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Snowcat [4.5K]
3 years ago
6

What is the degree of the equation?

Mathematics
2 answers:
wolverine [178]3 years ago
4 0

Answer:

1 (give brainliest plss)

Step-by-step explanation:

Step-1 : Multiply the coefficient of the first term by the constant 5 • -8 = -40

Step-2 : Find two factors of -40 whose sum equals the coefficient of the middle term, which is -3 .

-40 + 1 = -39

-20 + 2 = -18

-10 + 4 = -6

-8 + 5 = -3 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -8 and 5

5x2 - 8x + 5x - 8

Step-4 : Add up the first 2 terms, pulling out like factors :

x • (5x-8)

Add up the last 2 terms, pulling out common factors :

1 • (5x-8)

Step-5 : Add up the four terms of step 4 :

(x+1) • (5x-8)

Which is the desired factorization

Varvara68 [4.7K]3 years ago
4 0

\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Quadratic Equations.

Degree of a equation represents the highest power of the varible in the equation, For quadratic equation the degree is 2.

Leading Coefficient are the constant numbers that are significantly multiplied with the variables in the equation.

==> 5x^2 - 3x - 8 = 0

==> 5x^2 -8x +5x -8=0

==> x(5x-8) +1(5x-8)=0

==> (5x-8)(x+1) = 0

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Given that f(x) = 3x + 1 and g(x) = the quantity of 4x plus 2 divided by 3 , solve for g(f(0)). (5 points) 2 6 8 9
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3 years ago
How do you find the <br> y-intercept and the slope for number 11
Sophie [7]
The first step is to find the slope. Use the slope formula

m = (y2-y1)/(x2-x1)

The two points are (x1,y1) = (-1,5) and (x2,y2) = (2,-1)

So,
x1 = -1
y1 = 5
and
x2 = 2
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will be plugged into the slope formula to get...
m = (y2-y1)/(x2-x1)
m = (-1-5)/(2-(-1))
m = (-1-5)/(2+1)
m = (-6)/(3)
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The slope is -2

Use m = -2 and one of the points to find the y intercept b. I'll use the point (x,y) = (-1,5) ---> x = -1, y = 5

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The y intercept is 3

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

m = -2 is the slope
b = 3 is the y intercept

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6 0
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Find all values of the angle θ (in radians, with 0 ≤ θ &lt; 2π) for which the matrix a = cos θ −sin θ sin θ cos θ has real eigen
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\cos\theta-\lambda=\pm\sqrt{-\sin^2\theta}

\lambda=\cos\theta\pm\sqrt{-\sin^2\theta}

\sin^2\theta\ge0 for all values of \theta, so we need to have \sin\theta=0 in order for \lambda to be real-valued. This happens for

\sin\theta=0\implies\theta=n\pi

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Answer:

1 Mirror images across the y axis

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Step-by-step explanation:

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