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3 years ago

Do the ratios 8/3 and 18/7 form a proportion

Mathematics

2 answers:

r-ruslan [8.4K]3 years ago

7 0

Answer:

no 8/3 and 18/7 are not proportional

Step-by-step explanation:

8/3 = 2.667

18/7=2.571

2.571≠2.667

Anastaziya [24]3 years ago

4 0

Answer:

8 * 7 = 3 * 18

Simplifying

8 * 7 = 3 * 18

Multiply 8 * 7

56 = 3 * 18

Multiply 3 * 18

56 = 54

Solving

56 = 54

Step-by-step explanation:

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To rent a certain meeting room, a college charges a reservation fee of

Ann [662]

So this is an algebra question, where you have have to setup the equation. Let t represent the number of hours.
36 + 8.80t < 115.20 we can then setup the equation like this to solve for the max hours.
36 + 8.80t= 115.20

Now solve
8.80t = 115.2-36
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3 years ago

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saveliy_v [14]

Answer:

yes

Step-by-step explanation:

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3 years ago

What is 4/5 minus 2/3

Rzqust [24]

Answer:2

Step-by-step explanation:

Make the numbers whole by multiplying by 15 to both fractions. 4/5 would become 12 and 2/3 would become 10. Subtract 12-10 to get 2.

8 0

3 years ago

Read 2 more answers

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 $x_2$ .