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4 years ago

What are the values of x and y ? Log(3)x-log(3)3=log(3)y

Mathematics

1 answer:

Step2247 [10]4 years ago

8 0

Answer:

Step-by-step explanation:

$log_{3} x-log_{3}3=log_{3}y\\log_{3}(\frac{x}{3})=log_{3}y\\\frac{x}{3}=y\\x=3y\\y\geq 0,x \geq 0$

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QUESTION IS IN PICTURE, FIRST GETS BRAINLIEST

Irina18 [472]

Answer:

Step-by-step explanation:

65 + 70 + 73 + 49 +61 + 38 + 41 + 16 + 19 = 432

432 ÷ 9 = 48

8 0

3 years ago

Read 2 more answers

Answer all please 50 points

timofeeve [1]

Answer:

1/2 x 5= 2.5

6/10 x 8 = 4.8

3/6 x 9 = 4.5

1/6 x 5 = 0.8333333333 recuring

Step-by-step explanation:

1/2 x 5

1 ÷ 2 = 0.5

0.5 x 5 = 2.5

6/10 x 8

6 ÷ 10 = 0.6

0.6 x 8 = 4.8

3 ÷ 6 x 9

3 ÷ 6 = 0. 5

0. 5 x 9 = 4.5

1/6 x 5

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6 0

4 years ago

Read 2 more answers

Laryngeal cancer rates in smokers is 160.0 (per 100,000) and 25.0 (per 100,000) among nonsmokers. Among smokers, what percentage

Romashka [77]

Answer:

0.16%

Step-by-step explanation:

From the statement of the question;

Number of Laryngeal cancer due to smoking = 160

Population of smokers = 100,000

Hence the percentage of smokers liable to have Laryngeal cancer = 160/100000 ×100/1

=0.16%

Hence 0.16% of smokers are liable to Laryngeal cancer

4 0

3 years ago

What is the length of L

Elanso [62]

Answer:

Assuming the height is still 1 your answer would be 700

Step-by-step explanation:

6 0

3 years ago

Can you prove it??<br>it's hard,I tried but couldn't solve this.

BARSIC [14]

I'll abbreviate $s=\sin\theta$ and $c=\cos\theta$ , so the identity to prove is

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)$

On the left side, we can simplify a bit:

$\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}$

$\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}$

Then

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)$

So the establish the original equality, we need to show that

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s$

Combine the fractions:

$\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}$

We can rewrite the denominator as

$c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1$

then using the fact that $c^2+s^2=\cos^2\theta+\sin^2\theta=1$ , we get

$1-2s^2+2s-1=2s-2s^2$

so that we have

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s$

as desired.

6 0

4 years ago