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3 years ago

Verify identities Will give BRAINLIST cot^2(x) - csc^2(x) = -1 show steps

Mathematics

1 answer:

san4es73 [151]3 years ago

4 0

Answer:

Step-by-step explanation:

∫tan(x)dx∫cot(x)dx∫sec(x)dx∫csc(x)dx====−ln∣∣cos(x)∣∣+C=ln∣∣sec(x)∣∣+Cln∣∣sin(x)∣∣+C=−ln∣∣csc(x)∣∣+Cln∣∣sec(x)+tan(x)∣∣+C−ln∣∣csc(x)+cot(x)∣∣+C

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HELP!

konstantin123 [22]

Answer:

1. 10 square yards

2. $9.10

3. 36 girls

4. 20 cherries

5. 159 bracelets

Step-by-step explanation:

The area of this figure would be the area of the SQUARE minus the area of the RECTANGLE (taken out).

The middle part is taken out (this is a rectangle).

View this as a square with sides 4 by 4 with a rectangle portion taken out. So the area can be said as:

Area = Area of Whole Square - Area of Middle Rectangle Taken Out

Lets calculate:

Area of Whole Square:

side * side = 4 * 4 = 16

Now,

Area of Middle Rectangle Taken Out:

Length * Width

The length is given as 3

The width would be 4 - two 1 from two sides, that is 4 - 1 - 1 = 2

So, area = 3 * 2 = 6

Area of figure = 16 - 6 = 10 square yards

We have to find 20% of the bill.

The bill was 45.50 and we have to find 20% of 45.50.

We first convert percentage to decimal by dividing by 100:

20% = 20/100 = 0.20

Now we multiply this decimal with the bill to get our desired answer:

0.20 * 45.50 = $9.10

The tip is $9.10

The ratio of boys (b) to girls (g) would be 2 is to 3.

There are 24 boys.

We can set up a ratio (fraction) with boys on top and girls on bottom.

Then we equate it with actual number of boys and girls (let it be g). So the ratio set-up would be:

$\frac{2}{3}=\frac{24}{g}$

To solve for g, the number of girls, we have to cross multiply and use algebra to solve. The process is shown below:

$\frac{2}{3}=\frac{24}{g}\\2*g = 24*3\\2g=72\\g=\frac{72}{2}\\g=36$

There are 36 girls in sixth grade

For every "9" lemon, there are "4" cherries.

There are 45 lemons

If we think of "9" lemons as a "box", how many boxes of lemons are there?

45/9 = 5

Similarly, there would be "5" boxes of cherries. But there are "4" cherries per box. So there are:

4 * 5 = 20 cherries

We can also set up a ratio of cherry to lemon as 4 is to 9 and equate it with actual cherries (let it be "c") and lemons (45). Cross multiply and use algebra to solve for cherries ("c"). Algebra process shown below:

$\frac{Cherries}{Lemons}=\frac{4}{9}=\frac{c}{45}\\9*c=4*45\\9c=180\\c=\frac{180}{9}\\c=20$

See, we have the same answer, 20 cherries!

Number sold total = 14 + 27 + 45 = 86

She still has 73 in her stock

So, she basically started with whatever she now has (73) and whatever she has sold the 3 days (86), that would be:

73 + 86 = 159 bracelets

Tia started with 159 bracelets

4 0

3 years ago

Two decimal numbers have a difference of 0.7 And and a sum of 1.What are the numbers?

klio [65]

The two numbers are 0.85 and 0.15
if you add them together you get 1, 0.85+0.15=1
and if you subtract them you get 0.7, 0.85-0.15=0.7

3 0

3 years ago

Read 2 more answers

An article describes an experiment to determine the effectiveness of mushroom compost in removing petroleum contaminants from so

alexgriva [62]

Solution :

Let $p_1$ and $p_2$ represents the proportions of the seeds which germinate among the seeds planted in the soil containing $3\%$ and $5\%$ mushroom compost by weight respectively.

To test the null hypothesis $H_0: p_1=p_2$ against the alternate hypothesis $H_1:p_1 \neq p_2$ .

Let $\hat p_1, \hat p_2$ denotes the respective sample proportions and the $n_1, n_2$ represents the sample size respectively.

$$\hat p_1 = \frac{74}{155} = 0.477419$

$n_1=155$

$$p_2=\frac{86}{155}=0.554839$

$n_2=155$

The test statistic can be written as :

$$z=\frac{(\hat p_1 - \hat p_2)}{\sqrt{\frac{\hat p_1 \times (1-\hat p_1)}{n_1}} + \frac{\hat p_2 \times (1-\hat p_2)}{n_2}}}$

which under $H_0$ follows the standard normal distribution.

We reject $H_0$ at $0.05$ level of significance, if the P-value or if $|z_{obs}|>Z_{0.025}$

Now, the value of the test statistics = -1.368928

The critical value = $\pm 1.959964$

P-value = $P(|z|> z_{obs})= 2 \times P(z< -1.367928)$

$=2 \times 0.085667$

= 0.171335

Since the p-value > 0.05 and $|z_{obs}| \ngtr z_{critical} = 1.959964$ , so we fail to reject $H_0$ at $0.05$ level of significance.

Hence we conclude that the two population proportion are not significantly different.

Conclusion :

There is not sufficient evidence to conclude that the $\text{proportion}$ of the seeds that $\text{germinate differs}$ with the percent of the $\text{mushroom compost}$ in the soil.