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2 years ago

Find the coordinates of the midpoint of1. (6,3) and (4,5)

Mathematics

1 answer:

expeople1 [14]2 years ago

8 0

Answer:

(5,4)

Step-by-step explanation:

midpoint = (6 + 4 / 2 , 3 +5 / 2)

= (5,4)

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5<br> 3<br> 2<br> The total area of the rectangular prism shown is

marishachu [46]

Answer:

Step-by-step explanation:

You multipy 5 by 3 by 2

8 0

3 years ago

The coordinate −6 has a weight of 3, and the<br> coordinate 2 has a weight of 1

Damm [24]

The weight average of the coordinates is -4

<h3>How to determine the weight average?</h3>

The complete question is given as:

The coordinate -6 has a weight of 3 and the coordinate 2 has a weight of 1. And we need to calculate the weight average

The given parameters are:

Coordinate -6 has a weight of 3
Coordinate 2 has a weight of 1.

The weight average is then calculated as:

Weight average = Sum of (Weigh * Coordinate)/Sum of Weights

So, we have:

Weight average = (-6 * 3 + 2 * 1)/(3 +1)

Evaluate the products

Weight average = (-18 + 2)/(3 +1)

Evaluate the sum

Weight average = -16/4

Evaluate the quotient

Weight average = -4

Hence, the weight average of the coordinates is -4

Read more about average at

brainly.com/question/20118982

#SPJ1

<u>Complete question</u>

The coordinate -6 has a weight of 3 and the coordinate 2 has a weight of 1. Calculate the weight average

4 0

2 years ago

Combine like terms to create an equivalent expression: -n + (-4) - (-4n) + 6

Taya2010 [7]

Answer:

3n + 2

Step-by-step explanation:

-n + 4n -4 + 6

3n + 2

6 0

3 years ago

Find a rational number between 8 and 9

otez555 [7]

<h2>⟰ ANSWER ⟰</h2>

<h2> HOPE IT HELPS U!!!!(:</h2>

4 0

3 years ago

Prove whether the following are identities 2tanh 1/2x / 1−tanh^2 1/2 x = sinh x

Tanzania [10]

Recall that

$\cosh^2(x) - \sinh^2(x) = 1$

Dividing both sides by cosh²(x) gives

$1 - \tanh^2(x) = \mathrm{sech}^2(x)$

Also, recall the identity

$\sinh(2x) = 2\sinh(x)\cosh(x)$

Then

$\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \dfrac{2\tanh\left(\frac x2\right)}{\mathrm{sech}^2\left(\frac x2\right)} \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\tanh\left(\dfrac x2\right)\cosh^2\left(\dfrac x2\right) \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\sinh\left(\dfrac x2\right)\cosh\left(\dfrac x2\right) \\\\\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \sinh(x)$

4 0

3 years ago

Find the coordinates of the midpoint of1. (6,3) and (4,5)​

Find the coordinates of the midpoint of1. (6,3) and (4,5)