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Sunny_sXe [5.5K]
3 years ago
15

Tan 1.1 – tan 4.6/

Mathematics
1 answer:
VladimirAG [237]3 years ago
7 0

The value of the given equation is –0.37458.

Solution:

Given equation is:

$\frac{\tan 1.1-\tan 4.6}{1+\tan 1.1 \tan 4.6}

Let us first find the values.

The value of tan 1.1 = 1.96475

The value of tan 4.6 = 8.86017

Substitute these values in the given equation.

$\frac{\tan 1.1-\tan 4.6}{1+\tan 1.1 \tan 4.6}

            $=\frac{1.96475-8.86017}{1+1.96475 \times 8.86017}

            $=\frac{-6.89542}{1+17.4080}

            $=\frac{-6.89542}{18.4080}

            = –0.37458

$\frac{\tan 1.1-\tan 4.6}{1+\tan 1.1 \tan 4.6}=-0.37458

Hence the value of the given equation is –0.37458.

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Stephon has a square brick patio. He wants to reduce the width by 4 feet and increase the length by 4 feet.
alekssr [168]

Answer:  The expressions for the length and width of the new patio are

\ell=x+4,~~w=x-4.

And the area of the new patio is 384 sq. feet.

Step-by-step explanation:  Given that Stephen has a square brick patio. He wants to reduce the width by 4 feet and increase the length by 4 feet.

The length of one side of the square patio is represented by x.

We are to write the expressions for the length and width of the new patio and then to find the area of the new patio if the original patio measures 20 feet by 20 feet.

Since Stephen wants to reduce width of the patio by 4 feet, so the width of the new patio will be

w=(x-4)~\textup{feet}.

The length of the patio is increased by 4 feet, so the length of the new patio will be

\ell=(x+4)~\textup{feet}.

Now, if the original patio measures 20 feet by 20 feet, then we must have

w=x-4=20-4=16~\textup{feet}

and

\ell=x+4=20+4=24~\textup{feet}.

Therefore, the area of the new patio is given by

A_n=\ell \times w=24\times16=384~\textup{sq. feet}.

Thus, the expressions for the length and width of the new patio are

\ell=x+4,~~w=x-4.

And the area of the new patio is 384 sq. feet.

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4 years ago
What is the ratio table for 31​
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Answer:

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1    2   3

Step-by-step explanation:

5 0
4 years ago
Order 34 x 10^2, 1.2 x 10^7, 8.11 x 10^-3, and 435 from least to greatest.
Alex
The answer will be
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3 years ago
Alternate exterior angles are congruent.
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Answer:

<1 = <8

Step-by-step explanation:

An angle that will form a pair of alternate exterior angles with <1 will be the angle that is found outside of the parallel lines but on the opposite side of the transversal that cuts across the two parallel lines.

<8 is an exterior angle, and also is alternate angle to <1. It forms a pair of alternate exterior angles pair with <1.

Therefore, <1 = <8

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In a​ study, researchers wanted to measure the effect of alcohol on the hippocampal​ region, the portion of the brain responsibl
schepotkina [342]

Answer:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

Step-by-step explanation:

Information given

\bar X=8.19 cm^3 represent the sample mean

s=0.8 cm^3 represent the sample deviation

n=17 sample size  

\mu_o =9.02 represent the value to verify

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic

p_v represent the p value

System of hypothesis to check

We want to check if the true mean is less than the normal value of 9.02 cm^3, the system of hypothesis would be:  

Null hypothesis:\mu \geq 9.02  

Alternative hypothesis:\mu < 9.02  

The statistic for this case is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

Replacing the info given we got:

t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28    

The degrees of freedom are given by

df=n-1=17-1=16  

The p value would be given by this probability:

p_v =P(t_{(16)}  

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

6 0
3 years ago
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