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2 years ago

Find the 4th term of a proportion whose 1st, 2nd, and 3rd terms are 5, 10, and 7 respectively

Mathematics

1 answer:

fomenos2 years ago

6 0

12 is the 4th term? Dont quote me on that though. I am 90% sure.

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Which equation represents the solution of the equation 6x - 18 = 30?

defon

Answer:

x=8

Step-by-step explanation:

firstly take 30+18 which equals 48 and then divide that by six which will give you 8 as your answer

7 0

2 years ago

Read 2 more answers

OmZC > mZE > m2D

matrenka [14]

Answer:

Step-by-step explanation: Answer to ol in your answer.) What are mZA, mZB, and mZC? mZA= (Round to one decimal place as needed. Do not include the degree. I hope this either gives you a quick hint and a tip. Thank you and have a good day.

7 0

2 years ago

Please help with 2 Math questions. I dont really understand

lisabon 2012 [21]

For the first problem, the answer is D, because every year, the graph goes down by about $4,500.
For problem two,
a. It's located in quadrant one because x and y are both positive (I've attached a graph with labeled quadrants for reference)
I'm unsure about b and c but I hope I helped with the others!

3 0

3 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago

What is the length of the hypotenuse of the triangle

kotegsom [21]

On all sides of a triangle it’s about 7 inches

7 0

2 years ago