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Rashid [163]
2 years ago
15

I need help with this equation. Please solve for k

Mathematics
1 answer:
Levart [38]2 years ago
8 0

Hey there!

18 = -k + 3

-k + 3 = 18

-1k + 3 = 18

SUBTRACT 3 to BOTH SIDES

-1k + 3 - 3 = 18 - 3

SIMPLIFY IT!

-1k = 18 - 3

-1k = 15

DIVIDE -1 to BOTH SIDES

-1k/-1 = 15/-1

SIMPLIFY IT!
k = 15/-1

k = -15


Therefore, the answer is: k = -15



Good luck on your assignment & enjoy your day!


~Amphitrite1040:)

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4 0
3 years ago
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madam [21]
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8 0
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6 0
3 years ago
PLEASE HELP MEE I CAN'T SOLVE THIS BY MYSELF!
UNO [17]

Answer:

x = 22

Step-by-step explanation:

The figure is attached below.

Its given that SV is parallel to RU. We have to find the value of x.

Observe that, if we join point V to U, we will obtain a parallelogram RSVU. One of the properties of a parallelogram is that the sum of its two adjacent angles is always 180 degrees.

So, the two adjacent angles would be: ∠S and ∠R

From the figure:

∠S = 5x + 4

∠R = 44 + x

Since,

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6 0
2 years ago
On a coordinate plane, a curved line with a minimum value of (negative 2.5, negative 12) and a maximum value of (0, negative 3)
goldfiish [28.3K]

Answer:

f(x) > 0 over the interval (-\infty,-4)

Step-by-step explanation:

If  f(x)  is a continuous function,  and that all the critical points of behavior change are described by the given information, then we can say that the function crossed the x axis to reach a minimum value of -12 at the point x=-2.5, then as x increases it ascends to a maximum value of -3 for x = 0 (which is also its y-axis crossing) and therefore probably a local maximum.

Then the function was above the x axis (larger than zero) from - \infty, until it crossed the x axis (becoming then negative) at the point x = -4. So the function was positive (larger than zero) in such interval.

There is no such type of unique assertion regarding the positive or negative value of the function when one extends the interval from - \infty to -3, since between the  values -4 and -3 the function adopts negative values.

6 0
3 years ago
Read 2 more answers
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