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sattari [20]
2 years ago
12

10 points!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
hammer [34]2 years ago
5 0

Answer:

I thinks its d

Step-by-step explanation:

Hope this helps :)

konstantin123 [22]2 years ago
4 0

Answer:

I'm pretty sure it's D... ;)

Step-by-step explanation:

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The graph of f(x) = |x| is translated 6 units to the right and 2 units up to form a new function. Which statement about the rang
Novay_Z [31]

Answer:

Step-by-step explanation:

The answer is C, the range changes from (y | y > 0) to (y | y > 2)

sorry if i am wrong

7 0
3 years ago
YALL HELP ME<br> By the definition of similar polygons, the lengths of corresponding sides are_____
deff fn [24]
Similar Polygons

Similar polygons are polygons whose corresponding angles are congruent and their corresponding sides are proportional. In other words, Polygons that have the same shape but not necessarily the same size are called similar polygons. 

4 0
3 years ago
Find the area of each regular polygon. Round your answer to the nearest tenth if necessary.
tatuchka [14]

*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*

(21)

Area of a Regular Hexagon: \frac{3\sqrt{3}}{2}(side)^{2} = \frac{3\sqrt{3}}{2}*(\frac{20\sqrt{3} }{3} )^{2} =200\sqrt{3} square units

(22)

Similar to (21)

Area = 216\sqrt{3} square units

(23)

For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:

altitude=\frac{\sqrt{3}}{2}*side

side = \frac{36}{\sqrt{3}}

Hence, area of the hexagon will be: 648\sqrt{3} square units

(24)

Given is the inradius of an equilateral triangle.

Inradius = \frac{\sqrt{3}}{6}*side

Substituting the value of inradius and calculating the length of the side of the equilateral triangle:

Side = 16 units

Area of equilateral triangle = \frac{\sqrt{3}}{4}*(side)^{2} = \frac{\sqrt{3}}{4}*256 = 64\sqrt{3} square units

4 0
3 years ago
Thank you for anyone who knows this
katen-ka-za [31]

Answer:  

  • Part (a) Vertex = (-3, -1)
  • Part (b) Vertex = (4, 5)
  • Part (c) function 2; max value = 5

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

Explanations:

Part (a)

In this case, the vertex is the highest point, so the vertex of function 1 is (-3, -1)

If the parabola was flipped so that it opened upward, instead of downward, then the vertex would be the lowest point.

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

Part (b)

We could graph this function to see where the highest or lowest point is, but there's another way we could find the vertex using algebra.

The given equation is in the form y = ax^2 + bx + c where in this case

  • a = -2
  • b = 16
  • c = -27

The a and b values are plugged into the formula below to get

h = -b/(2a)

h = -16/(2*(-2))

h = -16/(-4)

h = 4

This is the x coordinate the vertex since the vertex is (h,k)

To find k, we plug that value of h into the original function to find its corresponding y value.

y = -2x^2 + 16x - 27

y = -2(4)^2 + 16(4) - 27

y = -2(16) + 16(4) - 27

y = -32 + 64 - 27

y = 5

We found that h = 4 leads to k = 5. The vertex is (h,k) = (4, 5)

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

Part (c)

We'll compare the y values of each vertex

  • Function 1 has a vertex of (-3, 1)
  • Function 2 has a vertex of (4, 5)

The vertex with the larger y value is the winner, so that would be function 2. This parabola reaches a higher peak compared to the other curve.

The max value here is y = 5. In other words, it's the maximum output the function can produce.

If you graphed functions 1 and 2 together on the same xy grid (see below), you'll see that the curve for function 2 reaches a higher peak.

7 0
2 years ago
IfA=−3n+2 andB=5n−7, what is the value of A−B, in simplest form?
alukav5142 [94]

Answer:

Either -4n or 0n but most likely -4n

4 0
3 years ago
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