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3 years ago

Find all solutions in the interval [0,2pi) sec^2x + 2tanx = 3

1 answer:

6 0

Tan^2x + 1 + tanx - 3 = 0 tan^2x + tanx - 2 = 0 (tanx + 2)(tanx - 1) = 0 tanx = -2 or tanx = 1 x = 2.034 radians or x = 5.176 radians or x = pi/4 radians or x = 5pi/4 radians

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The client is to receive cyclophosphamide (Cytoxan) 50 mg/kg intravenously in divided doses over 5 days. The client weighs 176 p

marishachu [46]

Answer:

800 mg

Step-by-step explanation:

As the excerpt states that the client weights 176 pounds and is to receive cyclophosphamide (Cytoxan) 50 mg/kg, we have to convert 176 pounds to kg:

1 kg → 2.20 pounds

x ← 176 pounds

x= (176 pounds * 1 kg)/2.20 pounds= 80 kg

Now, we have to determine the total amount of cyclophosphamide (Cytoxan) that the client has to receive:

50 mg/kg

For 80 kg: 80 kg *50 mg/kg= 4,000 mg

As the dose is divided over 5 days:

4,000 mg / 5= 800 mg

The client will receive 800 mg of cyclophosphamide each day.

6 0

3 years ago

Theresa pays $9.56 for 4 pounds of tomatoes. What is the cost of 1 pound of tomatoes?

photoshop1234 [79]

Answer:

$2.39

Step-by-step explanation:

to find the cost of 1 pound , divide the cost of 4 pounds by 4

$9.56 ÷ 4 = $2.39

8 0

2 years ago

Hey! I'v been struggling with this question lately. Can someone just lend me a hand?

Dominik [7]

So, this problem is asking us to find the value of our variable u.

How can you do this? By isolating the variable, or having only it on one side of the equals sign.

Here's the step by step approach. It's just simple algebra and reverse order of operation (which is what you should do whenever you are solving for a variable).

$\frac{u+88}{11} = 17$

Multiply both sides by 11 to get rid of the 11 denominator.
$u + 88 = 187$

Then, subtract both sides by 88 to get u by itself.
$u = 99$

Therefore, u is equal to 99.

6 0

3 years ago

Read 2 more answers

What is 0.04 divided by 23.6

Tcecarenko [31]

The answer is 0.0017. Do you not have a calculator?

5 0

3 years ago

Read 2 more answers

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago