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dem82 [27]
2 years ago
13

An equilateral triangle has an altitude of 21 inches. What is the side length of the triangle?

Mathematics
2 answers:
kvv77 [185]2 years ago
4 0
Hey, here’s your answer :)

Lena [83]2 years ago
3 0

here it is it is not my answer but i hope it helps!

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Can someone please help me I will be forever grateful
Crazy boy [7]

Answer:

5/8

Step-by-step explanation:

1 inches in fraction form would be 8/8 inches, and the button is 3/8 inches, so a 1 inch button is 5/8 inches longer than a 3/8 inch button

6 0
2 years ago
Read 2 more answers
Sam had some money in his pocket, and he found another $6. 50 in his dresser drawer. He then had a total of $19. 75. Let p repre
Anastasy [175]

Answer:

<h2>$26.25</h2>

<em><u>Solving steps:</u></em>

<em>Question:</em> <u>Sam had some money in his pocket, and he found another $6. 50 in his dresser drawer. He then had a total of $19. 75. Let p represent the amount of money Sam had in his pocket. Which equation can you use to find the amount of money Sam had in his pocket? How much money did Sam have in his pocket?.</u>

<em>Find: </em><em> </em><u>How much money did Sam have in his pocket?.</u>

<em>Solution:</em><em> </em>Let the equation be

<h3><em>=> P = T </em><em>+</em><em>F</em></h3>

<u>p represent amount of money</u>

<u>p represent amount of moneyt represent total</u>

<u>p represent amount of moneyt represent totalf represent money found</u>

<h3><em>=> P = T </em><em>+</em><em> </em><em>F</em></h3>

<u>insert the values</u>

<h3><em>=> P = $19.75 </em><em>+</em><em> </em><em>$6.50</em></h3>

add<u> 19.75 from 6.50 </u>

<h3><em>=> P = </em><em> </em><em>26.25</em></h3>

<em><u>THEREFORE THE AMOUNT OF MONEY </u></em><em><u>SAM</u></em><em><u> HAVE IN HIS POCKET</u></em><em><u> IS ABOUT</u></em><em><u> </u></em><em><u> </u></em><em><u>$</u></em><em><u>26.25</u></em>

8 0
2 years ago
Compare the scores: a score of 75 on a test with a mean of 65 and a standard deviation of 8 and a score of 75 on a test with a m
LiRa [457]

Answer:

The Zscore for both test is the same

Step-by-step explanation:

Given that :

TEST 1:

score (x) = 75

Mean (m) = 65

Standard deviation (s) = 8

TEST 2:

score (x) = 75

Mean (m) = 70

Standard deviation (s) = 4

USING the relation to obtain the standardized score :

Zscore = (x - m) / s

TEST 1:

Zscore = (75 - 65) / 8

Zscore = 10/8

Zscore = 1.25

TEST 2:

Zscore = (75 - 70) / 4

Zscore = 5/4

Zscore = 1.25

The standardized score for both test is the same.

7 0
2 years ago
What is the solution to the equation-0.2(x-20)=44-x
Black_prince [1.1K]

Answer:

x = 50

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
Find the line through (3, 1, −2) that intersects and is perpendicular to the line x = −1 + t, y = −2 + t, z = −1 + t. (HINT: If
mel-nik [20]

Answer:

( xo , yo , zo ) = ( 1 , 0 , 1 )

Step-by-step explanation:

Given:-

- A line passing through point (3, 1, −2) intersects and is perpendicular to line with coordinates:

                  x = −1 + t, y = −2 + t, z = −1 + t   .... t = arbitrary parameter.

Find:-

The coordinates for point of intersection.

Solution:-

- The line that passes through point (3, 1, −2) = ( a, b , c ) and an a arbitrary point on the given line have the following direction vector d2 :

                 d2 = ( x2 , y2 , z2 )

                 x2 = a - ( x ) = 3 - ( -1 + t ) = 4 - t

                 y2 = b - ( y ) = 1 - ( -2 + t ) = 3 - t

                 z2 = c - ( z )  = -2 - ( -1 + t ) = -1 - t

                d2 = (  4 - t , 3 - t , -1 - t )

- The direction vector d1 of the given line is:

                 d1 = ( x1 , y1 , z1 )

                 x1 = 1

                 y1 = 1

                 z1 = 1

                 d1 = ( 1 , 1 , 1 )

- The dot product of two orthogonal vectors is always equal to zero:

                 d1.d2 = 0

                 (  4 - t , 3 - t , -1 - t ) . ( 1 , 1 , 1 ) = 0

- Solve for parameter (t):

                 (4 - t) + (3 - t) + (-1 - t) = 0

                  6 -3t = 0

                  t = 2  

- The coordinates of the point of intersections can be evaluated by substituting the value of "t" into the given equation of line:

                 xo ( t = 2) = - 1 + 2 = 1

                 yo ( t = 2) = - 2 + 2 = 0

                 zo ( t = 2) = - 1 + 2 = 1

- The coordinates are:

                 ( xo , yo , zo ) = ( 1 , 0 , 1 )

8 0
3 years ago
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