<img src="https://tex.z-dn.net/?f=F%28X%29%20%3D%20%5Cfrac%7B%28X-3%29%28X%2B1%29%7D%7B%28X%2B3%29%7D" id="TexFormula1" title="F

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4 years ago

(X) = \frac{(X-3)(X+1)}{(X+3)}" alt="F(X) = \frac{(X-3)(X+1)}{(X+3)}" align="absmiddle" class="latex-formula">
What is the:
Domain:
Holes:
VA:
HA:
OA: Roots:
Y-Intercepts:

Mathematics

1 answer:

Nezavi [6.7K]4 years ago

7 0

Answer:

Step-by-step explanation:

Given function is

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

Domain of the function: (-∞,-3)∪(-3,∞) x | x ≠ -3

Holes: none. No common factors in numerator and denominator

VA: x = -3 as denominator is (x+3)

HA: No horizontal asymptote because degree of the denominator is less than numerator

OA: Roots are x = 3, -1 [(x-3) and (x+1) are the factors]

Y-Intercept: (0, -1) (by putting x = 0 in the given function)

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Answer:

Her goal : 1350

What she has now: 174

she earns 84$ a month

answer is 14 months because 84* 14 is 1176 and if u add 174 $ to that it will give u 1350.

Step-by-step explanation:

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Step-by-step explanation:

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The function, f(x) = –2x2 + x + 5, is in standard form. The quadratic equation is 0 = –2x2 + x + 5, where a = –2, b = 1, and c =

avanturin [10]

Given:

The function, f(x) = -2x^2 + x + 5

Quadratic equation: 0 = -2x^2 + x +5
where a = -2
b = 1
c = 5

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To solve for the zeros of the quadratic function, use this formula:

x = ( -b +-√ (b^2 - 4ac) ) / 2a

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3 years ago

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Find the lengths of the sides of the triangle PQR. P(2, −3, −4), Q(8, 0, 2), R(11, −6, −4) |PQ| = Incorrect: Your answer is inco

Aloiza [94]

Answer:

the length PQ is 9 units,the length QR is 9 units,the length PR is 9.48 units,the triangle is not a right triangle,this is a isosceles triangle

Step-by-step explanation:

Hello, I think I can help you with this

If you know two points, the distance between then its given by:

$P1(x_{1},y_{1},z_{1} ) \\P2(x_{2},y_{2},z_{2})\\\\d=\sqrt{(x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}+(z_{2}-z_{1} )^{2} }$

Step 1

use the formula to find the length PQ

Let

P1=P=P(2, −3, −4)

P2=Q=Q(8, 0, 2)

$d=\sqrt{(8-2)^{2} +(0-(-3))^{2}+(2-(-4))^{2}} \\ d=\sqrt{(6)^{2} +(3)^{2}+(6 )^{2}}} \\d=\sqrt{36+9+36}\\d=\sqrt{81} \\d=9\\$

the length PQ is 9 units

Step 2

use the formula to find the length QR

Let

P1=Q=Q(8, 0, 2)

P2=R= R(11, −6, −4)

$d=\sqrt{(11-8)^{2} +(6-0))^{2}+(-4-2 )^{2}} \\\\\\d=\sqrt{(3)^{2} +(6)^{2}+(-6 )^{2}}} \\d=\sqrt{9+36+36}\\d=\sqrt{81} \\d=9\\$

the length QR is 9 units

Step 3

use the formula to find the length PR

Let

P1=P(2, −3, −4)

P2=R= R(11, −6, −4)

$d=\sqrt{(11-2)^{2} +(-6-(-3)))^{2}+(-4-4 )^{2}} \\\\\\d=\sqrt{(9)^{2} +(-6+3)^{2}+(-4-(-4) )^{2}}} \\d=\sqrt{81+9+0}\\d=\sqrt{90} \\d=9.48\\$

the length PR is 9.48 units

Step 4

is it a right triangle?

you can check this by using:

$side^{2} +side^{2}=hypotenuse ^{2}$

Let

side 1=side 2= 9

hypotenuse = 9.48

Put the values into the equation

$9^{2} +9^{2} =9.48^{2}\\ 81+81=90\\162=90,false$

Hence, the triangle is not a right triangle

Step 5

is it an isosceles triangle?

In geometry, an isosceles triangle is a type of triangle that has two sides of equal length.

Now side PQ=QR, so this is a isosceles triangle

Have a great day

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3 years ago

2.2.17<br> Given the function H(v) = 10v +9, find each of the following.<br> H(3) =

Afina-wow [57]

Answer:

H(3)=39

Step-by-step explanation:

H(3) =10 (3) + 9

=30 +9

=39

Hope this helps...

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3 years ago