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3 years ago

Which of the following shows the number 7.12 x 10^-3 written in standard form? Answer choices : 0.00712 : 0.0712 : 0.712 : 7,120

Mathematics

1 answer:

svet-max [94.6K]3 years ago

7 0

written in standard form

7.12 x 10^-3 = 0.00712

hope it helps

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If 8 and 15 are sides to a right triangle, whats the missing side?

mr Goodwill [35]

Answer:

Step-by-step explanation:

a[2]+b[2]=c[2]

8*8+15*15=hypotenuse.

64+225=289

$\sqrt{x} 289$

hypotenuse=17

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3 years ago

KLMN is a parallelogram. Identify m∠N.

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3 years ago

N- 0.125= 4.1 what is N?

guajiro [1.7K]

Well, lets add 0.125 to 4.1.

We would get 4.225.

Finally we can check.

4.225-0.125= 4.1

How did we get it? What i did is I did the opposite. Instead of subtracting 0.125 I added it.

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3 years ago

Read 2 more answers

write the equaion of a line in slope-intercept form if the slope is 1/5 and the y-intercept is -9 explain please

faust18 [17]

Answer:

$y = \frac{1}{5} x - 9$

Step-by-step explanation:

The slope-intercept form is y = mx + b, where:

m = slope

b = y-intercept.

The slope (m) tells you the steepness of the line. It is the average rate of change which measures how the y-value changes for each one-unit change in the x-value. Hence, slope $= \frac{rise}{run}$ . So the given slope of 1/5 means that for every 1 unit change in the y-value, the x-value changes by 5 units (you go up 1 unit, and "run" 5 units to the right).

Next, the y-intercept is the point on the graph where it crosses the y-axis, and has the coordinates, (0, b). It is also the value of y when x = 0. Since you're given the y-intercept of -9, then that means that it is the y-coordinate of (0, b). So, it becomes (0, -9).

Now that we have our slope (m ) = 1/5, and the y-intercept (b ) = -9, we can write the equation of the line as:

$y = \frac{1}{5} x - 9$

(I'm also including a screenshot of the line where it shows the y-intercept of (0, -9) on the graph).

Please mark my answers as the Brainliest if you find my explanations helpful :)

7 0

2 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