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3 years ago

Which of the following vectors are orthogonal to (1,5)? Check all that apply

1 answer:

7 0

We know that
If the scalar product of two vectors is zero, both vectors are orthogonal

A. (-2,5)
(-2,5)*(1,5)-------> -2*1+5*5=23-----------> are not orthogonal

B. (10,-2)
(10,-2)*(1,5)-------> 10*1-2*5=0-----------> are orthogonal

C. (-1,-5)
(-1,-5)*(1,5)-------> -1*1-5*5=-26-----------> are not orthogonal

D. (-5,1)
(-5,1)*(1,5)-------> -5*1+1*5=0-----------> are orthogonal

the answer is
B. (10,-2) and D. (-5,1) are orthogonal to (1,5)

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Major league baseball salaries averaged $1.5 million with a standard deviation of $0.8 million in 1994. suppose a sample of 100

seropon [69]

Answer: The probability that the avg. salary of the 100 players exceeded $1 million is approximately 1.

Explanation:

Step 1: Estimate the standard error. Standard error can be calcualted by dividing the standard deviation by the square root of the sample size:

$SE=\frac{0.8}{\sqrt{100}}=\frac{0.8}{10} = 0.08$

So, Standard Error is 0.08 million or $80,000.

Step 2: Next, estimate the mean is how many standard errors below the population mean $1 million.

$\frac{1 - 1.5}{0.08}$

$=-6.250$

-6.250 means that $1 million is siz standard errors away from the mean. Since, the value is too far from the bell-shaped normal distribution curve that nearly 100% of the values are greater than it.

Therefore, we can say that because 100% values are greater than it, probability that the avg. salary of the 100 players exceeded $1 million is approximately 1.

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3 years ago

Which of the x-values are solutions to the following inequality? X < 365

e-lub [12.9K]

A and b because both values are less than 365

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2 years ago

Solve for X PLEASE HELP ITS URGENT

almond37 [142]

With pythagorean theorem and working backwards, you do 8100-1225 to get 6875, you then get the square root of that to get x=82.91 or rounded to 82.92

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3 years ago

Read 2 more answers

Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your

Darina [25.2K]

Answer:

Given definite integral as a limit of Riemann sums is:

$\lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Step-by-step explanation:

Given definite integral is:

$\int\limits^7_4 {\frac{x}{2}+x^{3}} \, dx \\f(x)=\frac{x}{2}+x^{3}---(1)\\\Delta x=\frac{b-a}{n}\\\\\Delta x=\frac{7-4}{n}=\frac{3}{n}\\\\x_{i}=a+\Delta xi\\a= Lower Limit=4\\\implies x_{i}=4+\frac{3}{n}i---(2)\\\\then\\f(x_{i})=\frac{x_{i}}{2}+x_{i}^{3}$

Substituting (2) in above

$f(x_{i})=\frac{1}{2}(4+\frac{3}{n}i)+(4+\frac{3}{n}i)^{3}\\\\f(x_{i})=(2+\frac{3}{2n}i)+(64+\frac{27}{n^{3}}i^{3}+3(16)\frac{3}{n}i+3(4)\frac{9}{n^{2}}i^{2})\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{3}{2n}i+\frac{144}{n}i+66\\\\f(x_{i})=\frac{27}{n^{3}}i^{3}+\frac{108}{n^{2}}i^{2}+\frac{291}{2n}i+66\\\\f(x_{i})=3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

Riemann sum is:

$= \lim_{n \to \infty} \sum^{n} _{i=1}3[\frac{9}{n^{3}}i^{3}+\frac{36}{n^{2}}i^{2}+\frac{97}{2n}i+22]$

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4 years ago

Write a division problem whose quotient has its first digit in the hundreds place

Katen [24]

Here's one --> 1000÷100=10

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3 years ago