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Serhud [2]
2 years ago
7

Need help with this question, please help!

Mathematics
1 answer:
lubasha [3.4K]2 years ago
6 0

Answer:

x = 1 in

Step-by-step explanation:

We can use ratios to solve since they are similar triangles

6         6+x

----- = ---------

5          6

Using cross products

6*6 = 5*(6+x)

36 = 30+6x

Subtract 30 from each side

36-30 = 30-30+6x

6 = 6x

Divide each side by 6

6/6 = 6x/6

1 =x

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Find the distance and midpoint for (2, 0, -2) and (5, -4, 6)
sweet-ann [11.9K]
Hi, It is well simples:

The distance between two points is :

d( A, B) = √[ (xb - xa)^2+(yb - ya)^2+(zb - za)^2]


Then,

A = (xa, ya, za )

B = (xb , yb , zb)

We know:

A = ( 2 ,0 ,-2)

And,

B = (5 , -4 , 6)
___________

xb - xa = 5-2 <=> 3

yb - ya = -4 - 0 <=> -4

zb - za = 6 - (-2) <=> 8

Then us stay:

d( A, B) = √[ (3)^2 + (-4)^2+(8)^2]

= √( 9 + 16 + 64)

= √( 25 + 64)

= √(89)

7 0
3 years ago
For each gym membership sold, the gym keeps $42, and the employee who sold it gets $8. What is the commission the employee earne
MArishka [77]
I’m going to go with that they make 21$ for each member
5 0
3 years ago
Read 2 more answers
Condense the following logs into a single log:
mamaluj [8]

QUESTION 1

The given logarithm is

8\log_g(x)+5\log_g(y)

We apply the power rule of logarithms; n\log_a(m)=\log_(m^n)

=\log_g(x^8)+\log_g(y^5)

We now apply the product rule of logarithm;

\log_a(m)+\log_a(n)=\log_a(mn)

=\log_g(x^8y^5)

QUESTION 2

The given logarithm is

8\log_5(x)+\frac{3}{4}\log_5(y)-5\log_5(z)

We apply the power rule of logarithm to get;

=\log_5(x^8)+\log_5(y^{\frac{3}{4}})-\log_5(z^5)

We apply the product to obtain;

=\log_5(x^8\times y^{\frac{3}{4}})-\log_5(z^5)

We apply the quotient rule; \log_a(m)-\log_a(n)=\log_a(\frac{m}{n} )

=\log_5(\frac{x^8\times y^{\frac{3}{4}}}{z^5})

=\log_5(\frac{x^8 \sqrt[4]{y^3} }{z^5})

7 0
2 years ago
Help!! I'll give points if anybody can answer these for me!
galina1969 [7]

Transformation involves changing the position of a shape.

The transformation from \triangle ABC to \triangle A'B'C' is 90 degrees clockwise rotation

Using the coordinates of A and A', we have:

A= (-6,3)

A'= (3,6)

Notice that the x-coordinate is <em>negated </em>and then <em>swapped </em>with the y-coordinate.

i.e.

(x,y) \to (y,-x)

The transformation of a point from <em>(x,y) to (y,-x)</em> is a 90 degrees clockwise rotation.

Hence, the transformation from \triangle ABC to \triangle A'B'C' is 90 degrees clockwise rotation

Read more about transformations at:

brainly.com/question/13801312

6 0
2 years ago
Help plz points 10 need help
Jobisdone [24]

Answer:

208

Step-by-step explanation:

6 0
2 years ago
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