Please help me with this math problem Evaluate C/2 - 3 + 6/D when c=14 and d=3

Mathematics

2 answers:

KATRIN_1 [288]3 years ago

8 0

I hope this helps you!

lbvjy [14]3 years ago

7 0

$\frac{c}{2} - 3 + \frac{6}{d } = \\ = \frac{14}{2} - 3 + \frac{6}{3} = \\ = 7 - 3 + 2 = \\ = 6$

You might be interested in

Can anyone help me out with this math problem?

inna [77]

The slope is negative

8 0

3 years ago

A case of 24 tennis balls weighs 3 pounds. How much would a shipment of 2560 tennis balls weigh? Please write

kolbaska11 [484]

The weight of a shipment of 2560 balls weighs 320 pounds.

<h3>How much would a shipment of 2560 tennis balls weigh?</h3>

We know that 24 tennis balls weigh 3 pounds. Then the weight of a single tennis ball is:

W = (3 lb)/(24 balls) = (1/8) pounds per ball.

To get the weight of 2560 tennis balls, we just need to multiply the number of tennis balls by the weight of a single ball, then we get:

Total weight = 2560*(1/8) pounds per ball = 320 pounds.

The weight of a shipment of 2560 balls weighs 320 pounds.

If you want to learn more about weight:

brainly.com/question/25973294

#SPJ1

8 0

2 years ago

Which equations for the measures of the unknown

MAXImum [283]

Answer: Option B, Option C, Option E

Step-by-step explanation:

The options written correctly, are:

$A)x = cos^{-1}(\frac{a}{c})\\\\B) x = sin^{-1}(\frac{c}{b})\\\\C)x = tan^{-1}(\frac{c}{a})\\\\D) y = sin^{-1}(\frac{a}{c})\\\\ E)y = cos^{-1}(\frac{c}{b})$

For this exercise you need to use the following Inverse Trigonometric Functions:

$1)\ sin^{-1}(x)\\\\2)\ cos^{-1}(x)\\\\3)\ tan^{-1}(x)\\\\$

When you have a Right triangle (a triangle that has an angle that measures 90 degrees) and you know that lenght of two sides, you can use the Inverse Trigonometric Functions to find the measure of an angle $\alpha$ :

$1)\alpha = sin^{-1}(\frac{opposite}{hypotenuse}) \\\\2)\ \alpha =cos^{-1}(\frac{adjacent}{hypotenuse})\\\\3)\ \alpha=tan^{-1}(\frac{opposite}{adjacent})$

Therefore, the conclusion is that the angles "x" and "y" can be found with these equations:

$x=sin^{-1}(\frac{c}{b})\\\\x= cos^{-1}(\frac{a}{b})\\\\x=tan^{-1}(\frac{c}{a})\\\\\\ y=sin^{-1}(\frac{a}{b})\\\\y=cos^{-1}(\frac{c}{b})\\\\y=tan^{-1}(\frac{a}{c})$