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2 years ago

Plz help me!!!!!!!!!!!!

Mathematics

1 answer:

Mashutka [201]2 years ago

7 0

Answer: 3, -3, 3i, -3i

Step-by-step explanation:

$x^4-81=0\\\\Factor:\\(x^2-9)(x^2+9)=0\\(x-3)(x+3)(x^2+9)=0\\\\\text{Apply the Zero Product Property:}\\x-3=0\qquad x+3=0\qquad x^2+9=0\\\boxed{x=3}\qquad \qquad \boxed{x=-3}\qquad \quad x^2=-9\\.\qquad \qquad \qquad \qquad \qquad \qquad x=\sqrt{-9}\\.\qquad \qquad \qquad \qquad \qquad \qquad \boxed{x=\pm 3i}$

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Use the Parabola tool to graph the quadratic function. f(x)=2x^2−12x+19 Graph the parabola by first plotting its vertex and then

Debora [2.8K]

To find the vertex of our parabola, we are going to use the vertex formula: For a quadratic of the form $f(x)=x^2+bx+c$ it vertex $(h,k)$ is given by $h= \frac{-b}{2a}$ , $k=f(h)$ .
We can infet from our parabola that $a=2$ and $b=-12$ . So lets replace those values in our formula:
$h= \frac{-b}{2a}$
$h= \frac{-(-12)}{2(2)}$
$h= \frac{12}{4}$
$h=3$
$k=f(h)=2(3)^2-12(3)+19$
$k=2(9)-36+19$
$k=18-17$
$k=1$
The vertex $(h,k)$ of our parabola is $(3,1)$

Now to find our second point, we are going to evaluate our parabola at $x=0$
$f(0)=2(0)^2-12(0)+19$
$f(0)=0+0+19$
$f(0)=19$
Our second point is $(0,19)$

We can conclude that we can graph the parabola $f(x)=2x^2-12x+19$ using its vertex $(3,1)$ and the point $(0,19)$ as follows:

7 0

2 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

4 0

2 years ago

What are the solutions? 8x^2-12x-23=-5

erastova [34]

Answer:

x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

Step-by-step explanation:

Solve for x over the real numbers:

8 x^2 - 12 x - 23 = -5

Divide both sides by 8:

x^2 - (3 x)/2 - 23/8 = -5/8

Add 23/8 to both sides:

x^2 - (3 x)/2 = 9/4

Add 9/16 to both sides:

x^2 - (3 x)/2 + 9/16 = 45/16

Write the left hand side as a square:

(x - 3/4)^2 = 45/16

Take the square root of both sides:

x - 3/4 = (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

x = 3/4 + (3 sqrt(5))/4 or x - 3/4 = -(3 sqrt(5))/4

Add 3/4 to both sides:

Answer: x = 3/4 + (3 sqrt(5))/4 or x = 3/4 - (3 sqrt(5))/4

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2 years ago

Hiroto paid $4.28 in sales tax for the items that he purchased.

Katena32 [7]

Answer:

$68.48

Step-by-step explanation:

You want to find cost such that ...

6.25% × cost = $4.28

Dividing by the coefficient of cost, we get ...

cost = $4.28/0.0625 = $68.48

8 0

2 years ago

The inequality describes the range of monthly average temperatures T in degrees Fahrenheit at a certain location. Find an equiva

kolezko [41]

Answer:

Now if the high and low monthly average temperatures satisfy the inequality, then the , monthly averages are always within 22 degrees of 43°F.

Step-by-step explanation:

The inequality describes the range of monthly average temperatures T in degrees Fahrenheit at a certain location.

The inequality expression is given as:

$|T-43|\leq 22$

now this expression could also be expressed as:

$-22\leq T-43\leq22\\\\-22+43 \leq T \leq 22+43\\\\21\leq T\leq 65$

Now if the high and low monthly average temperatures satisfy the inequality, then the , monthly averages are always within 22 degrees of 43°F.

( As the difference is 22 degrees to the left and right)

5 0

3 years ago