The sine of 72 degrees is 0.95. What's the cosine of 18 degrees?

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

3 years ago

Can somebody answer this ?? :)

Alexxandr [17]

Answer:

Because the dashed triangle is smaller than the original it's a reduction. The scale factor is 1/3 because if we look at the altitudes of the triangle, the original is 6 and the dilated is 2 and 2 / 6 = 1 / 3.

4 0

3 years ago

What is the solution to the system of equations graphed below?

MA_775_DIABLO [31]

B because that is where they intersect

5 0

3 years ago

Read 2 more answers

Need help with these table what do I do?

Greeley [361]

Answer:

just ask it again

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The rate of change of the number of mountain lions N(t) in a population is directly proportional to 725 - N(t), where t is the t

wolverine [178]

Answer:

a) The implied differential equation is $\frac{dN}{725 - N} = kdt$

b) The general equation is $N = 725 - C e^{-kt}$

c) The particular equation is $N = 725 - 325 e^{-0.49t}$

d) The population when t = 5, N(5) = 697 = 700( to the nearest 50)

Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, $\frac{dN}{dt} \alpha (725 - N(t))$

$\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt$

Integrating both sides of the equation

$\int {\frac{1 }{725 - N}} \, dN = \int {k} \, dt\\- ln (725 - N) = kt + C\\ ln (725 - N) = -(kt + C)\\725 - N = e^{-(kt + C)} \\725 - N = e^{-kt} * e^{-C} \\725 - N = C e^{-kt}\\N = 725 - C e^{-kt}$