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3 years ago

Ahhahanvm? shushaha bruh what do i do

Mathematics

2 answers:

Basile [38]3 years ago

6 0

Uh I’m not even gonna question it but thanks for the free points ig? Lol

kirill115 [55]3 years ago

3 0

\u gotta ask ur question

Step-by-step explanation:

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3z-4=6z-17 how do solve

Maru [420]

3z-4=6z-17
-3z-4=-17 (subtracted the 6z from both sides)
-3z=-13 (added the 4 to both sides)
z= $\frac{-13}{-3z}$ (divided both sides by -3z)
z= $\frac{13}{3}$

5 0

4 years ago

Read 2 more answers

Suppose that the average number of accidents occurring on a highway each day is 3.5. Find the probability that at least two acci

irina1246 [14]

Answer:

48 yards

Step-by-step explanation:

3 0

4 years ago

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Answers are<br> 126in cubed<br> 377in cubed<br> 565in cubed<br> 1131in cubed

suter [353]

126

Step-by-step explanation:

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3 0

3 years ago

Evaluate exactly without the use of a calculator (remember to rationalize any denominators). a.(cos 60o)(sin 270o) + tan 225o b.

Rom4ik [11]

Given:

The trigonometric expressions are given as,

$\begin{gathered} a)\text{ }(\cos 60\degree)(\sin 270\degree)+\tan 225\degree \\ b)-\tan 240\degree+(cos45\degree)(\sec 135\degree) \end{gathered}$

Explanation:

The given expression can be rewritten as,

$=(\cos 60\degree)(\sin (360\degree-90\degree))+\tan (270\degree-45\degree)\text{ . . . . .(1)}$

Since, from the trigonometric ratios,

$\begin{gathered} \sin (360\degree-90\degree)=-\sin 90\degree \\ \tan (270\degree-45\degree)=\cot 45\degree \end{gathered}$

On plugging the obtained ratios in equation (1),

$=(\cos 60\degree)(-\sin 90\degree)+\cot 45$

Substitute the trigonometric values in the above equation.

$\begin{gathered} =\frac{1}{2}(-1)+1 \\ =-\frac{1}{2}+1 \\ =\frac{1}{2} \end{gathered}$

Hence, the exact value of the expression is 1/2.

The given expression can be rewritten as,

$=-\tan (270\degree-30\degree)+(\cos 45\degree)(\sec (90\degree+45\degree))\text{ . . . ..(2)}$

Since, from the trigonometric ratios,

$\begin{gathered} \tan (270\degree-30\degree)=\cot 30\degree \\ \sec (90\degree+45\degree)=-\csc 45\degree \end{gathered}$

On plugging the obtained ratios in equation (2),

$=-\cot 30\degree+(\cos 45\degree)(-\csc 45)$

Substitute the trigonometric values in the above equation.

$\begin{gathered} =-\sqrt[]{3}+\frac{1}{\sqrt[]{2}}(-\sqrt[]{2}) \\ =-\sqrt[]{3}-1 \end{gathered}$

Hence, the exact value of the expression is -√3-1.

7 0

2 years ago

There are 17 portable mini suites (a.k.a. cages) in a row at the paws and claws holiday pet resort. they are neatly labeled with

DiKsa [7]

Given that there are 17 portable mini suites (a.k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with their guests' names. There are 8 poodles and 9 tabbies. How many ways can the "suites" be arranged if: a) there are no restrictions.

b) cats and dogs must alternate.

c) dogs must be next to each other.

d) dogs must be next to each other and cats must be next to each other.

Part A:

If there are no restrictions, then the number of ways the "suites" can be arranged is given by:

$17!=17\times16\times15\times14\times13\times12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2\times1=355,687,428,096,000 \ ways$

Part B:

If cats and dogs must alternate, then the number of ways the "suites" can be arranged is given by:

$9!\times8!=9\times8\times7\times6\times5\times4\times3\times2\times1\times8\times7\times6\times5\times4\times3\times2\times1=14,631,321,600 \ ways$

Part C:

If dogs must be next to each other, then the group of dogs are taken as 1 object and there are now 10 objects to arrange. The number of ways to arrange 10 objects is given by:

$10!=10\times9\times8\times7\times6\times5\times4\times3\times2\times1=3,628,800
 \ ways$

Also, the group of dogs can be arranged in:

$8!=8\times7\times6\times5\times4\times3\times2\times1=40,320
 \ ways$

Therefore, the total number of ways the "suites" can be arranged is given by:

$10!\times8!=3,628,800\times40,320=146,313,216,000 \ ways$

Part D:

If dogs must be next to each other and cats must be next to each other, then the group of dogs are taken as 1 object and the group of cats are taken as 1 object, then there are now 2 objects to arrange.

The number of ways to arrange two objects is given by:

$2!=2\times1=2$

The number of ways the 9 dogs are to be arranged is given by:

$9!=9\times8\times7\times6\times5\times4\times3\times2\times1=362,880$

The number of ways the 8 cats are to be arranged is given by:
$8!=8\times7\times6\times5\times4\times3\times2\times1=40,320$

Therefore, the total number of ways the "suites" can be arranged is given by:

$2!\times9!\times8!=2\times362,880\times40,320=29,262,643,200 \ ways$

4 0

3 years ago