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2 years ago

Reflect the triangle (-2,-2) (-6,-8) (-8,-8) over the x-axis. What are the new vertices?

1 answer:

5 0

Reflecting across the x axis changes the y value to the opposite, but doesn't change the x value, so the new values are:
(-2,2), (-6, 8), (-8, 8)

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A ramp is used to load suitcases on an airplane. If the cargo door is 7 feet from the ground and the angle formed by the end of

melisa1 [442]

Answer: $16.56\ ft$

Step-by-step explanation:

You must draw a right triangle as the one attached, where "x" is the lenght in feet of the ramp.

You need to use the following Trigonometric Identity:

$sin\alpha=\frac{opposite}{hypotenuse}$

In this case you can identify that:

$\alpha =25\°\\\\opposite=7\\\\hypotenuse=x$

Now you must substitute these values into $sin\alpha=\frac{opposite}{hypotenuse}$ :

$sin(25\°)=\frac{7}{x}$

Finally, you must solve for "x" in order to find its value.

This is:

$sin(25\°)=\frac{7}{x}\\\\xsin(25\°)=7\\\\x=\frac{7}{sin(25\°)}\\\\x=16.56$

6 0

2 years ago

There are 56 workers during each shift. The workers are split into teams of 14 each. How many work during a shift?

Over [174]

Answer:

4 teams

Step-by-step explanation:

To evenly distribute 14 into 56, you divide. 56/14 is 4.

Therefore the correct answer is 4 teams.

Hope I helped!!!

5 0

3 years ago

1/2 (6x-10) - x =<br><br> What's the other side of the equation?

LiRa [457]

Answer:

Step-by-step explanation:

1/2 (6x-10) - x

Opening bracket

= 6x/2 - 10/2 - x

= 3x - 5 - x

= 2x - 5

4 0

2 years ago

If, for all x, f'(x)=(x-2)^4(x-1)^3, it follows that funtaion f has

Elenna [48]

Answer:

We say that f(x) has an absolute (or global) minimum at x=c if f(x)≥f(c) f ( x ) ≥ f ( c ) for every x in the domain we are working on. We say that f(x) has a relative (or local) minimum at x=c iff(x)≥f(c) f ( x ) ≥ f ( c ) for every x in some open interval around x=c .

4 0

2 years ago

Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

ddd [48]

Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$