All categories

3 years ago

Please help with this question

Mathematics

1 answer:

natita [175]3 years ago

8 0

Answer:

$12\sin{(2\theta)}\cos{(\theta)}$

Step-by-step explanation:

The appropriate trig identity is ...

$\sin{a}+\sin{b}=2\sin{\left(\dfrac{a+b}{2}\right)}\cos{\left(\dfrac{a-b}{2}\right)}$

Here, you have a scale factor of 6 and a=3θ, b=θ. Filling in these values gives ...

$6\sin{3\theta}+6\sin{\theta}=12\sin{\left(\dfrac{3\theta+\theta}{2}\right)}\cos{\left(\dfrac{3\theta-\theta}{2}\right)}\\\\=12\sin{(2\theta)}\cos{(\theta)}$

You might be interested in

The run is the blank

scoray [572]

Answer:

rise over run

Step-by-step explanation:

5 0

3 years ago

A rectangular box has a square base. The combined length of a side of the square base, and the height is 20 in. Let x be the len

aniked [119]

Answer:

a. V = (20-x) $x^{2}$ $in^{3}$

b . 1185.185 $in^{3}$

Step-by-step explanation:

Given that:

The height: 20 - x (in )
Let x be the length of a side of the base of the box (x>0)

a. Write a polynomial function in factored form modeling the volume V of the box.

As we know that, this is a rectangular box has a square base so the Volume of it is:

V = h * $x^{2}$ $in^{3}$

<=> V = (20-x) $x^{2}$ $in^{3}$

b. What is the maximum possible volume of the box?

To maximum the volume of it, we need to use first derivative of the volume.

<=> dV / Dx = -3 $x^{2}$ + 40x

Let dV / Dx = 0, we have:

-3 $x^{2}$ + 40x = 0

<=> x = 40/3

=>the height h = 20/3

So the maximum possible volume of the box is:

V = 20/3 * 40/3 *40/3

= 1185.185 $in^{3}$

7 0

3 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago

HELP PLZ PICK ONE OF THE ONES SHOWN:

defon

I am doing with Question Number 2

Let the two numbers be 'x' and 'y'.

According to the question, the product of two numbers is 14.

$x \times y=14$

So, xy=14 (Equation 1)

Now, one of the numbers is 3 ½ times the other.

So, $x = 3\frac{1}{2}\times y$

$x=\frac{7}{2}y$ (Equation 2)

Substituting the value of 'x' from equation 2 in equation 1, we get

$\frac{7}{2}y \times y = 14$

$\frac{7}{2}y^{2} = 14$

$y^{2} = 4$

So, y= 2 and -2

So, $x=\frac{14}{2}$ and $x=\frac{14}{-2}$

x=7 and -7

Numbers are 2,7 and -2,-7.

Sum of numbers 2 and 7 = 9

Sum of numbers -2 and -7 = -9.

6 0

3 years ago

Is 16=3 one, no, or many solutions?

elixir [45]

Answer:

No solutions; parallel

Step-by-step explanation:

The equations x=16 and x=3 both have a slope of 0. So when graphed, they would be parallel and there are no solutions between parallel lines.

3 0

3 years ago