In the diagram below, BD is parallel to XY. What is the value of x?

Mathematics

2 answers:

alexira [117]3 years ago

3 0

Answer:

A. 55

Step-by-step explanation:

dybincka [34]3 years ago

3 0

Answer: A. 55

Step-by-step explanation:

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PLEASE HELP! WILL UPVOTE!

bagirrra123 [75]

Just "plug n chug" as my teacher would say. In other words, replace the letters with the values.

Choice A is (7,19) which means x = 7 and y = 19. Let's replace the x and y value with these numbers in both equations. If we get a true equation, then we found the answer.

2x-y = -5
2*7-19 = -5
14-19 = -5
-5 = -5 ... true
So that works. Let's try the other equation
x+3y = 22
7+3*19 = 22
7+57 = 22
64 = 22 ... false
So choice A is NOT the answer because of this. Let's try out choice B
2x-y = -5
2*4-9 = -5
-1 = -5 ... false
so B doesn't work either. How about C?
2x-y = -5
2*8-21 = -5
-5 = -5 ... that works
x+3y = 22
8+3*21 = 22
71 = 22 ... but this is false
So D has to be the answer. Let's check to confirm or not
2x-y = -5
2*1-7 = -5
-5 = -5 ... true
x+3y = 22
1+3*7 = 22
22 = 22 ... true
BOTH equations are true at the same time so (1,7) is definitely the final answer

6 0

3 years ago

R^2 = 4r-4<br><br> Please show steps

natita [175]

Answer:

r = 2

Step-by-step explanation:

$r^{2}$ = 4r-4

square root both sides

r = $\sqrt{}$ 4r-4

r = 2r-2

2r-r = 2

r = 2

8 0

3 years ago

Solve this equation pls

Aleksandr-060686 [28]

Answer:

${ \rm{y = \frac{3 \sqrt{x} + 2x}{ {4x}^{2} } }} \\$

• We are gonna use the quotient rule

${ \boxed{ \bf{ \frac{dy}{dx} = \frac{ {\huge{v} } \frac{du}{dx} - { \huge{u}} \frac{dv}{dx} }{ {v}^{2} } }}} \\$

u is (3√x + 2x)
v is 4x²
du/dx is 3/2√x
dv/dx is 8x

• Therefore:

${ \rm{ \frac{dy}{dx} = \frac{(4 {x}^{2})( \frac{3}{2 \sqrt{x} }) - (3 \sqrt{x} + 2x)(8x) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{(6 {x}^{2})( {x}^{ - \frac{1}{2} } ) - (24 {x}^{ \frac{3}{2} } + 16 {x}^{2} ) }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ 6 {x}^{ \frac{3}{2} } - {24x}^{ \frac{3}{2} } - {16x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{ - 18 {x}^{ \frac{3}{2} } - 16 {x}^{2} }{16 {x}^{4} } }} \\ \\ { \rm{ \frac{dy}{dx} = - 2 {x}^{ \frac{3}{2} } (9 + 8 {x}^{ \frac{4}{3} } ) \div 16 {x}^{4} }} \\ \\ { \boxed{ \boxed{ \rm{ \: \: \frac{dy}{dx} = - \frac{8 \sqrt{ {x}^{ \frac{8}{3} }} + 9}{8x {}^{ \frac{8}{3} } } \: \: }}}}$