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3 years ago

10

in one baseball season, peter hit twice the difference of the number of home runs alice hit and 6. altogether, they hit 18 home

runs. hohw many home runs did each player hit that season?

1 answer:

Vedmedyk [2.9K]3 years ago

6 0

I think it would be 3 because if you divide 18/6=3

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I need help pls and thank you

Aleks [24]

The fourth table is the correct answer

6 0

3 years ago

What is 5/7*w=40 1/2

sukhopar [10]

Answer:

56 7/10

Step-by-step explanation:

convert 40 1/2 to a fraction 40 1/2 = 81/2

multiply both sides by 7/5

this gets w by itself on the left

81/2 x 7/5 = 567/10 = 56 7/10

7 0

3 years ago

25 pints of apple juice are poured into 3/4 pint bottles, how many bottles are filled

Westkost [7]

Answer:

33 1/3

Step-by-step explanation:

25/1 * 4/3 =

100/3 = 33.3... or 33 1/3

6 0

3 years ago

What is the length of the segment with endpoints (-3, 4) and (5, 4)? A) 2 B) 8 C) 16 D) 64 E) 2 17

Cerrena [4.2K]

Answer:

The correct answer is B) 8

Step-by-step explanation:

$\sqrt{(4-4)^{2}+(5+3)^{2} } \\\\\\\sqrt{8^{2} } \\\\8$

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 $x_2$ .

Remembering,

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

The two vectors we are looking for are:

$v_1=(\frac{1}{10},-\frac{3}{10})\\v_2=(-\frac{1}{10},\frac{3}{10})$

5 0

3 years ago