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Blababa [14]
3 years ago
12

Any help on the PreCalc question above? I am horrible at this subject...

Mathematics
1 answer:
Volgvan3 years ago
3 0

Answer: Domain = [-4, ∞)    Range = [-1, ∞)

<u>Step-by-step explanation:</u>

Domain represents all of the x-values in the graph.

The smallest x-value is -4.  

The biggest x-value is infinity <em>(see the arrow on the far right line)</em>

Interval notation is: [-4, ∞).

NOTE: -4 has a bracket because it is an included value (closed dot)

Range represents all of the y-values in the graph.

The smallest y-value is -1.

The biggest x-value is infinity <em>(see the arrow on the far right line)</em>

Interval notation is: [-1, ∞).

NOTE: -1 has a bracket because it is an included value (closed dot in at least one place on the line)

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Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point
Maksim231197 [3]

Answer: (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

or  (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]

Taking square on both the sides , we get

(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0

Using quadratic formula : x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}

so, (6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0) are the required points.

since \sqrt{3}=1.732

so, (6+7(1.732),0)\text{ and }(6-7(1.732),0) are the required points.

i.e. (18.124,0) and ( -6.124,0) are the required points.

3 0
3 years ago
Can you answer #1 and #3 please (find the slope) ASAP
daser333 [38]

To calculate the slope of a line, you only need two points from that line, (x1, y1) and (x2, y2).

The equation used to calculate the slope from two points is:

(y2 - y1) / (x2 - x1)

#1:

First point : (5 ,7) = (x1, y1)

Second point : (2, -2) = (x2, y2)

Equation : (-2 - 7) / (2 - 5)

Solve the equation:

(-2 - 7) = -9

(2 - 5) = -3

-9/-3 = 3 (a negative number divided by a negative number gives you a positive number)

The slope of the line in #1 is : 3


#3:

First point : (5, 9) = (x1, y1)

Second point : (5, -2) = (x2, y2)

Equation : (-2 - 9) / (5 - 5)

Solve the equation:

In this case, you don't need to solve anything, because anything divided by zero will give you 0 (5 minus 5 = 0).

The slope of the line in #3 is : 0


8 0
3 years ago
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What is<br> h(-3) given<br> h(x) = -12 – xl?
Flauer [41]

Answer:

= -6

Step-by-step explanation:

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4 0
4 years ago
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I need help on this equation i need to pass the test
emmasim [6.3K]

Answers: x=2 and x=-3

6 0
4 years ago
51 ≤ 15 +6w<br> Solve the inequality for W .<br> Simplify your answer as much as possible.
maks197457 [2]

Answer:

w ≥ 6

Step-by-step explanation:

51 ≤ 15 +6w

51-15 ≤ 6w

36 ≤ 6w

6w ≥ 36

w ≥ 36/6

w ≥ 6

3 0
3 years ago
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