What is 16+2 and 18-3 and 2*3?

Mathematics

1 answer:

stepladder [879]3 years ago

5 0

Answer:

16+2=18

18-3=15

2*3=6

Step-by-step explanation:

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Find the coordinates of the point 7/10 of the way from A to B. a=(-3,-6) b=(12,4)

Artemon [7]

Answer:

The coordinates of $M$ are $x = \frac{15}{2}$ and $y = 1$ .

Step-by-step explanation:

Let be $A = (-3,-6)$ and $B = (12, 4)$ endpoints of segment AB and $M$ a point located 7/10 the way from A to B. Vectorially, we get this formula:

$\overrightarrow {AM} = \frac{7}{10}\cdot \overrightarrow {AB}$

$\vec M - \vec A = \frac{7}{10}\cdot (\vec B - \vec A)$

By Linear Algebra we get the location of $M$ :

$\vec M = \vec A + \frac{7}{10}\cdot (\vec B - \vec A)$

$\vec M = \vec A +\frac{7}{10}\cdot \vec B - \frac{7}{10}\cdot \vec A$

$\vec M = \frac{3}{10}\cdot \vec A + \frac{7}{10}\cdot \vec B$

If we know that $\vec A = (-3,-6)$ and $\vec B = (12, 4)$ , then:

$\vec M = \frac{3}{10}\cdot (-3,-6)+\frac{7}{10}\cdot (12,4)$

$\vec M = \left(-\frac{9}{10},-\frac{9}{5} \right)+\left(\frac{42}{5} ,\frac{14}{5} \right)$

$\vec M =\left(-\frac{9}{10}+\frac{42}{5} ,-\frac{9}{5}+\frac{14}{5} \right)$

$\vec M = \left(\frac{15}{2} ,1\right)$

The coordinates of $M$ are $x = \frac{15}{2}$ and $y = 1$ .

6 0

4 years ago

Verify that the following equation is an identity. (1/sinx) - (1/cscx) = cscx - sinx

Anarel [89]

Manipulating the left side of the equation, we obtain:

$\dfrac{1}{\sin x}-\dfrac{1}{\csc x}=\dfrac{\csc x-\sin x}{\sin x\cdot\csc x}$

Using that $\csc x=\dfrac{1}{\sin x}$ :

$\dfrac{\csc x-\sin x}{\sin x\cdot\csc x}=\dfrac{\csc x-\sin x}{\sin x\cdot\dfrac{1}{\sin x}}=\dfrac{\csc x-\sin x}{1}=\csc x-\sin x\\\\\boxed{\dfrac{1}{\sin x}-\dfrac{1}{\csc x}=\csc x-\sin x}~~\blacksquare$