All categories

2 years ago

Given right triangle jkl, what is the value of cos(l)? five-thirteenths five-twelfths twelve-thirteenths twelve-fifths

Mathematics

1 answer:

kykrilka [37]2 years ago

3 0

The value of the cosine ratio cos(L) is 5/13

<h3>How to determine the cosine ratio?</h3>

The complete question is added as an attachment

Start by calculating the hypotenuse (h) using

h^2 = 5^2 + 12^2

Evaluate the exponent

h^2 = 25 + 144

Evaluate the sum

h^2 = 169

Evaluate the exponent of both sides

h = 13

The cosine ratio is then calculated as:

cos(L) = KL/h

This gives

cos(L) =5/13

Hence, the value of the cosine ratio cos(L) is 5/13

You might be interested in

Ugh I’m kinda stuck someone can help me ?

lara31 [8.8K]

Answer:

To graph this line, move one of the points to positive three on the horizontal (side to side) line and negative six on the vertical (up and down) line. To get a slope of negative 1/2 you move down one point and right two points, and you continue until you reach the edge of the graph. Then you do the opposite on the way up (up one and left two).

Hope this helps!

3 0

3 years ago

Read 2 more answers

A 10-pound block of ice melts at a rate of 3% per hour. Find the weight of the block of ice after 1 day.

mylen [45]

Answer:

The weight of the block of ice after 1 day is of 4.81 pounds.

Step-by-step explanation:

Equation for exponential decay:

The equation that models an amount after t hours, subject to exponential decay, is given by:

$A(t) = A(0)(1 - r)^t$

In which A(0) is the initial amount and r is the decay rate, as a decimal.

A 10-pound block of ice melts at a rate of 3% per hour.

This means that $A(0) = 10, r = 0.03$

$A(t) = A(0)(1 - r)^t$

$A(t) = 10(1 - 0.03)^t$

$A(t) = 10(0.97)^t$

Find the weight of the block of ice after 1 day.

One day has 24 hours, so this is A(24).

$A(24) = 10(0.97)^{24} = 4.81$

The weight of the block of ice after 1 day is of 4.81 pounds.

6 0

3 years ago

Point C has the coordinates (–1, 4) and point D has the coordinates (2, 0). What is the distance between points C and D?

QveST [7]

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
C&({{ -1}}\quad ,&{{ 4}})\quad 
% (c,d)
D&({{ 2}}\quad ,&{{ 0}})
\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
CD=\sqrt{[2-(-1)]^2+(0-4)^2}\implies CD=\sqrt{(2+1)^2+(-4)^2}
\\\\\\
CD=\sqrt{9+16}\implies CD=\sqrt{25}\implies CD=5$