Solve for Y <br> 4x + 8y = 16

An employee currently makes $10 per hour and will be given yearly raises if performance evaluations are satisfactory. Assuming a

OleMash [197]

Answer:

C) If the yearly raise is $3 per hour then the relationship is linear and the hourly rate "n" years from now will be h0 = 10 and hn = (hn − 1)(3) for n = {1,2,3, ...}.

Step-by-step explanation:

j put c

6 0

3 years ago

Read 2 more answers

A triangle has side lengths of 12 millimeters, 17 millimeters, and 13 millimeters. Is this triangle equilateral?

nordsb [41]

No, this triangle is not equilateral. This triangle is scalene. All three sides are of different lengths, which is why it's scalene. In order to have an equilateral triangle, all three sides need to be the same length.

7 0

3 years ago

Find f(5) for f(x) = 1/4(2)x^<br> A.8<br> B.2<br> C.4<br> D.32

Katena32 [7]

A. 8 HOPE THIS IS CORRECT!

7 0

3 years ago

HELP PLEASE ANYONE STEP BY STEP TIMED

UNO [17]

Answer:

$\huge\boxed{\text{B) 4}}$

Step-by-step explanation:

We can use the properties of some of these circles to solve and find the value of x.

FIrst off - we know that to find the side of a right triangle we can use the Pythagorean Theorem, which states that $a^2 + b^2 = c^2$ (a and b being legs, c being the hypotenuse.)

<h2><u>Finding the hypotenuse</u></h2>

We also know that this triangle is in a circle. Part of the hypotenuse and the altitude of the triangle are in the circle.

Point W is the center of the circle. Therefore, any points stretching to the edge from it will be equal (since this is a circle).

With this knowledge - UW and WZ are equal.

Therefore, since UW is $x-1$ , WZ will be too.

We also know the length of VZ. We can add this to WZ to find the length of the whole hypotenuse.

$\displaystyle x-1+2 \\ x+1$

So the length of the hypotenuse, WV, is $x+1$ .

<h2><u>Finding the value of x</u></h2>

Now that we know the length of the hypotenuse, we can use the Pythagorean Theorem like we stated before to find the value of x.

We know the two legs are $x-1$ and $2x-4$ , while the hypotenuse is

$\displaystyle (x-1)^2 + (2x-4)^2 = (x+1)^2$
$(x^2 - 2x + 1) + (4x^2-16x+16) = (x^2 + 2x + 1)$ (use FOIL to find the value of each expression squared)
$5x^2 - 18x + 17 = x^2 + 2x + 1$ (simplify the left side)
$4x^2 -20x + 16 = 0$ (subtract both sides by $x^2 +2x + 1$ )
$x = \frac{{ -b \pm \sqrt {b^2 - 4ac} }}{{2a}}$ (Quadratic formula)
$\frac{{ -(-20) \pm \sqrt {(-20)^2 - 4 \cdot 4 \cdot 16} }}{{2 \cdot 4}}$ (plug in values from equation)
$\frac{{ 20 \pm \sqrt {400 - 256} }}{8}}$ (simplify)
$\frac{{ 20 \pm \sqrt {144} }}{8}}$ (simplify)
$\frac{{ 20 \pm 12 }}{8}}$ (simplify)
$\displaystyle x = \frac{20+12}{8} \ \text{and} \ \frac{20-12}{8}$ (use plus/minus to find two roots)
$\displaystyle x = \frac{32}{8} \ \text{and} \ \frac{8}{8}$ (simplify)
$\displaystyle x = 4 \ \text{and} \ 1$