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2 years ago

Find the angle between the given vector in degrees. u=<6, 4>, v=<7, 5> HELPPPPPPPPPP

Mathematics

1 answer:

Paladinen [302]2 years ago

7 0

Recall that

$\vec a\cdot\vec b=\|\vec a\|\|\vec b\|\cos\theta$

where $\theta$ is the angle between the vectors $\vec a,\vec b$ whose magnitudes are $\|\vec a\|,\|\vec b\|$ , respectively.

We have

$\langle6,4\rangle\cdot\langle7,5\rangle=\sqrt{6^2+4^2}\sqrt{7^2+5^2}\cos\theta$

$\cos\theta=\dfrac{31}{\sqrt{962}}\implies\theta\approx88.2^\circ$

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Determine the length of the unknown sides for the right triangle

alisha [4.7K]

Answer:

Below in bold.

Step-by-step explanation:

sin 66.4 = opposite side / hypotenuse

sin 66.4 = 21.3 / y

y = 21.3 / sin 66.4

y = 23.24.

tan 66.4 = 21.3 / x

x = 21.3 / tan 66.4

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3 years ago

May you please help me with Highschool Ap Calculus.

nadya68 [22]

Answer:

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2 years ago

Which point is in the solution set of the given system of inequalities 3x+y>-3,x+2y<4

Dima020 [189]

Step 1. Solve both inequalities for $y$ :
$3x+y\ \textgreater \ -3$
$y\ \textgreater \ -3x-3$

$x+2y\ \textless \ 4$
$2y\ \textless \ -x+4$
$y\ \textless \ - \frac{1}{2} x+2$

Step 2. To check a point in the solution of the given system of inequalities, look for the intercepts of the lines $-3x-3$ and $- \frac{1}{2} x+2$ :

$y=-3x-3$ (1)
$y=- \frac{1}{2} x+2$ (2)

Replace (1) in (2):
$-3x-3=- \frac{1}{2} x+2$
Solve for $x$ :
$\frac{5}{2} x=-5$
$x=-2$ (3)

Replace (3) in (1):
$y=-3x-3$
$y=-3(-2)-3$
$y=6-3$
$y=3$

We can conclude that the point (-2,3) is in the solution of the system if <span>inequalities</span>; also any point inside the dark shaded area of the graph of the system of inequalities is also a solution of the system.