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2 years ago

Please help me with precal

Mathematics

1 answer:

Doss [256]2 years ago

6 0

Consider the function $f(x)=\sqrt{x}.$ This function has:

the domain $x\in [0,\infty);$
the range $y\in [0,\infty).$

If the domain of unknown function is $[a,\infty),$ then $x\ge a$ or $x-a\ge 0.$ This means that you have $\sqrt{x-a}$ as a part of needed function.

If the range of unknown function is $[b,\infty),$ then $y\ge b.$ This means that you have to translate function b units up and then the expression of the function is

$y=\sqrt{x-a}+b.$

Answer: correct choice is B.

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Find the area of the following figure.

Zigmanuir [339]

Answer:

312

Step-by-step explanation:

Triangles

1/2 12 *10 = 6*10 = 60

there are two triangles meaning we have to multiply 60 by 2 = 120

Rectangle

12*16 = 192

Add the two and you have the area 120+192=312

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2 years ago

1 is 25% of what number

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Answer: 4

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What units of distance are in the metric system? How are they related?

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Answer:

The most common units of distance In the metric system of measurement, are millimeters, centimeters, meters, and kilometers.

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2 years ago

. Two rectangles have a ratio of similarity of 4/7. If one side of the small rectangle is 12m long, how long is the same side of

Akimi4 [234]

Answer:

For two rectangles to be similar, their sides have to be proportional (form equal ratios). The ratio of the two longer sides should equal the ratio of the two shorter sides.

Step-by-step explanation:

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2 years ago

How many permutations can be formed from all the letters in the word engineering?

laila [671]

Answer:

277,200

Step-by-step explanation:

To find the number of permutation we can form from the letters of the word "engineering", we first need to find the frequencies of the different letters present.

E = 3

G = 2

N= 3

I = 2

R = 1

Now that we have the frequencies, we count the number of letters in the word "engineering".

E N G I N E E R I N G

11 letters

Now we take the factorial of total number of letters and divide it by the number of repeats and their factorial

So we get:

$\dfrac{11!}{3!2!3!2!1!}$

We remove the 1! because it will just yield 1.

$\dfrac{11!}{3!2!3!2!}$

So the total number of permutations from the letters of the word "engineering" will be:

Total number of permutations = $\dfrac{39,916,800}{144}$

Total number of permutations = 277,200

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3 years ago