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mrs_skeptik [129]
2 years ago
12

Please i need help !!!​

Mathematics
1 answer:
In-s [12.5K]2 years ago
4 0

Answer:

Jenna is correct.

Step-by-step explanation:

See attached image.

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What’s the answer to this problem?
Delicious77 [7]

Answer:

f (-4)=0

f (-1)=3

f (0)=0

Step-by-step explanation:

The answers are the y value for the given x values on the graph.

3 0
3 years ago
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WHAT ARE THE THREE DIFFERENT TYPES OF ANGLES!!!!!!??
Vikentia [17]
<em>Hello there, and thank you for asking your question here on brainly.

<u>Answer: The three different types of angles are an acute angle, which is an angle that is an angle that measures from 1</u></em>° <em><u>to 89°. A right angle, that specifically measures to 90</u></em>°. <em><u>Finally, an obtuse angle that measures from 91° to 179°. An angle that also measures 180° is a straight line, and is called a straight angle. 

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6 0
3 years ago
Can someone help please!!!!
Nuetrik [128]

66

Step-by-step explanation:

pretty sure it's 66 because the two lengths added together is 66

7 0
3 years ago
HELP!! Algebra help!! Will give stars thank u so much &lt;333
Anna35 [415]

Answers:

  • Part a)  \bf{\sqrt{x^2+(x^2-3)^2}
  • Part b)  3
  • Part c)   2.24
  • Part d)  1.58

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

Work Shown:

Part (a)

The origin is the point (0,0) which we'll make the first point, so let (x1,y1) = (0,0)

The other point is of the form (x,y) where y = x^2-3. So the point can be stated as (x2,y2) = (x,y). We'll replace y with x^2-3

We apply the distance formula to say...

d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d = \sqrt{(0-x)^2+(0-y)^2}\\\\d = \sqrt{(0-x)^2+(-y)^2}\\\\d = \sqrt{x^2 + y^2}\\\\d = \sqrt{x^2 + (x^2-3)^2}\\\\

We could expand things out and combine like terms, but that's just extra unneeded work in my opinion.

Saying d = \sqrt{x^2 + (x^2-3)^2} is the same as writing d = sqrt(x^2-(x^2-3)^2)

-------------------------------------------

Part (b)

Plug in x = 0 and you should find the following

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(0) = \sqrt{0^2 + (0^2-3)^2}\\\\d(0) = \sqrt{(-3)^2}\\\\d(0) = \sqrt{9}\\\\d(0) = 3\\\\

This says that the point (x,y) = (0,3) is 3 units away from the origin (0,0).

-------------------------------------------

Part (c)

Repeat for x = 1

d(x) = \sqrt{x^2 + (x^2-3)^2}\\\\d(1) = \sqrt{1^2 + (1^2-3)^2}\\\\d(1) = \sqrt{1 + (1-3)^2}\\\\d(1) = \sqrt{1 + (-2)^2}\\\\d(1) = \sqrt{1 + 4}\\\\d(1) = \sqrt{5}\\\\d(1) \approx 2.23606797749979\\\\d(1) \approx 2.24\\\\

-------------------------------------------

Part (d)

Graph the d(x) function found back in part (a)

Use the minimum function on your graphing calculator to find the lowest point such that x > 0.

See the diagram below. I used GeoGebra to make the graph. Desmos probably has a similar feature (but I'm not entirely sure). If you have a TI83 or TI84, then your calculator has the minimum function feature.

The red point of this diagram is what we're after. That point is approximately (1.58, 1.66)

This means the smallest d can get is d = 1.66 and it happens when x = 1.58 approximately.

6 0
3 years ago
Determine which relation is a function.
Nezavi [6.7K]
Functions can only have 1 value of 'x'.

Option A is not a function because it has two -5's

Option B is not a function because it has two -2's.

Option C is a function.

Option D is not a function because it has two 6's.
5 0
3 years ago
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