^{4x}} dx$

The indefinite integral is mathematically represented as

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx$

Now let $e^{2x} = u$

=> $\frac{du}{dx} 2e^{2x}$

=> $2 e^{2x}dx = du$

$\int\limits {\frac{e^{2x}}{ 25 + e^{4x}}} \, dx = \int\limits {\frac{1}{ 2(25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 25 + u^2)} } \, du$

$= \frac{1}{2} \int\limits {\frac{1}{ 5^2 + u^2)} } \, du$

$= \frac{1}{2} \frac{tan^{-1} [\frac{u}{5} ]}{5} + C$

Now substituting for u

$\frac{1}{10} * tan^{-1}[\frac{e^{2x}}{5} ] + C$

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The area of a rectangular soccer field is 5000 yd^2. The length of the field is twice the length. Find the dimensions of the fie

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You can divide the field into two squares of area 2500 yd^2. The side of each of those will be √2500 = 50 yd.

The dimensions of the field are 50 yd by 100 yd.

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Step-by-step explanation:

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Which expression is equivalent to 1/4 Y - 1/2? PLEASE HELP!!!

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Answer:

Option 2

Step-by-step explanation:

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2 years ago

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Determine the measure of each segment then indicate whether the statements are true or false

kupik [55]

Answer:

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

Step-by-step explanation:

Considering the graph

Given the vertices of the segment AB

A(-4, 4)

B(2, 5)

Finding the length of AB using the formula

$d_{AB}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(2-\left(-4\right)\right)^2+\left(5-4\right)^2}$

$=\sqrt{\left(2+4\right)^2+\left(5-4\right)^2}$

$=\sqrt{6^2+1}$

$=\sqrt{36+1}$

$=\sqrt{37}$

$d_{AB}\:=\sqrt{37}$

$d_{AB}=6.08$ units

Given the vertices of the segment JK

J(2, 2)

K(7, 2)

From the graph, it is clear that the length of JK = 5 units

$d_{JK}=5$ units

Given the vertices of the segment GH

G(-5, -2)

H(-2, -2)

Finding the length of GH using the formula

$d_{GH}\:=\:\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$=\sqrt{\left(-2-\left(-5\right)\right)^2+\left(-2-\left(-2\right)\right)^2}$

$=\sqrt{\left(5-2\right)^2+\left(2-2\right)^2}$

$=\sqrt{3^2+0}$

$=\sqrt{3^2}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$d_{GH}\:=\:3$ units

Thus, from the calculations, it is clear that:

$d_{AB}=6.08$

$d_{JK}=5$

$d_{GH}\:=\:3$

Thus,

$d_{AB}\ne d_{JK}$

$d_{AB}\ne \:d_{GH}$

$d_{GH}\ne \:d_{JK}$

Therefore,

Option (A) is false

Option (B) is false

Option (C) is false

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