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3 years ago

If (x+12y)/x=19 find (144x^2 +24xy + y^2) /y^2

Mathematics

1 answer:

choli [55]3 years ago

7 0

Answer:

Step-by-step explanation:

Given:

(x+12y)/x=19

Simplifying:

x+ 12y = 19x
12y=18x
x/y = 2/3

Finding the value of:

(144x^2 +24xy + y^2) /y^2 =
144 (x/y)^2 + 24 (x/y) + 1 =

Substituting x/y with 2/3:

144 (2/3)^2 + 24 (2/3) + 1 =
144 (4/9) + 16 + 1 =
64 + 16 + 1 =
81

Answer is 81

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The sum of two consecutive integers is 7. find the numbers

faust18 [17]

X + (x + 1) = 7

2x + 1 (-1) = 7 (-1)

2x = 6

2x/2 = 6/2

x = 3

x +1 = 3 + 1

x + 1 = 4

The two integers are 3, 4

hope this helps

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Answer:

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Step-by-step explanation:

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vladimir2022 [97]

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Step-by-step explanation:

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3 years ago

Mai and Jada are both running on a treadmill. Mai travels 5 miles in 15 minutes. Jada travels 4 miles in 14 minutes. Who ran at

sasho [114]

Answer:

Mai

Step-by-step explanation:

To find the rate, you must determine who is going faster per minute. To do this, you must divide the total number of miles by the total number of minutes.

Mai: 5 miles / 15 minutes = 1/3 miles per minute

Jada: 4 miles / 14 minutes = 2/7 miles per minute

Then, use a common denominator to make calculations easier:

Mai: 1/3 * 7/7

Jada: 2/7 * 3/3

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3 0

3 years ago

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

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3 years ago