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3 years ago

Identify the diameter of a circle o.

Mathematics

2 answers:

ella [17]3 years ago

8 0

Answer:

Step-by-step explanation:

What is a diameter?

It is a line going straight through the center of the circle to the opposite side

So it is AC

Andrej [43]3 years ago

6 0

It’s AC, cause it passes through the center and intersects it at two points

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A new bank customer with 3,000 wants to open a money market account the bank is offering a simple interest rate of 1.8% how much

Tcecarenko [31]

Answer:

$3540

Step-by-step explanation:

See explanation

Step-by-step explanation:

The Simple Interest will be calculated as:

= P × R × T

= $3000 × 1.8% × 10.

= $3000 × 0.018 × 10

= $540

The interest that the customer earn in 10 years is $540.

Then, the account balance after 10 years will be:

= Principal + Interest

= $3000 + $540

= $3540

Therefore, the account balance will be $3540

5 0

3 years ago

A triangular prism is 20 millimeters long and has a triangular face with a base of 24 millimeters and a height of 16 millimeters

uysha [10]

Answer:

The surface area of the prism is $1,664\ mm^{2}$

Step-by-step explanation:

we know that

The surface area of the triangular prism is equal to

$SA=2B+PL$

where

B is the area of the triangular face

P is the perimeter of the triangular face

L is the length of the triangular prism

<em>Find the area of the triangular face B</em>

$B=\frac{1}{2}(24)(16)= 192\ mm^{2}$

<em>Find the perimeter of the triangular face P</em>

$P=(24+20+20)= 64\ mm$

we have

$L=20\ mm$

substitute the values

$SA=2(192)+(64)(20)=1,664\ mm^{2}$

7 0

3 years ago

Sum and difference tan(135+30)

Mumz [18]

$\tan(x+y)=\dfrac{\sin(x+y)}{\cos(x+y)}=\dfrac{\sin x\cos y+\sin y\cos x}{\cos x\cos y-\sin x\sin y}=\dfrac{\tan x+\tan y}{1-\tan x\tan y}$

So,

$\tan(135^\circ+30^\circ)=\dfrac{\tan135^\circ+\tan30^\circ}{1-\tan135^\circ\tan30^\circ}$

You should know that $\tan135^\circ=-1$ and $\tan30^\circ=\dfrac1{\sqrt3}$

Putting everything together,

$\tan(135^\circ+30^\circ)=\dfrac{-1+\dfrac1{\sqrt3}}{1-(-1)\left(\dfrac1{\sqrt3}\right)}=\dfrac{1-\sqrt3}{\sqrt3+1}=\sqrt3-2$