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2 years ago

X^(2)-4x-17=0 Please show work step by step

Mathematics

1 answer:

Dimas [21]2 years ago

7 0

Step-by-step explanation:

It seems here that they are asking us to solve for x

to do this we first need to factor

Since we can't factor this using the normal method we can instead do this

x^2 -4x-17=0

add 4 to both sides as it is a perfect square

x^2 - 4x + 4 = 17 +4

(x-2)^2 = sqrt 21

x-2 = ± 4.58

x -2 = 4.58 x-2 = -4.58

x= 6.58 x=-2.58

Or just say x=2+√21 or x=2−√21

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(2.3×10^4)×(1.5×10^-2)

Wewaii [24]

Answer:

3450

Step-by-step explanation:

(2.3x10^4)x(1.5x10^-2)

(2.3x10,000)x(1.5x.1)

23,000x.15

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2 years ago

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Explain why f(x) = x^2+4x+3/x^2-x-2 is not continuous at x = -1.

liberstina [14]

Answer:

The value of x = -1 makes the denominator of the function equal to zero. That is why this value is not included in the domain of f(x)

Step-by-step explanation:

We have the following expression

$f(x) = \frac{x^2+4x+3}{x^2-x-2}$

Since the division between zero is not defined then the function f(x) can not include the values of x that make the denominator of the function zero.

Now we search that values of x make 0 the denominator factoring the polynomial $x^2-x-2$

We need two numbers that when adding them get as a result -1 and when multiplying those numbers, obtain -2 as a result.

These numbers are -2 and 1

Then the factors are:

$(x-2) (x + 1)$

We do the same with the numerator

$x^2+4x+3$

We need two numbers that when adding them get as a result 4 and when multiplying those numbers, obtain 3 as a result.

These numbers are 3 and 1

Then the factors are:

$(x+3)(x + 1)$

Therefore

$f(x) = \frac{(x+3)(x+1)}{(x-2)(x+1)}$

Note that $\frac{(x+1)}{(x+1)}=1$ only if $x \neq -1$

So since $x = -1$ is not included in the domain the function has a discontinuity in $x = -1$

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3 years ago

Each vertical cross-section of the triangular

Blizzard [7]

Answer:

The vertical height is $h=3\ units$

Step-by-step explanation:

we know that

In the right triangle ABC

see the attached figure

To find the height h apply the Pythagoras Theorem

$c^{2}=a^{2}+b^{2}$

we have

$c=AB=3\sqrt{2}\ units$

$a=BC=6/2=3\ units$

$b=AC=h\ units$

substitute

$(3\sqrt{2})^{2}=3^{2}+h^{2}$

$18=9+h^{2}$

$h^{2}=18-9=9$

$h=3\ units$

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prisoha [69]

blue pencils to yellow pencils is

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