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3 years ago

What are the solutions to x^2-5x+3=0?

Mathematics

1 answer:

Viktor [21]3 years ago

4 0

Answer:

Step-by-step explanation:

You need to use the quadratic equation

x = $\frac{-b +/- \sqrt{b^2 - 4ac} }{2a}$

Givens

a = 1
b = - 5
c = 3

x = $\frac{5 +/- \sqrt{5^2 - 4a*1*3} }{2} \\$

x =( 5 +/- sqrt(25 - 12) ) / 2

x = (5 +/- sqrt(13) )/2

x = (5 + sqrt(13) / 2

x = 4.303 rounded

x = (5 - sqrt(13) ) /2

x = .6972

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Find e^cos(2+3i) as a complex number expressed in Cartesian form.

ozzi

Answer:

The complex number $e^{\cos(2+31)} = \exp(\cos(2+3i))$ has Cartesian form

$\exp\left(\cosh 3\cos 2\right)\cos(\sinh 3\sin 2)-i\exp\left(\cosh 3\cos 2\right)\sin(\sinh 3\sin 2)$ .

Step-by-step explanation:

First, we need to recall the definition of $\cos z$ when $z$ is a complex number:

$\cos z = \cos(x+iy) = \frac{e^{iz}+e^{-iz}}{2}$ .

Then,

$\cos(2+3i) = \frac{e^{i(2+31)} + e^{-i(2+31)}}{2} = \frac{e^{2i-3}+e^{-2i+3}}{2}$ . (I)

Now, recall the definition of the complex exponential:

$e^{z}=e^{x+iy} = e^x(\cos y +i\sin y)$ .

So,

$e^{2i-3} = e^{-3}(\cos 2+i\sin 2)$

$e^{-2i+3} = e^{3}(\cos 2-i\sin 2)$ (we use that $\sin(-y)=-\sin y)$ .

Thus,

$e^{2i-3}+e^{-2i+3} = e^{-3}\cos 2+ie^{-3}\sin 2 + e^{3}\cos 2-ie^{3}\sin 2)$

Now we group conveniently in the above expression:

$e^{2i-3}+e^{-2i+3} = (e^{-3}+e^{3})\cos 2 + i(e^{-3}-e^{3})\sin 2$ .

Now, substituting this equality in (I) we get

$\cos(2+3i) = \frac{e^{-3}+e^{3}}{2}\cos 2 -i\frac{e^{3}-e^{-3}}{2}\sin 2 = \cosh 3\cos 2-i\sinh 3\sin 2$ .

Thus,

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2-i\sinh 3\sin 2\right)$

$\exp\left(\cos(2+3i)\right) = \exp\left(\cosh 3\cos 2\right)\left[ \cos(\sinh 3\sin 2)-i\sin(\sinh 3\sin 2)\right]$ .

5 0

3 years ago

5.3.PS-/ Simplify the expression. - -2.8f + 0.8f – 16 - 9

bezimeni [28]

=-2f-25 is your answer

8 0

3 years ago

5x³+8x²-7x-6

Lady_Fox [76]

Answer:

Step-by-step explanation:

Whether we divide using long division or using synthetic division, the rule is the same: If, after division, there is no remainder (i. e., the remainder is zero), the divisor binomial is a factor or the associated root is indeed a root/zero/solution.

Divide 5x³+8x²-7x-6 by (x+2) using synthetic division. Use the divisor -2 (which comes from letting x+2 = 0):

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

-2 / 5 8 -7 -6

-10 4 6

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

5 -2 -3 0 Since the remainder here is 0, we know that

-2 is a root of 5x³+8x²-7x-6 and that (x+2) is

a factor of 5x³+8x²-7x-6.

Now check out the possibility that (x+1) is a factor of 5x^3 + 8x^2 - 7x - 6:

Use -1 as the divisor in synthetic division:

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

-1 / 5 8 -7 -6

-5 -3 10

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

5 3 -10 4

Since there is a non-zero remainder (4), we can conclude that (x + 1) is NOT a factor of the given polynomial expression.

5 0

3 years ago

A projectile is fired from a cliff feet above the water at an inclination of 45° to the horizontal, with a muzzle velocity of

Anastasy [175]

Answer:

$170 Feet

Step-by-step explanation:

It is very long process

4 0

3 years ago

Help me please!!! i cant do these

LuckyWell [14K]

Answer:

sec(B)=b. ab/bc

sin(B)=c. ac/ab

csc(B)=f. ab/ac

cot(B)=a. bc/ac

tan(B)=e. ac/bc

cos(B)=d. bc/ab

Step-by-step explanation:

SOHCAHTOA

Sine=Opposite/Hypotenuse

Cosine=Adjacent/Hypotenuse

Tangent=Opposite/Adjacent

SHACHOCotAO

Secant=Hypotenuse/Adjacent

Cosecant=Hypotenuse/Opposite

Cotangent=Adjacent/Opposite

6 0

2 years ago