Solve P=Kt/v for k

Find x, y, and z. Multiple choice answers are below!

labwork [276]

Answer:

The correct answer is B. x ≈ 4.5, y = 5, z = 9

Step-by-step explanation:

1. We will use the Pythagorean theorem for solving for x, this way:

x² = 6² - 4²

x² = 36 - 16

x² = 20

x = √20

x = 2 √5 = 4.47 ≈ 4.5

2. For solving for y, we will do it this way:

There are only two possible options: y = 5, included in answer B or y = 9 as it is defined on answer C.

If we review it carefully, we will realize that y = 9 is not accurate because as we can see in the graph y + 4 = z and in this specific case, 9 + 4 ≠ 5 but it's true that 5 + 4 = 9.

Because of this reason, the correct answer is B. x ≈ 4.5, y = 5, z = 9

7 0

3 years ago

Two less than 2 times a number is the same as the number plus 64. What is the number?

sineoko [7]

What is 3 times the number?

3*66 = 198

4 0

2 years ago

Read 2 more answers

Find the four rational numbers and irrational numbers between 2/5 and 5/6

Savatey [412]

The four?

Of course there are an infinite number of rationals between any two different real numbers, as well as an even bigger infinite number of irrationals.

The average will be between the numbers:

$\frac 1 2 (\frac 2 5 + \frac 5 6) = \dfrac{37}{60}$

That's a lot of work to get a number in between. We can just see

$\frac 1 2$
is between the two numbers.

The mediant, or freshman addition, will always be in between:

$\dfrac{2 + 5}{5 + 6} = \dfrac{7}{11}$

2/5=.4 and 5/6 is about .83, so

$.6 = \frac{6}{10}$

is in between as is .7, .41, .530940394 and as many rationals as we care to generate.

To get irrationals we could just add a teeny irrational to the ones we just generated, like $\frac{6}{10} + \frac{\pi}{100}$

We could just change that denominator a bit and get as many as we like.

But let's get some square roots. The geometric mean will be between

$\sqrt{ \dfrac 2 5 \cdot \dfrac 5 6 } = \dfrac{1}{\sqrt{3}}
$

That's the tangent from one of trig's biggest cliches, but I digress. It's in between.

While we're on trig cliches,

$\dfrac{1}{\sqrt{2}}$

is in between as well.

Keeping with the trig theme, also in between are

$\dfrac{\pi} 6$

and

$\dfrac{\pi}{4}$

the angles associated with the some of the above trig function values.

We could obviously go on as long as we cared to.

7 0

3 years ago

Evaluate: 3y + 1 = 13 + 4y **What do you move first?!**

yan [13]

Answer: isolate the variable

Step-by-step explanation:

5 0

2 years ago

How many different combinations are possible if each lock contains the numbers 0 to 39, and each combination contains three dist

Georgia [21]

(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 95

8 0

2 years ago