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Rasek [7]
3 years ago
9

Factor each quadratic and find the roots, Show all work

Mathematics
1 answer:
mojhsa [17]3 years ago
8 0
Don’t check those links
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Need an answer for question 7 HURRY PLEASE 20 POINTS
Andrei [34K]

Answer:

2x^2 = 6x - 5.

-x^2 - 10x = 34.

These have only complex roots/

Step-by-step explanation:

3x^2 - 5x = -8

3x^2 - 5x + 8 = 0

There are complex roots if the discriminant  9b^2 - 4ac) is negative.

Here the discriminant D = (-5)^2 - 4*-5*8 = 25 + 160

This is positive so the roots are real.

2x^2 = 6x - 5

2x^2 - 6x + 5 = 0

D = (-6)^2 - 4*2*5 =  36 - 40 = -4

So this has no real roots only complex ones.

12x = 9x^2 + 4

9x^2 - 12x + 4 = 0

D = (-12)^2 - 4*9 * 4 = 144 - 144 = 0.

- Real roots.

-x^2 - 10x = 34

x^2 + 10x + 34 = 0

D = (10)^2 - 4*1*34 = 100 - 136 = -36.

No real roots = only complex roots.

7 0
3 years ago
In the triangle pictured, let A, B, C be the angles at the three vertices, and let a,b,c be the sides opposite those angles. Acc
Troyanec [42]

Answer:

Step-by-step explanation:

(a)

Consider the following:

A=\frac{\pi}{4}=45°\\\\B=\frac{\pi}{3}=60°

Use sine rule,

\frac{b}{a}=\frac{\sinB}{\sin A}
\\\\=\frac{\sin{\frac{\pi}{3}}
}{\sin{\frac{\pi}{4}}}\\\\=\frac{[\frac{\sqrt{3}}{2}]}{\frac{1}{\sqrt{2}}}\\\\=\frac{\sqrt{2}}{2}\times \frac{\sqrt{2}}{1}=\sqrt{\frac{3}{2}}

Again consider,

\frac{b}{a}=\frac{\sin{B}}{\sin{A}}
\\\\\sin{B}=\frac{b}{a}\times \sin{A}\\\\\sin{B}=\sqrt{\frac{3}{2}}\sin {A}\\\\B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Thus, the angle B is function of A is, B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

Now find \frac{dB}{dA}

Differentiate implicitly the function \sin{B}=\sqrt{\frac{3}{2}}\sin{A} with respect to A to get,

\cos {B}.\frac{dB}{dA}=\sqrt{\frac{3}{2}}\cos A\\\\\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos A}{\cos B}

b)

When A=\frac{\pi}{4},B=\frac{\pi}{3}, the value of \frac{dB}{dA} is,

\frac{dB}{dA}=\sqrt{\frac{3}{2}}.\frac{\cos {\frac{\pi}{4}}}{\cos {\frac{\pi}{3}}}\\\\=\sqrt{\frac{3}{2}}.\frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}\\\\=\sqrt{3}

c)

In general, the linear approximation at x= a is,

f(x)=f'(x).(x-a)+f(a)

Here the function f(A)=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{A}]

At A=\frac{\pi}{4}

f(\frac{\pi}{4})=B=\sin^{-1}[\sqrt{\frac{3}{2}}\sin{\frac{\pi}{4}}]\\\\=\sin^{-1}[\sqrt{\frac{3}{2}}.\frac{1}{\sqrt{2}}]\\\\\=\sin^{-1}(\frac{\sqrt{2}}{2})\\\\=\frac{\pi}{3}

And,

f'(A)=\frac{dB}{dA}=\sqrt{3} from part b

Therefore, the linear approximation at A=\frac{\pi}{4} is,

f(x)=f'(A).(x-A)+f(A)\\\\=f'(\frac{\pi}{4}).(x-\frac{\pi}{4})+f(\frac{\pi}{4})\\\\=\sqrt{3}.[x-\frac{\pi}{4}]+\frac{\pi}{3}

d)

Use part (c), when A=46°, B is approximately,

B=f(46°)=\sqrt{3}[46°-\frac{\pi}{4}]+\frac{\pi}{3}\\\\=\sqrt{3}(1°)+\frac{\pi}{3}\\\\=61.732°

8 0
3 years ago
I need help with this because I'm not very good at understanding this.
HACTEHA [7]

Answer:

the initial value is 2

Step-by-step explanation:

"y=mx+b" is the equation

"b" is the initial value, and there is no "b" value

4 0
3 years ago
The product of two and x plus the quotient of w and five
Nataly [62]

Answer:

2x+w/5

Step-by-step explanation:

(2*x)+(w/5)

2x +w/5

8 0
3 years ago
What is the value of x?
IrinaK [193]

Answer:

27

Step-by-step explanation:

104-77=27

3 0
3 years ago
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