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2 years ago

Please help!!! I will vote BRAINLIEST What is the value of k?

Mathematics

1 answer:

Natali [406]2 years ago

4 0

<h3>Answer: k = 10</h3>

======================================================

Work Shown:

Use the remote interior angle theorem.

The interior angles (4k+5) and (6k+10) add to the exterior angle 115. Note the interior angles are not adjacent to the exterior angle. This is where the "remote" aspect comes in.

So,

(4k+5) + (6k+10) = 115

10k + 15 = 115

10k = 115 - 15

10k = 100

k = 100/10

k = 10

We can stop here.

If you want to keep going, then,

4k+5 = 4*10+5 = 45

6k+10 = 6*10+10 = 70

Note how 45+70 = 115 to help confirm our answer.

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List seven guidelines that will help you plan a working budget.

Nutka1998 [239]

Gather financial details, list all sources of income, categorize expenses, list supplies, other expenses, calculate overall cost and subtract from your income, if needed tweak budget.

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3 years ago

Given that f.x 3x-2 over x+1 g[x] x +5 evaluate f[-4] and gf [-2]

Jobisdone [24]

The value of f[ -4 ] and g°f[-2] are $\frac{14}{3}$ and 13 respectively.

<h3>What is the value of f[-4] and g°f[-2]?</h3>

Given the function;

$f(x) = \frac{3x-2}{x+1}$
$g(x)=x+5$
f[ -4 ] = ?
g°f[ -2 ] = ?

For f[ -4 ], we substitute -4 for every variable x in the function.

$f(x) = \frac{3x-2}{x+1}\\\\f(-4) = \frac{3(-4)-2}{(-4)+1}\\\\f(-4) = \frac{-12-2}{-4+1}\\\\f(-4) = \frac{-14}{-3}\\\\f(-4) = \frac{14}{3}$

For g°f[-2]

g°f[-2] is expressed as g(f(-2))

$g(\frac{3x-2}{x+1}) = (\frac{3x-2}{x+1}) + 5\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2}{x+1} + \frac{5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{3x-2+5(x+1)}{x+1}\\\\g(\frac{3x-2}{x+1}) = \frac{8x+3}{x+1}\\\\We\ substitute \ in \ [-2] \\\\g(\frac{3x-2}{x+1}) = \frac{8(-2)+3}{(-2)+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-16+3}{-2+1}\\\\g(\frac{3x-2}{x+1}) = \frac{-13}{-1}\\\\g(\frac{3x-2}{x+1}) = 13$

Therefore, the value of f[ -4 ] and g°f[-2] are $\frac{14}{3}$ and 13 respectively.

Learn more about composite functions here: brainly.com/question/20379727

#SPJ1

6 0

1 year ago

-2+0=? pleaseee help 10pointsss

Luba_88 [7]

-2 is the correct answer.

8 0

2 years ago

Read 2 more answers

8) through: (-3, -2), perp. to y = x – 1<br> A) y=-5x – 1 B) y=-4x – 5<br> C) y=-x – 5 D) y=-5x – 4

nexus9112 [7]

<u>Answer:</u>

The equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

<u>Solution:</u>

Given, line equation is y = x – 1 ⇒ x – y – 1 = 0. And a point is (-3, -2)

We have to find the line equation which is perpendicular to above given line and passing through the given point.

Now, let us find the slope of the given line equation.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-1}=1$

We know that, <em>product of slopes of perpendicular lines is -1. </em>

So, 1 $\times$ slope of perpendicular line = -1

slope of perpendicular line = -1

Now let us write point slope form for our required line.

$\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

y – (-2) = -1(x – (-3))

y + 2 = -1(x + 3)

y + 2 = -x – 3

x + y + 2 + 3 = 0

x + y + 5 = 0

y = -x -5

Hence the equation through (-3, -2) and perpendicular to y = x – 1 is y = -x -5 and option c is correct.

8 0

3 years ago

A game is being played with two players. There are 21 balloons and the player who pops the last balloon wins. One or two can be

34kurt

Answer:

be the second player, and always leave a multiple of 3 balloons

Step-by-step explanation:

In order to win, a player must force the other player to leave one or two balloons. To do that, the winning player must leave one more balloon than the maximum number that can be popped. That is, the winner will be the player who leaves 3 balloons,

Working backward, we find that the winner must leave a multiple of 3 after each turn. Since the starting number is a multiple of 3, the first player must lose if the second player plays optimally.

The winning strategy is ...

be the second player
always leave a multiple of 3 balloons.

4 0

2 years ago