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3 years ago

HELP ASAP!!find the midpoint of the line segment MN, where M is (16,-4) and N is (2,10).

Mathematics

1 answer:

Leya [2.2K]3 years ago

5 0

2+ 16= 18/2= 9. x= 9

10+(-4)= 6/2 = 3. y= 3.

the mid point is D. (9,3)

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Use the diagram to complete the statement. Triangle J K L is shown. Angle K J L is a right angle. Angle J K L is 52 degrees and

zzz [600]

Answer:

$\bold{sin(38^\circ)=cos(52^\circ)}$

Step-by-step explanation:

Given that $\triangle KJL$ is a right angled triangle.

$\angle JKL = 52^\circ\\\angle KLJ = 38^\circ$

and

$\angle KJL = 90^\circ$

Kindly refer to the attached image of $\triangle KJL$ in which all the given angles are shown.

To find:

sin(38°) = ?

a) cos(38°)

b) cos(52°)

c) tan(38°)

d) tan(52°)

Solution:

Let us use the trigonometric identities in the given $\triangle KJL$ .

We have to find the value of sin(38°).

We know that sine trigonometric identity is given as:

$sin\theta =\dfrac{Perpendicular}{Hypotenuse}$

$sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}$ ....... (1)

Now, let us find out the values of trigonometric functions given in options one by one:

$cos\theta =\dfrac{Base}{Hypotenuse}$

$cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}$ ....... (2)

By (1) and (2):

sin(38°) $\neq$ cos(38°).

$cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL}$ ...... (3)

Comparing equations (1) and (3):

we get the both are same.

$\therefore \bold{sin(38^\circ)=cos(52^\circ)}$

6 0

2 years ago

What is 'a)' on question 4?

Nikolay [14]

A = 4b

b = 7

a = 28

7 * 4 = 28

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2 years ago

Students are selling grapefruits and nuts for a fundraiser. The grapefruits cost $1 each and a bag of nuts cost $10 each. They s

NNADVOKAT [17]

Answer:

100(g+b)=307

Step-by-step explanation:

equation

g=307÷1000

3 0

2 years ago

PLEASE HELP ITS MY TRIMESTER TEST MY MOM WILL KILL ME IF I FAIL! <3 WILL MARK BRAINLIST!

sineoko [7]

Answer:

The quadrilateral is a trapezoid

6 0

2 years ago

Mai biked 5 1/4 miles today, and Noah biked 2 1/2 miles. How many times the length of Noah’s bike ride was Mai’s bike ride?

OlgaM077 [116]

Answer:

The length of Mai's bike ride was 2.1 times the length of Noah's ride.

Step-by-step explanation:

Mai biked 5 1/4 miles today

So he biked, in miles:

$5 + \frac{1}{4} = \frac{5*4 + 1}{4} = \frac{20+1}{4} = \frac{21}{4}$

Noah biked 2 1/2 miles.

So, in miles, he biked:

$2 + \frac{1}{2} = \frac{2*2 + 1}{2} = \frac{4+1}{2} = \frac{5}{2}$

How many times the length of Noah’s bike ride was Mai’s bike ride?

We divide the Mai distance by Noah's distance. In a division of fractions, we multiply the numerator by the inverse of the denominator. So

$\frac{\frac{21}{4}}{\frac{5}{2}} = \frac{21}{4} \times {2}{5} = \frac{21*1}{2*5} = \frac{21}{10} = 2.1$

The length of Mai's bike ride was 2.1 times the length of Noah's ride.

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2 years ago