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Artist 52 [7]
3 years ago
8

HELPP IM ALMOST DONE PLEASEEEE

Mathematics
2 answers:
Anna71 [15]3 years ago
8 0

Answer:

number A

the template does not have the needed triangle for the other numvers

sdas [7]3 years ago
4 0
The answer is AAAAAAAAAA
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PLEASE HELP ASAP!!!!! THANK YOU~
VashaNatasha [74]

Answer:

y=4x^2-8x-4

Step-by-step explanation:

<u>Quadratic Function</u>

The quadratic function can be expressed in the following form:

y=a(x-x_1)(x-x_2)

Where a is a real number different from 0, and x1, x2 are the roots or zeroes of the function.

From the conditions stated in the problem, we know

x_1=1+\sqrt{2}, \ x_1=1-\sqrt{2}

Substitute in the general formula above:

y=a[x-(1+\sqrt{2})][x-(1-\sqrt{2})]

Operate the indicated product

y=a(x^2-2x-1)

To find the value of a, we use the y-intercept which is the value of y when x=0, thus

y=a(0^2-2(0)-1)=-4

It follows that

a=4

Thus, the required quadratic function is

y=4(x^2-2x-1)

Or, equivalently

\boxed{y=4x^2-8x-4}

5 0
4 years ago
The cross section of a square pyramid taken perpendicular to the base that passes through the top vertex produces which two-dime
4vir4ik [10]
The area of the cross section is in the shape of a triangle.
8 0
4 years ago
Read 2 more answers
I need help please!!!
blondinia [14]

Answer:

the correct answerr is b i can see it clearly but from what i see it s b

Step-by-step explanation:

6 0
3 years ago
What is the value of x?​
vladimir1956 [14]

Answer:

Step-by-step explanation:

x + 112 + 133+128+100+120= 720

x + 593 = 720

x = 127

3 0
3 years ago
Solve x^3 = 125/27.
cupoosta [38]
C. 5/3

5^3= 125
3^3 = 27
7 0
3 years ago
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