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aksik [14]
3 years ago
6

Multiply and simplify tan θ(cos θ − csc θ)

Mathematics
1 answer:
Karo-lina-s [1.5K]3 years ago
3 0

Answer:

Here

Step-by-step explanation:

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Write an expression that represent the following. Add the sum by 5. Next, add 15 to that product,then triple the results.
Irina-Kira [14]
5+15=20
20x3=60

I'm probably wrong, but thats how I understood the question.
5 0
3 years ago
Please do this math for me or else i will steal your cookies
kompoz [17]

Answer:

Step-by-step explanation:

1.

-7x+16=58\\\\\mathrm{Subtract\:}16\mathrm{\:from\:both\:sides}\\-7x+16-16=58-16\\\\-7x=42\\\\\mathrm{Divide\:both\:sides\:by\:}-7\\\\\frac{-7x}{-7}=\frac{42}{-7}\\\\x =-6

2.

-2x+15=-9\\\\\mathrm{Subtract\:}15\mathrm{\:from\:both\:sides}\\\\-2x+15-15=-9-15\\\\-2x=-24\\\\\mathrm{Divide\:both\:sides\:by\:}-2\\\\\frac{-2x}{-2}=\frac{-24}{-2}\\\\x=12\\

3.

5x-4=36\\\\\mathrm{Add\:}4\mathrm{\:to\:both\:sides}\\\\5x-4+4=36+4\\\\5x=40\\\\\mathrm{Divide\:both\:sides\:by\:}5\\\\\frac{5x}{5}=\frac{40}{5}\\\\x=8

4.

25-3x=88\\\\\mathrm{Subtract\:}25\mathrm{\:from\:both\:sides}\\\\25-3x-25=88-25\\\\\mathrm{Divide\:both\:sides\:by\:}-3\\\\\frac{-3x}{-3}=\frac{63}{-3}\\\\x=-21

5.

-11=7-x\\\\\mathrm{Add\:}x\mathrm{\:to\:both\:sides}\\\\-11+x=7-x+x\\\\-11+x=7\\\\\mathrm{Add\:}11\mathrm{\:to\:both\:sides}\\\\-11+x+11=7+11\\\\x=18

6.

65+15x=35\\\\\mathrm{Subtract\:}65\mathrm{\:from\:both\:sides}\\\\65+15x-65=35-65\\\\\mathrm{Divide\:both\:sides\:by\:}15\\\\\frac{15x}{15}=\frac{-30}{15}\\\\x=-2

7.

\frac{1}{2} x-18=2\\\\\mathrm{Add\:}18\mathrm{\:to\:both\:sides}\\\\\frac{1}{2}x-18+18=2+18\\\\\frac{1}{2}x=20\\\\\mathrm{Multiply\:both\:sides\:by\:}2\\\\2\times\frac{1}{2}x=20\times \:2\\\\x=4

8.

\frac{2}{3}x-10=-12 \\\\\mathrm{Add\:}10\mathrm{\:to\:both\:sides}\\\\\frac{2}{3}x-10+10=-12+10\\\\\frac{2}{3}x=-2\\\\Divide\:both\:sides\:by\: 2/3\\\\\frac{2}{3}x\div \frac{2}{3}  =-2\div \frac{2}{3}  \\\\\frac{2}{3}  x\times \frac{3}{2} =-2\times\frac{3}{2} \\\\x =-3

9.

6-\frac{1}{3}x=-1 \\\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\\\6-\frac{1}{3}x-6=-1-6\\\\-\frac{1}{3}x=-7\\\\\mathrm{Multiply\:both\:sides\:by\:}-3\\\\\left(-\frac{1}{3}x\right)\left(-3\right)=\left(-7\right)\left(-3\right)\\\\x=21

10.

4-9x=-14\\\\\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}\\\\4-9x-4=-14-4\\\\-9x=-18\\\\\mathrm{Divide\:both\:sides\:by\:}-9\\\\\frac{-9x}{-9}=\frac{-18}{-9}\\\\x=2

11.

11-x=29\\\\\mathrm{Subtract\:}11\mathrm{\:from\:both\:sides}\\\\-x=18\\\\\mathrm{Divide\:both\:sides\:by\:}-1\\\\\frac{-x}{-1}=\frac{18}{-1}\\\\x=-18

12.

-9-11x=68\\\\\mathrm{Add\:}9\mathrm{\:to\:both\:sides}\\\\-9-11x+9=68+9\\\\-11x=77\\\\\mathrm{Divide\:both\:sides\:by\:}-11\\\\\frac{-11x}{-11}=\frac{77}{-11}\\\\x=-7

13.

45+\frac{5}{6}x =50\\\\\mathrm{Subtract\:}45\mathrm{\:from\:both\:sides}\\\\45+\frac{5}{6}x-45=50-45\\\\\frac{5}{6}x=5\\\\\mathrm{Divide\:both\:sides\:by\:}5/6\\\\\frac{5}{6} x\div\frac{5}{6}=5\div\frac{5}{6}\\\\\frac{5}{6}x\times \frac{6}{5}= 5\times\frac{6}{5}\\\\ x=6

14.

-5x+17=-33\\\\\mathrm{Subtract\:}17\mathrm{\:from\:both\:sides}\\\\-5x+17-17=-33-17\\\\\mathrm{Divide\:both\:sides\:by\:}-5\\\\\frac{-5x}{-5}=\frac{-50}{-5}\\\\x=10

15.

95=-4+33x\\\\-4+33x=95\\\\\mathrm{Add\:}4\mathrm{\:to\:both\:sides}\\\\-4+33x+4=95+4\\\\\mathrm{Divide\:both\:sides\:by\:}33\\\\\frac{33x}{33}=\frac{99}{33}\\\\x=3

3 0
3 years ago
Wright the inequality that represents the sentence.
Serga [27]
X-5 is greater than or equal to -28
X*4 is less than or equal to -10
(x/3) + 6 is greater than 14
7 0
3 years ago
. Find the inverse of the function below on the given interval and write it in the form yequalsf Superscript negative 1 Baseline
Elena L [17]

Answer:

The inverse of the function is f^{-1}(x)=\frac{x-5}{3}.

Step-by-step explanation:

The function provided is:

f (x)=3x+5

Let f(x)=y.

Then the value of <em>x</em> is:

y=3x+5\\\\3x=y-5\\\\x=\frac{y-5}{3}

For the inverse of the function, x\rightarrow y.

⇒ f^{-1}(x)=\frac{x-5}{3}

Compute the value of f[f^{-1}(x)] as follows:

f[f^{-1}(x)]=f[\frac{x-5}{3}]

               =3[\frac{x-5}{3}]+5\\\\=x-5+5\\\\=x

Hence proved that f[f^{-1}(x)]=x.

Compute the value of f^{-1}[f(x)] as follows:

f^{-1}[f(x)]=f^{-1}[3x+5]

               =\frac{(3x+5)-5}{3}\\\\=\frac{3x+5-5}{3}\\\\=x

Hence proved that f^{-1}[f(x)]=x.

8 0
3 years ago
HELP ASAP!!!! EXPLAIN
Amanda [17]

Answer:

i think the answer is 50

Step-by-step explanation:

6 0
3 years ago
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