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3 years ago

Can someone tell me… Given the right triangle ABC, what is the value of tan(A)?

Mathematics

1 answer:

sergey [27]3 years ago

6 0

Answer:

Step-by-step explanation:

tanA = $\frac{opposite}{adjacent}$ = $\frac{BC}{AC}$ = $\frac{24}{10}$ = $\frac{12}{5}$ → C

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Please help me simplify these expressions

Fantom [35]

Answer:

1. -18x¹¹

2. 3n⁷

Step-by-step explanation:

For these problems, there are two things you need to worry about: negative signs and exponents.

1. Let's look at the signs first. There is only one value with a negative sign, meaning that the negative sign will stay.

When multiplying with exponents, you have to add up the exponents. Don't forget the numerical coefficients.

-3x² · 3x · 2x³ · x⁵ = -18x¹¹

2. There are two negative signs in this probem, meaning that they will cancel out. Multiply the rest like we did in the first problem.

3n² · -n² · -n³ = 3n⁷

7 0

3 years ago

Please help, ty if you do :D

Dennis_Churaev [7]

Answer:

Your Answer is going to be C

3 0

3 years ago

Identify the slope of the line for the equation y = 10x + 3.

Iteru [2.4K]

Answer:324

Step-by-step explanation:

int i = 42.7; /* konwersja z double do int */

float f = i; /* konwersja z int do float */

double d = f; /* konwersja z float do double */

unsigned u = i; /* konwersja z int do unsigned int */

f = 4.2; /* konwersja z double do float */

i = d; /* konwersja z double do int */

char *str = "foo"; /* konwersja z const char* do char* [1] */

const char *cstr = str; /* konwersja z char* do const char* */

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Podcza

7 0

3 years ago

The tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm2 and a stand

Elanso [62]

Answer:

95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment.

Step-by-step explanation:

We are given that the tensile strength of stainless steel produced by a plant has been stable for a long time with a mean of 72 kg/mm 2 and a standard deviation of 2.15.

A machine was recently adjusted and a sample of 50 items were taken to determine if the mean tensile strength has changed. The mean of this sample is 74.28. Assume that the standard deviation did not change because of the adjustment to the machine.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean strength of 50 items = 74.28

$\sigma$ = population standard deviation = 2.15

n = sample of items = 50

$\mu$ = population mean tensile strength after machine was adjusted

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ 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 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $74.28-1.96 \times {\frac{2.15}{\sqrt{50} } }$ , $74.28+1.96 \times {\frac{2.15}{\sqrt{50} } }$ ]

= [73.68 kg/mm2 , 74.88 kg/mm2]

Therefore, 95% confidence interval for the mean of tensile strength after the machine was adjusted is [73.68 kg/mm2 , 74.88 kg/mm2].

Yes, this data suggest that the tensile strength was changed after the adjustment as earlier the mean tensile strength was 72 kg/mm2 and now the mean strength lies between 73.68 kg/mm2 and 74.88 kg/mm2 after adjustment.

8 0

3 years ago

{0, 2, -2} number line

ser-zykov [4K]

I drew a picture for you. It's below ^.^

4 0

3 years ago