C. tan(450) Help please!

Which is greater 1) 14/28 or 17/28 2)26/35 or 26/38 3)18/18 or 12/12

Sergio039 [100]

Answer:

1. 17/28

2. 26/35

3. they're equal

4 0

2 years ago

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Find the inverse of the function. f(x) = 7x + 8

lana66690 [7]

Answer:

$\frac{x-8}{7}$

Step-by-step explanation:

A way to find the inverse function is to swap the x and y (f(x)) in the equation.

$y = 7x + 8\\x = 7y + 8\\x-8 = 7y\\y = \frac{x-8}{7}$

We can check that it is an inverse function by using a property of inverse functions:

$f(g(x)) = x$

If we plug in the f(x) function into the variable x of the inverse function, we should end up with x

$y = \frac{(7x + 8) - 8}{7}\\y = \frac{7x}{7} \\y = x\\$

It checks out!

4 0

3 years ago

Angel and Jayden were at track practice. The track is 2/5 kilometers around.  Angel ran 1 lap in 2 minutes.  Jayden ran 3 laps

choli [55]

Answer:

6/25 km in 1 min; or 0.24 km/min

Step-by-step explanation:

The info about Angel is not necessary.

Jayden runs 3 laps in 5 mins

Each lap is 2/5 km, so Jayden runs 3 x (2/5 km) in 5 min = 6/5 km in 5 min.

Distance = rate x time

6/5 km = rate (5 min)

We want to isolate r (rate), so divide both sides by 5 min

6/5 km ÷ 5 min = r

6/5 km (1/5 min) = r

6/25 km/min = r (notice how the units worked out correctly to km/min)

So, Jayden runs 6/25 km in 1 min.

7 0

3 years ago

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If the sum of the zereos of the quadratic polynomial is 3x^2-(3k-2)x-(k-6) is equal to the product of the zereos, then find k?

lys-0071 [83]

Answer:

2

Step-by-step explanation:

So I'm going to use vieta's formula.

Let u and v the zeros of the given quadratic in ax^2+bx+c form.

By vieta's formula:

1) u+v=-b/a

2) uv=c/a

We are also given not by the formula but by this problem:

3) u+v=uv

If we plug 1) and 2) into 3) we get:

-b/a=c/a

Multiply both sides by a:

-b=c

Here we have:

a=3

b=-(3k-2)

c=-(k-6)

So we are solving

-b=c for k:

3k-2=-(k-6)

Distribute:

3k-2=-k+6

Add k on both sides:

4k-2=6

Add 2 on both side:

4k=8

Divide both sides by 4:

k=2

Let's check:

$3x^2-(3k-2)x-(k-6) \text{ with }k=2$ :

$3x^2-(3\cdot 2-2)x-(2-6)$

$3x^2-4x+4$

I'm going to solve $3x^2-4x+4=0$ for x using the quadratic formula:

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

$\frac{4\pm \sqrt{(-4)^2-4(3)(4)}}{2(3)}$

$\frac{4\pm \sqrt{16-16(3)}}{6}$

$\frac{4\pm \sqrt{16}\sqrt{1-(3)}}{6}$

$\frac{4\pm 4\sqrt{-2}}{6}$

$\frac{2\pm 2\sqrt{-2}}{3}$

$\frac{2\pm 2i\sqrt{2}}{3}$

Let's see if uv=u+v holds.

$uv=\frac{2+2i\sqrt{2}}{3} \cdot \frac{2-2i\sqrt{2}}{3}$

Keep in mind you are multiplying conjugates:

$uv=\frac{1}{9}(4-4i^2(2))$

$uv=\frac{1}{9}(4+4(2))$

$uv=\frac{12}{9}=\frac{4}{3}$

Let's see what u+v is now:

$u+v=\frac{2+2i\sqrt{2}}{3}+\frac{2-2i\sqrt{2}}{3}$

$u+v=\frac{2}{3}+\frac{2}{3}=\frac{4}{3}$

We have confirmed uv=u+v for k=2.

4 0

3 years ago

if angle 1 and angle 4 are supplementary which theorem leads to the conclusion that a is parallel to s

Lubov Fominskaja [6]

Answer:

Supplementary Angle's Theorem

Step-by-step explanation:

The supplementary angle theorem states that if two angles are said to be supplementary to the same angle, then the two angles are said to be congruent.

7 0

3 years ago