All categories

2 years ago

5( x + 9) What is the answer to that

Mathematics

2 answers:

emmasim [6.3K]2 years ago

7 0

Answer:

5x + 45

Step-by-step explanation:

distribute 5 to both x and 9 so multiply them both

andre [41]2 years ago

5 0

Answer:

qwertyuioplkjhgfdsazxcvbnm

Step-by-step explanation:

You might be interested in

A) Convert 23ten to binary.

WITCHER [35]

Answer:

answer : 10111

Step-by-step explanation:

3 0

2 years ago

Which statements are true about polygons? Select three options.

Assoli18 [71]

Answer:

A polygon is any 2-dimensional shape formed with straight lines.

Step-by-step explanation:

5 0

2 years ago

Judy is selling small candles at the fair for $4 each and large candles for $6 each. She collected $144 selling a total of 28 ca

Pepsi [2]

Answer:

Number of small candles = 12

Number of large candles = 16

Step-by-step explanation:

Total candles = 28

x + y = 28 ---------------(I)

Cost of 'x' small candles = 4 *x = 4x

Cost of 'y' large candles = 6 *y = 6y

Total amount = $144

4x + 6y = 144 -------------(II)

Multiply equation (I) by (-4) and then add the equation. So, x will be eliminated and we can find the value of 'y'

(I) * (-4) -4x - 4y = -112

(II) 4x + 6y = 144 {Now, add}

2y = 32

y = 32/2

y = 16

Plug in y = 16 in equation (I)

x + 16 = 28

x = 28 - 16

x = 12

4 0

2 years ago

Read 2 more answers

Pls help! Identify the range of the function shown in the graph. Thanks

s2008m [1.1K]

Answer:

Step-by-step explanation:

Range is all y values of a graph and seeing that it has no arrows means its not infinite so its not B. The graph starts at 0 and ends at 5 so the smallest value is 0 and the largest is 5 therefore making the answer A.

6 0

2 years ago

Find two vectors in R2 with Euclidian Norm 1 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

2 years ago