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3 years ago

15

Is f(x) = 10x2 + 10, x > 0 the inverse of g(x) =

exFormula1" title="\sqrt \frac{x-10}{10}" alt="\sqrt \frac{x-10}{10}" align="absmiddle" class="latex-formula">

1 answer:

agasfer [191]3 years ago

6 0

Step-by-step explanation:

$x > - \frac{ 1}{5}$

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Sort the following numbers from least to greatest.<br> 18.5635, 18 13/16, 18.3125

Vikki [24]

Answer:

18.3125

18.5635

18 13/16

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Please help me and show your work

morpeh [17]

Answer:

Step-by-step explanation:

y = 3x + 4

Plugin x = 3 in the equation,

y = 3*3 + 4

= 9 + 4

y = 13

Plugin x = 4 in equation,

y = 3*4 + 4

= 12 + 4

y = 16

Plugin x = 5 in the equation,

y = 3*5 + 4

= 15 + 4

y = 19

2) y =x - 7

Plugin x = 10 in the equation

y = 10 - 7

y = 3

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2 years ago

Find the polynomial of minimum degree, with real coefficients, zeros at

drek231 [11]

Answer:

$\huge\boxed{p(x)=4x^3-20x^2+4x+300}$

Step-by-step explanation:

$\text{If}\ x=4\pm3i\ \text{and}\ x=-3\ \text{are the zeros of a polynomial, then it has a form:}\\\\p(x)=\bigg(x-(4-3i)\bigg)\bigg(x-(4+3i)\bigg)\bigg(x-(-3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x-4+3i)(x-4-3i)(x+3)\bigg(r(x)\bigg)\\\\p(x)=\underbrace{\bigg((x-4)+3i\bigg)\bigg((x-4)-3i\bigg)}_{\text{use}\ (a+b)(a-b)=a^2-b^2}(x+3)\bigg(r(x)\bigg)\\\\p(x)=\bigg((x-4)^2-(3i)^2\bigg)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2$

$p(x)=(x^2-2(x)(4)+4^2-3^2i^2)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ i^2=-1\\\\p(x)=(x^2-8x+16-9(-1))(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+16+9)(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+25)(x+3)\bigg(r(x)\bigg)\qquad\text{use FOIL}:\ (a+b)(c+d)=ac+ad+bc+bd\\\\p(x)=\bigg((x^2)(x)+(x^2)(3)+(-8x)(x)+(-8x)(3)+(25)(x)+(25)(3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x^3+3x^2-8x^2-24x+25x+75)\bigg(r(x)\bigg)\qquad\text{combine like terms}\\\\p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)$

$\text{The y-intercept is at 300}.\\\\\text{For}\ w(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0\\\\\text{y-intercept is}\ a_0\\\\\text{Therefore for}\ p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)\\\\\text{y-intercet is}\ 75\bigg(r(x)\bigg)\\\\75\bigg(r(x)\bigg)=300\qquad\text{divide both sides by 75}\\\\r(x)=4\\\\\text{Finally:}\\\\p(x)=(x^3-5x^2+x+75)(4)\qquad\text{use the distributive property}\\\\p(x)=(x^3)(4)+(-5x^2)(4)+(x)(4)+(75)(4)\\\\p(x)=4x^3-20x^2+4x+300$

7 0

3 years ago

The function h(x)=x^2+3 and g(x)=x^2-6 if g(x)=h(x)+k what is the value of k

riadik2000 [5.3K]

Answer:

hello : k = - 9

Step-by-step explanation:

h(x) =x²+3 and g(x) = x² - 6.....(*)

g(x) = x² +3 + k.....(**)

by (*) and (**) : 3+k = -6

3+k - 3 = - 6 - 3

k = - 9

8 0

3 years ago

Hi what is the answer for this?

saw5 [17]

This is a common factor problem.

Pencils come in a pack of 12
Erasers come in a pack of 10

First, break the number into their prime factors(the idea is that we will break the number down into its smallest multiples, which are prime numbers):
10 = 2 * 5
12 = 2 * 2 *3

So now we take the unique multiples of each number, and when we multiply them together, we will get the smallest number that both 10 and 12 can be divided into(this is what the problem is asking for)
We have (2*2*3) that comes from 12, and the only unique number that comes from the 10 is (5)
So now, we multiply:
2*2*3*5=60
However, this isn't exactly out answer. Now we have to divide our answer by the number of each this in the pack to know how many packs to buy.
60/12=5 packs of pencils
60/10= 6 packs of erasers

I hope this helps. Let me know if you have any questions!!

5 0

3 years ago

Read 2 more answers