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3 years ago

If sin yº = 7/q and tan yº = 7/r , what is the value of cos yº? (6 points)

Mathematics

1 answer:

olya-2409 [2.1K]3 years ago

4 0

Answer:

cos r/q is the answer,

Step-by-step explanation:

this can be seen when you elaborate sin=opposite/hypotenus and tan=opposite/adjacent by referring to the right angle triangle .

then, apply values of adjacent dan hypotenuse to cos

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I think I know how to solve this, but could someone please check my work for me?

Fiesta28 [93]

Answer is :

cos⁻¹[(-1-4√7)/(√17, √8) = 173.33° ≈ 173°

5 0

3 years ago

Find the sum of the series if n = 6, r = 1/4, a = 2.

emmasim [6.3K]

Given =

n=6,

r=1/4

a=2

Sn = a+ar+ar²+ar³+ar⁴+ar⁵= 2+0.5+0.125+0.03125+0.0078125+0.00195~ 2.67

4 0

3 years ago

Can anyone help me pleassseeee!

Xelga [282]

Answer:

baba booey

Step-by-step explanation:

3 0

3 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago

Water is leaking out of a tank at a steady rate the tank started with 75 gallons after 15 hours the tank has 33 gallons of water

Natasha_Volkova [10]

Answer:

The equation will be " $y=-2.8 x+75$ ".

Step-by-step explanation:

The given values are:

Started with,

= 75 gallons

After 15 hrs,

Tank has,

= 33 gallons

Now,

The slope will be:

= $\frac{33-75}{15}$

= $\frac{-42}{15}$

= $-2.8$

Thus,

The equation is " $y=-2.5x+75$ ".

5 0

3 years ago