4 tan 45° +(2 sin 45°) (6 cos 45°)

Mathematics

1 answer:

Paladinen [302]3 years ago

4 0

Answer:

value of 4 tan 45° +(2 sin 45°) (6 cos 45°) is 10.

Step-by-step explanation:

To solve this question we need to know value of

tan 45° = 1

sin 45° = $1/\sqrt{2}$

cos 45° = $1/\sqrt{2}$

we will be using above value sin the given expression as shown

$4 tan 45 +(2 sin 45) (6 cos 45) = 4*1 + (2*1/\sqrt{2} )(6*1/\sqrt{2})\\ = 4 + (2/\sqrt{2}) (6/\sqrt{2})\\= 4 + (2*6)/(\sqrt{2}*\sqrt{2})\\= 4 + (12/2)\\= 4 + 6 = 10$

Thus, value of 4 tan 45° +(2 sin 45°) (6 cos 45°) is 10.

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Find the midpoint of the line segment joining the points (10,7) and (negative 2,negative 7).

Basile [38]

The midpoint of the line segment joining the points (10,7) and (negative 2,negative 7) is (4, 0)

<h3><u>Solution:</u></h3>

Given that line segment joining the points (10,7) and (negative 2, negative 7)

Thus the two points are (10, 7) and (-2, -7)

The midpoint of two points $(x_1, y_1)$ and $(x_2, y_2)$ is given as:

$M(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

$\text {Here } x_{1}=10 \text { and } y_{1}=7 \text { and } x_{2}=-2 \text { and } y_{2}=-7$

Substituting the values in formula, we get

$\begin{array}{l}{M(x, y)=\left(\frac{10+(-2)}{2}, \frac{7+(-7)}{2}\right)} \\\\ {M(x, y)=\left(\frac{10-2}{2}, \frac{7-7}{2}\right)} \\\\ {M(x, y)=(4,0)}\end{array}$