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3 years ago

Use the information to answer the question.

Mathematics

1 answer:

stealth61 [152]3 years ago

6 0

Answer:

before the sale, each poster was $8.49

Step-by-step explanation:

27.36 divided by 4 = 6.84 + 1.65 = 849

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HELLP!! U WILL GET I,000 POINTS Alistair writes the system of equations below. 2x+3y=−60 2x+3y=120 -----------------------------

GaryK [48]

Answer:

Step-by-step explanation:

Does it say what x and y equals

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3 years ago

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A linear model is given for the data in the table: y=1.25x+2.

Fed [463]

Answer:

Option A

Step-by-step explanation:

Given that A linear model is given for the data in the table: y=1.25x+2.

Let us write observed values for each x and also the predicted values as per equation.

x 2 3 4 8 10 16 20 24 Total

y((O) 3 4 7 12 16 22 28 30

y(P) 4.5 5.75 7 12 14.5 22 27 32

DEv 1.5 1.75 0 0 1.5 0 1 2 7 75

where y(0) represents observed y or y in the table given

y(P) gives values of y predicted as per the equation 1.25x+2

Dev represents the absolute difference

Hence answer is option

A.7.75

8 0

3 years ago

Which juice container is larger: 2l or 1 gal?

tangare [24]

The conversion between Litres and gallons is about 4L/gal,
so 2L * 1gal/ 4L = 1/2 gal
The 1 gal is about twice as large

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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 $x_2$ .