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3 years ago

Please help me thank you

Mathematics

1 answer:

Phantasy [73]3 years ago

3 0

The answer is B. The second one

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9x - 5 = 2x + 8<br><br><br> PLSSSS Help. Algebra 1

drek231 [11]

Step-by-step explanation:

9x - 2x = 8 + 5

7x = 13

x =13÷7

x = 1.8571

approximately = 1.86

I think this is it

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3 years ago

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A person jogs 1/2 miles in 1/12 hours. The person's speed is how many miles per hour?

ipn [44]

Answer:

6 miles per hour

Step-by-step explanation:

speed = distance/time

speed = 1/2÷1/12 which is the same as

1/2 x 12/1 = 6

6 miles per hour.

or ou could change the 1/2 to become 0.5

and the 1/12 to become 0.083333333

and divide 0.5 ÷ 0.083333333 = 6.000000024

rounded to one decima place become 6.0

6 miles per hour

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3 years ago

What is central tendency??

astra-53 [7]

DescriptionIn statistics, a central tendency is a central or typical value for a probability distribution. It may also be called a center or location of the distribution. Colloquially, measures of central tendency are often called averages

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3 years ago

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Find two vectors in R2 with Euclidian Norm 1<br> whoseEuclidian inner product with (3,1) is zero.

alina1380 [7]

Answer:

$v_1=(\frac{1}{10},-\frac{3}{10})$

$v_2=(-\frac{1}{10},\frac{3}{10})$

Step-by-step explanation:

First we define two generic vectors in our $\mathbb{R}^2$ space:

$v_1 = (x_1,y_1)$
$v_2 = (x_2,y_2)$

By definition we know that Euclidean norm on an 2-dimensional Euclidean space $\mathbb{R}^2$ is:

$\left \| v \right \|= \sqrt{x^2+y^2}$

Also we know that the inner product in $\mathbb{R}^2$ space is defined as:

$v_1 \bullet v_2 = (x_1,y_1) \bullet(x_2,y_2)= x_1x_2+y_1y_2$

So as first condition we have that both two vectors have Euclidian Norm 1, that is:

$\left \| v_1 \right \|= \sqrt{x^2+y^2}=1$

and

$\left \| v_2 \right \|= \sqrt{x^2+y^2}=1$

As second condition we have that:

$v_1 \bullet (3,1) = (x_1,y_1) \bullet(3,1)= 3x_1+y_1=0$

$v_2 \bullet (3,1) = (x_2,y_2) \bullet(3,1)= 3x_2+y_2=0$

Which is the same:

$y_1=-3x_1\\y_2=-3x_2$

Replacing the second condition on the first condition we have:

$\sqrt{x_1^2+y_1^2}=1 \\\left | x_1^2+y_1^2 \right |=1 \\\left | x_1^2+(-3x_1)^2 \right |=1 \\\left | x_1^2+9x_1^2 \right |=1 \\\left | 10x_1^2 \right |=1 \\x_1^2= \frac{1}{10}$

Since $x_1^2= \frac{1}{10}$ we have two posible solutions, $x_1=\frac{1}{10}$ or $x_1=-\frac{1}{10}$ . If we choose $x_1=\frac{1}{10}$ , we can choose next the other solution for $x_2$ .