Work out the surface area?

Calculate the interest produced by a principal of $ 4,500 at 5% annual simple interest in 8 months.

enyata [817]

Answer:

4,500 x 5 = 22,500

4,500 divided by 5 = 900

4,500 plus 5 = 4,505

4,500 minus 5 = 4,495

Step-by-step explanation:

it is either one of those that u have to choose from good luck

5 0

4 years ago

PLEASE HELP ASAP

KATRIN_1 [288]

Answer:

752.884 cubic feet

Step-by-step explanation:

Brainliest?

5 0

3 years ago

Find the center and radius of the circle x^2+y^2-10x-12y+45=0.

BARSIC [14]

Answer:

The center is (5, -6) with a radius of r

Step-by-step explanation:

First put the equation in Standard Form for a circle:

(x + a)² + (y + b)² = r²

Now, Group the term

x² + y² - 10x - 12y + 45 = 0

x² - 10x + y² - 12y = - 45

Then, The x and y terms must be perfect squares

x² - 10x + 25 + y² - 12y + 36 = - 45 + 25 + 36

add constant both sides we get,

x² - 10x + 25 + y² - 12y + 36 = 16

Now, Simplify

(x - 5)² + (y + 6)² = 16

√(x - 5)² + (y + 6)² = √16

x - 5 + y + 6 = 4

Here, (a,b) = (5, -6) and r² = 16, with r = 4

Thus, The center is (5, -6) with a radius of r

-TheUnknownScientist 72

6 0

3 years ago

PLS HELP WILL MARK YOU BRAINLIEST

Sunny_sXe [5.5K]

Answer:

a2 + 4a + 16

Step-by-step explanation:

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value

BRAINLIEST?

4 0

3 years ago

Consider the planes 4x+3y+4z=1 and 4x+4z=0. (A) Find the unique point P on the y−axis which is on both planes. ( 0 equation edit

Nana76 [90]

Answer:

Step-by-step explanation:

A) From the order of the exercise we already know that the intersection points lies on the Y-axis, so its coordinates are P(0;y;0). In order to find it, we only need to substitute the equation 4x+4z=0 into the equation 4x+3y+4z=1. Then,

1=4x+3y+4z = 3y + (4x+4z)= 3y+0.

From the expression above it is easy to obtain that y=1/3, and the intersection point is P(0;1/3;0).

B) To obtain the parallel vector to both planes we use the cross product of the normal vector of the planes.

$\left[\begin{array}{ccc}i&j&k\\4&3&4\\4&0&4\end{array}\right] = 12i-0j+12k$

As we want a unit vector, we must calculate the modulus of u:

$|u|=\sqrt{12^2+0^2+12^2} = \sqrt{2\cdot 12^2}=12\sqrt{2}$ .

Thus, the wanted vector is $\frac{u}{12\sqrt{2}}$ . Therefore,

$u = \frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}k = \frac{\sqrt{2}}{2}i-\frac{\sqrt{2}}{2}k$ .

C) In order to obtain the vector equation of the intersection line of both planes, we just need to put together the above results.

$r = \frac{1}{3}j +\lambda \left( \frac{\sqrt{2}}{2}i- \frac{\sqrt{2}}{2}k \right)$

where $\lambda$ is a real number.

8 0

4 years ago