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Thepotemich [5.8K]

3 years ago

11

Trigonometry please help #2 ASAP

1 answer:

lana [24]3 years ago

5 0

1. cos (angle) = adjacent leg / hypotenuse

cos(50) = 10/X

x = 10/cos(50)

x = 15.6 feet

2.

tan(angle) = Opposite leg / adjacent leg

tan(5.5) = 600/x

x = 600/tan(5.5)

x = 6,231.2 feet.

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One number is 4 times as large another. The sum of their reciprocals is 45/4. Find the two numbers

Reptile [31]

Let,
the numbers be "x" and "y"
According to the question,
x = 4y ......................................................equation (1)

$\frac{1}{x} + \frac{1}{y}= \frac{45}{4}$ ...................................equation (2)

Taking equation (2)
$\frac{1}{x} + \frac{1}{y}= \frac{45}{4}$

Substituting the value of "x" from equation (1), we get,

$\frac{1}{(4y)} + \frac{1}{y}= \frac{45}{4}$

$\frac{1}{4y} + \frac{1}{y}* \frac{4}{4} = \frac{45}{4}$

$\frac{1}{4y} + \frac{4}{4y}= \frac{45}{4}$

$\frac{1+4}{4y} = \frac{45}{4}$

$\frac{5}{4y} = \frac{45}{4}$

Cross multiplying, we get,

$5*4 = 45*4y$

$20 = 180y$
<u />
$\frac{20}{180} = y$

$\frac{1}{9} = y$

$y = \frac{1}{9}$

Now,
Taking equation (1)
x = 4y
Substituting the value of "y", we get

$x = 4( \frac{1}{9} )$

$x= \frac{4}{9}$

So, the numbers are $\frac{1}{9}$ and $\frac{4}{9}$

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What is the value of y in the solution to the system of equations?

Yuliya22 [10]

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Step-by-step explanation:

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trapecia [35]

$\huge \bf༆ Answer ༄$

Let the capacity of bus be x students

And van be y students, now ;

From the given statements we get two equations ~

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260 \: \: \: \: \: (1)$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:13x + 5y = 670 \: \: \: \: \: (2)$

multiply the equation (2) with 2 [ it won't change the values ]

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260 \: \: \: \: \: \: \: \: \: (1)$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:26x + 10y = 1340 \: \: \: \: \: (3)$

Now, deduct equation (1) from equation (3)

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:26x + 10y - 2x - 10y = 1340 - 260$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:24x = 1080$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 1080 \div 24$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:x = 45$

Therefore each bus can carry (x) = 45 students

Now, plug the value of x in equation (1) to find y ~

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:2x + 10y = 260$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:(2 \times 45) + 10y = 260$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:90 + 10y = 260$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:10y = 260 - 90$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:10y = 170$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = 170 \div 10$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:y = 17$

Hence, each van can carry (y) = 17 students in total.

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Casey is completing an experiment that requires 300 mL of potassium chloride (KCl) at a concentration of 0.2 M. But Casey's lab

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