If f(x) = 6x - 4, what is f(x) when x=8

Svetach [21]

3px = 2q(r-5x)
distribute
2qr - 10qx
3px = 2qr - 10qx

To solve for x
add 10qx to both sides
3px + 10qx = 2qr
factor out x
x(3p+10q) = 2qr
divide both sides by 3p+10q to isolate x
x = 2qr / (3p+10q)

Hope this helps :)

3 0

3 years ago

HELP PLEASE DESPERATE!!

devlian [24]

Answer:

A,B,C

Step-by-step explanation:

in these quadrilaterals, the diagonals bisect each other paralellogram, rectangle, rhombus, square

in these quadrilaterals, the diagonals are congruent rectangle, square, isosceles trapezoid

in these quadrilaterals, each of the diagonals bisects a pair of opposite angles rhombus, square

5 0

3 years ago

Read 2 more answers

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$tan(\alpha)=cot(\beta)$

Find the value of the length side adjacent to the angle alpha

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

What is the completely factored form of x3-64?

nataly862011 [7]

a^3 - b^3 = (a - b) (a^2 + ab + b^2)
so
x^3 - 4^3 = (x - 4)(x^2 + 4x + 16)

hope it helps

7 0

3 years ago

Read 2 more answers

PLEASE HELP WITH SIGNIFICANT FIGURES *AWARDING BRAINEST ANSWER* ASAP

marishachu [46]

1) 0.07 zeros only matter if there is a non-zero digit, as well as the decimal to the left of it
2)90,000 it would have been 89000 if the 972 had been an 872
3).00420 must include the last zero, it is considered significant because of the reason in number one

4 0

3 years ago