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2 years ago

Alexa uses the binomial theorem to correctly expand (x+2)^4. which is one of the terms in alexa's answer?

2 answers:

8 0

^0 ...............1
^1..............1 1
^2............1 2 1
^3...........1 3 3 1
^4..........1 4 6 4 1

(x+2)^4 = 1(x)^4(2)^0 + 4(x)^3(2)^1 + 6(x)^2(2)^2 + 4(x)^1(2)^3 + 1(x)^0(2)4
= $x^{4} + 8 x^{3} + 24 x^{2} + 32x + 16$

Aloiza [94]2 years ago

6 0

The first term is x^4.
The second term is 8x^3

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381.10-214.43 equals

dmitriy555 [2]

Answer:

166.67

Step-by-step explanation:

3 0

2 years ago

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ?

KIM [24]

Answer:

The answer is

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$tan(\theta)=-1$

Step-by-step explanation:

we know that

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

$sin^{2}(\theta)+cos^{2}(\theta)=1$

In this problem we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

$\frac{3\pi}{2}$

The angle $\theta$ belong to the third or fourth quadrant

The value of $sin(\theta)$ is negative

Step 1

Find the value of $sin(\theta)$

Remember

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$sin^{2}(\theta)+(\frac{\sqrt{2}}{2})^{2}=1$

$sin^{2}(\theta)=1-\frac{1}{2}$

$sin^{2}(\theta)=\frac{1}{2}$

$sin(\theta)=-\frac{\sqrt{2}}{2}$ ------> remember that the value is negative

Step 2

Find the value of $tan(\theta)$

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

we have

$sin(\theta)=-\frac{\sqrt{2}}{2}$

$cos(\theta)=\frac{\sqrt{2}}{2}$

substitute

$tan(\theta)=\frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}$

$tan(\theta)=-1$

8 0

2 years ago

Read 2 more answers

The net of a triangular prism is shown below. What is the surface area of the prism? A. 128 cm^2 B. 152 cm^2 C. 176 cm^2 D. 304

shtirl [24]

Answer:

B. 152 cm²

Step-by-step explanation:

To find the surface area using a net, do this:

Take apart the figure. We see that there are three rectangles and two triangles. Find the area of each ( $A=l*w$ ) and then add the values together:

The first rectangle on the left is the same as the one on the right.

$5*8=40$

Two measures are 40 cm².

The middle rectangle is:

$6*8=48$

48 cm²

The formula for the area of a triangle is $A=\frac{1}{2}*b*h$ :

$A=\frac{1}{2}*6*4\\\\A=\frac{1*6*4}{2}\\\\A=\frac{24}{2}\\\\ A=12$

The area of the two triangles is 12 cm².

Now add the values:

$40+40+48+12+12=152$

The area of the figure is 152 cm².

:Done

4 0

2 years ago

Tan 235° = 2tan20°+ tan215°

Mariulka [41]

Given : tan 235 = 2 tan 20 + tan 215

To Find : prove that

Solution:

tan 235 = 2 tan 20 + tan 215

Tan x = Tan (180 + x)

tan 235 = tan ( 180 + 55) = tan55

tan 215 = tan (180 + 35) = tan 35

=> tan 55 = 2tan 20 + tan 35

55 = 20 + 35

=> 20 = 55 - 35

taking Tan both sides

=> Tan 20 = Tan ( 55 - 35)

=> Tan 20 = (Tan55 - Tan35) /(1 + Tan55 . Tan35)

Tan35 = Cot55 = 1/tan55 => Tan55 . Tan35 =1

=> Tan 20 = (Tan 55 - Tan 35) /(1 + 1)

=> Tan 20 = (Tan 55 - Tan 35) /2

=> 2 Tan 20 = Tan 55 - Tan 35

=> 2 Tan 20 + Tan 35 = Tan 55

=> tan 55 = 2tan 20 + tan 35

=> tan 235 = 2tan 20 + tan 215

QED

Hence Proved

5 0

1 year ago

Tesla wanted to determine the average miles per kWh that their vehicles get across all models and variations. They took a sample

LiRa [457]

Answer:

$\mu_{\bar x} = \mu = E(X) =30KWh$

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

For this case we select a sample of n =100

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

So then the sample mean would be:

$\mu_{\bar x} = \mu = E(X) =30KWh$

And the standard deviation would be:

$\sigma_{\bar X}= \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{100}}=0.3$

6 0

1 year ago