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3 years ago

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Which of the following is a correct tangent ratio for the figure? A) tan (24) 76 B) tan(76°) °= 2 C) tan(76°) = D) tan(8") = 24

76

1 answer:

Sphinxa [80]3 years ago

6 0

Answer:

C

Step-by-step explanation:

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Write an equation in slope intercept form that describes a line with a slope of 3 and containing the point (5,-1)

soldier1979 [14.2K]

Answer:

<h2> y = 3x - 16</h2>

Step-by-step explanation:

Equation of line with a given slope of a and containing a given point (x₀, y₀):

y - y₀ = a(x - x₀)

a = 3, x₀ = 5, y₀ = -1

So:

y - (-1) = 3(x - 5)

y = 3x - 15 - 1

<u>y = 3x - 16</u>

3 0

3 years ago

Mariulka [41]

Answer:

$A = 74.7^\circ$

$B = 42.5^\circ$

$C = 62.8^\circ$

Step-by-step explanation:

Given

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

Required

The measure of each angle

First, we calculate the length of the three sides of the triangle.

This is calculated using distance formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

For AB

$A = (-1,2) \to (x_1,y_1)$

$B = (2,8) \to (x_2,y_2)$

$d = \sqrt{(-1 - 2)^2 + (2 - 8)^2$

$d = \sqrt{(-3)^2 + (-6)^2$

$d = \sqrt{45$

So:

$AB = \sqrt{45$

For BC

$B = (2,8) \to (x_2,y_2)$

$C = (4,1) \to (x_3,y_3)$

$BC = \sqrt{(2 - 4)^2 + (8 - 1)^2$

$BC = \sqrt{(-2)^2 + (7)^2$

$BC = \sqrt{53$

For AC

$A = (-1,2) \to (x_1,y_1)$

$C = (4,1) \to (x_3,y_3)$

$AC = \sqrt{(-1 - 4)^2 + (2 - 1)^2$

$AC = \sqrt{(-5)^2 + (1)^2$

$AC = \sqrt{26$

So, we have:

$AB = \sqrt{45$

$BC = \sqrt{53$

$AC = \sqrt{26$

By representation

$AB \to c$

$BC \to a$

$AC \to b$

So, we have:

$a = \sqrt{53$

$b = \sqrt{26$

$c = \sqrt{45$

By cosine laws, the angles are calculated using:

$a^2 = b^2 + c^2 -2bc \cos A$

$b^2 = a^2 + c^2 -2ac \cos B$

$c^2 = a^2 + b^2 -2ab\ cos C$

$a^2 = b^2 + c^2 -2bc \cos A$

$(\sqrt{53})^2 = (\sqrt{26})^2 +(\sqrt{45})^2 - 2 * (\sqrt{26}) +(\sqrt{45}) * \cos A$

$53 = 26 +45 - 2 * 34.21 * \cos A$

$53 = 26 +45 - 68.42 * \cos A$

Collect like terms

$53 - 26 -45 = - 68.42 * \cos A$

$-18 = - 68.42 * \cos A$

Solve for $\cos A$

$\cos A =\frac{-18}{-68.42}$

$\cos A =0.2631$

Take arc cos of both sides

$A =\cos^{-1}(0.2631)$

$A = 74.7^\circ$

$b^2 = a^2 + c^2 -2ac \cos B$

$(\sqrt{26})^2 = (\sqrt{53})^2 +(\sqrt{45})^2 - 2 * (\sqrt{53}) +(\sqrt{45}) * \cos B$

$26 = 53 +45 -97.67 * \cos B$

Collect like terms

$26 - 53 -45= -97.67 * \cos B$

$-72= -97.67 * \cos B$

Solve for $\cos B$

$\cos B = \frac{-72}{-97.67}$

$\cos B = 0.7372$

Take arc cos of both sides

$B = \cos^{-1}(0.7372)$

$B = 42.5^\circ$

For the third angle, we use:

$A + B + C = 180$ --- angles in a triangle

Make C the subject

$C = 180 - A -B$

$C = 180 - 74.7 -42.5$

$C = 62.8^\circ$

8 0

3 years ago

Evaluate cos0 if sin0= sqrt(5)/3<br><br> a. 9/5<br> b. 2/3<br> c. 5/9<br> d. 3/2

Evgesh-ka [11]

Solution given:

$\sin( \alpha ) = \frac{\sqrt{5} }{3}$

so

P=√5

h=3

b= $\sqrt{3 {}^{2} - \sqrt{5} {}^{2} }$

b=2

so

$\cos( \alpha ) = \frac{b}{h} = \frac{2}{3}$

Note

$\alpha = 0$

<u>b</u><u>:</u><u>⅔</u>

8 0

3 years ago

Mr.Martinez rents a car for one day.The charge is $36 plus $0.18 per mile.Mr.Martinez has a budget $60.How many miles can he dri

kondor19780726 [428]

1... the wording is a little confusing.
Is the charge $0.18 per mile or is it actually $36.18 per mile?

8 0

4 years ago

Can someone help me with this ?

LuckyWell [14K]

Answer:

X is 65.7 hope this helps!

3 0

3 years ago