WILL GIVE EXTRA POINTS assignment name: transformations

An electrician charges a set fee of $50 for every house call and then charges an additional hourly rate. One day, the electricia

Serhud [2]

Answer:

$50+5t=C$

Step-by-step explanation:

Suppose the electrician spends t hours at the house working.

Amount charged for every house call by electrician = $50

The electrician earned a total of $400 for a 5 hour job.

Also, C denotes total cost of the electrician's services.

So, the equation for C in terms of t is given below.

$50+5t=C$

Here, $C=400$

7 0

4 years ago

Which triangles could you use the HL Theorem to prove that triangle 1 is congruent to triangle 2?

suter [353]

Answer: D) III only

With figure I, we can't use HL (hypotenuse leg) theorem because we don't know if the two hypotenuses are the same length. We know that one pair of legs (the vertical legs) are congruent, but we don't have enough info

With figure II, we have two pairs of congruent legs. The tickmarks tell us this. However we don't know if the hypotenuses are congruent. Note: we can use the LL (leg leg) theorem, which is related to the HL theorem, but that's not what the teacher wants

Figure III is the only figure where we know that a pair of hypotenuses are the same length as shown by the similar tickmarks. The vertical legs are congruent as well for the upper pair of triangles. We have enough info to use HL.

8 0

3 years ago

Solve the inequality x > 2 - |2x-1|

never [62]

Answer:

$x 1$

Step-by-step explanation:

It can be solved analytically or graphically. Graphically is the fastest way if you know how to handle the two lines. Let's rearrange terms in either case:

$x-2>-|2x-1| \rightarrow |2x-1| >2-x$

Graphic: Red line is $y=2-x$ , solid blue line is $y=|2x-1|$ , the thin extension is the parent line without the absolute value.

Solving the equation is equivalent to answer "when is the blue line higher than the red one?" Which obviously happens outside the two intercept points. Let's find the first coordinate of both - we don't really care what's the y coordinate of either point.

For the left point, the x coordinate is given by $1-2x = 2-x \rightarrow 2x-1=x-2 \rightarrow x=-1$

For the right point, we have

$2x-1=2-x \rightarrow 3x=3 \rightarrow x=1$

Analytical:

We split the problem in two parts: while the content absolute value is non negative, we have

$2x-1 \geq 0\\2x-1 >2-x\\\\x\geq \frac12\\3x > 3 \rightarrow x>1$

Which admits as solutions all valeus of x greater than 1 (since both have to be satisfied at once, and values between 1/2 and 1 will not work for the second.

OR, with a negative quantity inside the absolute value, we get

$2x-1 2-x\\\\x$