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3 years ago

Write the equations in graphing form, then state the vertex of the parabola or the center and radius of the circle.

1 answer:

6 0

All we need is to put this form in the vertex form f(x) = (ax+b)^2 + c

So we have f (x)= 3x^2+12x+11 ....

Let's complete the square (if you aware of it)

f(x)= 3x^2+12x+11 = 3(x^2+4x)+11 = 3(x^2+4x+4-4)+11
= 3([x^2+4x+4]-4)+11 = 3[(x+2)^2-4]+11 =3(x+2)^2 - 12 +11 = 3(x+2)^2 -1

so our form would be:

$f(x)=3(x+2)^2-1$

Here is a parabola with vertex of (-2,-1) and with positive slope (concave up)

I hope that helps!

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Remember to show work and explain. Use the math font.

MrMuchimi

Answer:

$\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}$

Step-by-step explanation:

$\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================$

$1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}$

$\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)$

$--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)$

$--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}$

6 0

3 years ago

Are these ratios equivalent?

vaieri [72.5K]

Answer:

Yes

Step-by-step explanation:

3 medium cars to 1 small car can be made into a fraction like this: $\frac{3}{1}$ or $\frac{1}{3}$ . But you must be consistent if the denominator represents medium cars or small cars.

For this, I'm just going to make medium cars the numerator.

18 medium cars to 6 cars can be made into a fraction like this : $\frac{18}{6}$ .

When you divide both fractions in a calculator (or in your head), you will realize that they both are the same value (equivalent).

$\frac{3}{1}=\frac{18}{6} = 3$

$\frac{1}{3} = \frac{6}{18} = .3333333$

These ratios are equivalent.

4 0

3 years ago

The perimeter of this triangle

Irina-Kira [14]

Step-by-step explanation:

Perimeter of the given triangle = x + x + x + 7 = (3x + 7) units.

8 0

3 years ago

What are the domain and range of x= -2?

nikdorinn [45]

Answer:

The range is all real numbers, because it's a vertical line, and the domain is -2 because it's a vertical line with -2 as it's only x.

Step-by-step explanation:

3 0

2 years ago

6, 9, 12, 15 which pair of numbers has an odd product and an even sum?

amid [387]

The only way for two integers to have an odd product is each integer is odd.
For example, 1*3=3 (odd), 1 and 3 are both odd.
Or,
5*11=55, all of 5,11,55 are odd.
The sum of two odd integers is always even, so the condition of even sum is automatically satisfied when the product is odd.
Out of the four integers, there are only two odd numbers, so choose the pair to be these two odd numbers and you'd get the right answer.

8 0

3 years ago