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beks73 [17]
3 years ago
15

What is the range of the function represented by this graph? the graph of a quadratic function with a minimum value at the point

(-3,1)
A. y ≤ 1
B. -4 ≤ y ≤ -2
C. y ≥ 1
D. all real numbers
Mathematics
1 answer:
hammer [34]3 years ago
5 0

Answer:

I think the answer is C......

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Ben ordered a shipping contained so he could ship a model car to Europe.
juin [17]

Answer:

52 ft²

Step-by-step explanation:

Given :

Dimension : 2 by 3 by 4

Surface area of a rectangular prism :

2(wl + lh + wh)

Length = l

Width = w

Height = h

Surface Area = 2[(2*3)+(2*4)+(4*3)]

Surfave Area = 2[6+8+12]

Surface Area = 2(26)

Surface Area = 52 ft²

5 0
3 years ago
2. In AABC, m < B = 22°, m < C = 52° and a = 30. Find the length of b to the nearest tenth.
miskamm [114]

Answer:

232

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Well I'm having troubles with something I have no idea what it is or how to do it. The problem is x^2 -18x+8=0. I only know how
Lubov Fominskaja [6]

Answer:

x = \frac{-(-18) -\sqrt{(-18)^2 -4*1*8}}{2} = \frac{18 -2 \sqrt{73}}{2}= 9 -\sqrt{73}

x = \frac{-(-18) +\sqrt{(-18)^2 -4*1*8}}{2} = \frac{18 +2 \sqrt{73}}{2}= 9 +\sqrt{73}

Step-by-step explanation:

We have the following equation:

x^2 -18 x +8 =0

We can use the quadratic formula given by:

x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}

Where:

a = 1 , b=-18, c =8

And replacing we got:

x = \frac{-(-18) -\sqrt{(-18)^2 -4*1*8}}{2} = \frac{18 -2 \sqrt{73}}{2}= 9 -\sqrt{73}

x = \frac{-(-18) +\sqrt{(-18)^2 -4*1*8}}{2} = \frac{18 +2 \sqrt{73}}{2}= 9 +\sqrt{73}

6 0
3 years ago
Describe the transformations (from the parent function) of this exponential function : y=3(2^x-1)+1
viktelen [127]
For the answer to the question above asking to d<span>escribe the transformations (from the parent function) of this exponential function: y=3(2^x-1)+1

I think</span> <span>"(2 IS THE POWER OF X-1 NOT JUST X)" 

Then why didn't you just write in in parenthesis, like Y=3(2^(X-1))+1 or Y=3 * 2^(X-1)+1? 

The successive transformations are: 
x --> -1 --> 2^ --> *3 --> +1</span>
3 0
4 years ago
HELP please urgent!! - PLEASE CLICK, NEED HELP -
Makovka662 [10]

Answer:

y=3x^2-12x-135

Step-by-step explanation:

The standard form of a quadratic is  y=ax^2+bx+c

We will use the x and y values from each of our 3 points to find a, b, and c.  Filling in the x and y values from each point:

First point (-5, 0):

0=a(-5)^2+b(-5)+c and

0 = 25a - 5b + c

Second point (9, 0):

0=a(9)^2+b(9)+c and

0 = 81a + 9b + c

Third point (8, -39):

-39=a(8)^2+b(8)+c and

-39 = 64a + 8b + c

Use the elimination method of solving systems on the first 2 equations to eliminate the c.  Multiply the first equation by -1 to get:

-25a + 5b - c = 0

81a + 9b + c = 0

When the c's cancel out you're left with

56a + 14b = 0

Now use the second and third equations and elimination to get rid of the c's.  Multiply the second equation by -1 to get:

-81a - 9b - c = 0

64a + 8b + c = -39

When the c's cancel out you're left with

-17a - 1b = -39

Between those 2 bolded equations, eliminate the b's.  Do this by multiplying the second of the 2 by 14 to get:

56a + 14b = 0

-238a - 14b = -546

When the b's cancel out you're left with

-182a = -546 and

a = 3

Use this value of a to back substitute to find b:

56a + 14b = 0 so 56(3) + 14b = 0 gives you

168 + 14b = 0 and 14b = -168 so

b = -12

Now back sub in a and b to find c:

0 = 25a - 5b + c gives you

0 = 75+ 60 + c so

0 = 135 + c and

c = -135

Put that all together into the standard form equation to get

y=3x^2-12x-135

6 0
4 years ago
Read 2 more answers
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