Pls help me with this!!!!!!!!!

Liono4ka [1.6K]

First simplify the section in the parenthesis.
-1/6 + 2/3(8 1/4) + -1/2
Then multiply 2/3 by 8 1/4.
-1/6 + 5 1/2 + -1/2
Add -1/2 to 5 1/2.
-1/6 + 5
Add 5 to -1/6.
4 5/6 is the fully simplified answer.
Hope this helps!

7 0

3 years ago

Find the value of p in the figure below.

nirvana33 [79]

From the chord theorem we have:

$4\cdot4.5=6\cdot(2p-3)\\\\18=12p-18\\\\12p=18+18\\\\12p=36\qquad|:12\\\\\boxed{p=3}$

6 0

3 years ago

How do you model a decimal using a fraction

stiks02 [169]

Okie dokie,
When converting a decimal, you use the place the decimal is in...
Let's review the places: tenths, hundredths, thousandths, etc.
You look at the last number to determine what place you're using!
-------------------
Now here's an example for the tenths place:
.5
it's in the tenths place, right? so put it over that number (with no decimal) over a 10.
(mobile) 5/10 or $\frac{5}{10}$
Now, decide if you can simplify. You can! 5/10 simplifies down to 1/2.
$\frac{1}{2}$ is your answer!
--------
Example for the hundredths place:
.26
the last number is in the hundredths place, so put the number (without a decimal) over 100!
26/100 or $\frac{26}{100}$
You can simplify this!
13/50 is your answer!

Calculators can also come in handy!
Good luck!

4 0

3 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= 9<span>5</span>

8 0

3 years ago

If you are driving on the freeway at 55 mph and you travel for 5 hours, how far will you have gone?

alina1380 [7]

275 because 55 times 5 equals 275 ! Hope this helps

8 0

3 years ago

Read 2 more answers

Pls help me with this!!!!!!!!!​

Pls help me with this!!!!!!!!!