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Daniel [21]
3 years ago
9

F4f follow me on ig Fendimulatto Juicyminkz

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
4 0

Answer:

ok

Step-by-step explanation:

ok i will follow because I can get points

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a family pays 45 dollars each month for cable television.the cost increase 7 percent.how many dollars is the increase?
sineoko [7]
7% of $45:
              = 0.07 * 45
              = 3.15
Increase = $3.15
8 0
3 years ago
A sales agent makes 40 calls a day. About 15% he calls result in a sale. How many days
Tems11 [23]

Answer: 20 days

Step-by-step explanation:

Since the sale agent makes 40 calls a day and about 15% he calls result in a sale. This means the number of sales will be:

= 15% × 40

= 0.15 × 40

= 6

6 sales out of every 40 calls per day.

The number of calls to make 120 sales will be:

= 15% of x = 120

0.15 × x = 120

0.15x = 120

x = 120/0.15

x = 800 calls

Since he needs 800 calls and he makes 40 calls per day, the number of days needed will be:

= 800/40

= 20 days

6 0
3 years ago
What equation is solved by the graphed systems of equations?
vodka [1.7K]

Answer:

simultaneous linear equations

Step-by-step explanation:

y=7x+3

y=x-3

8 0
3 years ago
Find the exact value of the trigonometric expression.
Dmitrij [34]
\bf \textit{Half-Angle Identities}
\\\\
sin\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1-cos(\theta)}{2}}
\qquad 
cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}}\\\\
-------------------------------\\\\
\cfrac{1}{12}\cdot 2\implies \cfrac{1}{6}\qquad therefore\qquad \cfrac{\pi }{12}\cdot 2\implies \cfrac{\pi }{6}~thus~ \cfrac{\quad \frac{\pi }{6}\quad }{2}\implies \cfrac{\pi }{12}

\bf sec\left( \frac{\pi }{12} \right)\implies sec\left( \cfrac{\frac{\pi }{6}}{2} \right)\implies \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\impliedby \textit{now, let's do the bottom}
\\\\\\
cos\left( \cfrac{\frac{\pi }{6}}{2} \right)=\pm\sqrt{\cfrac{1+cos\left( \frac{\pi }{6} \right)}{2}}\implies \pm\sqrt{\cfrac{1+\frac{\sqrt{3}}{2}}{2}}\implies \pm\sqrt{\cfrac{\frac{2+\sqrt{3}}{2}}{2}}

\bf \pm \sqrt{\cfrac{2+\sqrt{3}}{4}}\implies \pm\cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}\implies \pm \cfrac{\sqrt{2+\sqrt{3}}}{2}
\\\\\\
therefore\qquad \cfrac{1}{cos\left( \frac{\frac{\pi }{6}}{2} \right)}\implies \cfrac{2}{\sqrt{2+\sqrt{3}}}


which simplifies thus far to 

\bf \cfrac{2}{\sqrt{2+\sqrt{3}}}\cdot \cfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\implies \cfrac{2\sqrt{2+\sqrt{3}}}{2+\sqrt{3}}\cdot \stackrel{conjugate}{\cfrac{2-\sqrt{3}}{2-\sqrt{3}}}
\\\\\\
\cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{2^2-(\sqrt{3})^2}\implies \cfrac{2\sqrt{2+\sqrt{3}}(2-\sqrt{3})}{1}

\bf 2\sqrt{2+\sqrt{3}}(2-\sqrt{3})\impliedby \stackrel{\textit{keep in mind that}}{(2-\sqrt{3})=(\sqrt{2-\sqrt{3}})(\sqrt{2-\sqrt{3}})}
\\\\\\
2\sqrt{2+\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}\cdot \sqrt{2-\sqrt{3}}
\\\\\\
2\sqrt{(2+\sqrt{3})(2-\sqrt{3})\cdot (2-\sqrt{3})}
\\\\\\
2\sqrt{[2^2-(\sqrt{3})^2]\cdot (2-\sqrt{3})}\implies 2\sqrt{1(2-\sqrt{3})}
\\\\\\
2\sqrt{2-\sqrt{3}}
7 0
4 years ago
WORTH 30 POINTS. Properties of Logarithm: Please help! I do not understand how to solve this kind of problem.
horrorfan [7]
Logb1=0 log b 1 = 0 . This follows from the fact that b0=1 b 0 = 1 .
logbb=1 log b b = 1 . This follows from the fact that b1=b b 1 = b .
logbbx=x log b b x = x . This can be generalized out to logbbf(x)=f(x) log b b f ( x ) = f ( x ) .
7 0
3 years ago
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