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Galina-37 [17]
3 years ago
14

Find the axis of symmetry for this parabola:

Mathematics
1 answer:
lara31 [8.8K]3 years ago
6 0

Answer:

x = - 1

Step-by-step explanation:

The equation of the axis of symmetry for a parabola in standard form

y = ax² + bx + c : a ≠ 0 is found using

x = - \frac{b}{2a}

y = - x² - 2x - 5 ← is in standard form

with a = - 1 and b = - 2, thus equation of axis of symmetry is

x = - \frac{-2}{-2} = - 1

Equation of axis of symmetry is x = - 1

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How many real and complex roots exists for a polynomial?
solmaris [256]

Step-by-step explanation:

whenever a complex number is a root of a polynomial with real coefficients, its complex conjugate is also a root of that polynomial. as an example, we'll find the roots of the polynomial..

x^5 - x^4 + x^3 - x^2 - 12x + 12.

the fifth-degree polynomial does indeed have five roots; three real, and two complex.

7 0
3 years ago
What two numbers add to 8 and multiply to 28
anygoal [31]
Find numbers that multiply to 28 and add them to see if they add to 8
28=
1 and 28=29 not 8
2 and 14=16 not 8
4 and 7=11 not 8
that's it'
no 2 numbers
we must use quadratic formula

x+y=8
xy=28

x+y=8
subtract x fromb oths ides
y=8-x
subsitute
x(8-x)=28
distribute
8x-x^2=28
add x^2 to both sides
8x=28+x^2
subtract 8x
x^2-8x+28=0

if you have
ax^2+bx+c=0 then x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}
so if we have
1x^2-8+28=0 then
a=1
b=-8
c=28

x=\frac{-(-8)+/- \sqrt{(-8)^{2}-4(1)(28)} }{2(1)}
x=\frac{8+/- \sqrt{-48} }{2}
x=\frac{8+/- 4 \sqrt{-3} }{2}
x=4+/- 2 \sqrt{-3}
x=4+/- 2i \sqrt{3}
there are no real numbers that satisfy this
5 0
3 years ago
Please answer this correctly for a brainliest ☺️
Dimas [21]
Hello I think the answer would be three.
5 0
3 years ago
Each home room has to have the same number of girls and boys in each home room.
agasfer [191]
4 for part A 
17 for part B
3 0
3 years ago
A particle moves along the curve y = 5x^2 – 1 in such a way that the y value is decreasing at the rate of 2 units per second. At
Shtirlitz [24]

Answer:

The correct option is;

Increasing one fifth unit/sec

Step-by-step explanation:

The equation that gives the curve of the particle of the particle is y = 5·x² - 1

The rate of decrease of the y value dy/dt = 2 units per second

We have;

dy/dx = dy/dt × dt/dx

dy/dx = 10·x

dy/dt = 2 units/sec

dt/dx = (dy/dx)/(dy/dt)

dx/dt = dy/dt/(dy/dx) = 2 unit/sec/(10·x)

When x = 1

dx/dt = 2/(10·x) = 2 unit/sec/(10 × 1) = 1/5 unit/sec

dx/dt = 1/5 unit/sec

Therefore, x is increasing one fifth unit/sec.

8 0
3 years ago
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