All categories

3 years ago

Need help asap!!! please

Mathematics

1 answer:

Mama L [17]3 years ago

7 0

Answer:

x = -8

Step-by-step explanation:

g(x) = 8x + 2

given: g(x) = 8x + 2 = -62

8x = -62 -2

8x = -64

x = -8

You might be interested in

Please help me with the one that says convince me !

lora16 [44]

Answer:

Cory is correct

Step-by-step explanation:

When you divide 3.9 by 0.52 the quotient is 7.5. 7.5 will round up to 8.

4 0

2 years ago

Which expression is equivalent to 3xy + 10xy - 5x + 2x - 2yx? A) 15xy + 7x Eliminate B) 13xy - 7x - 2yx C) -8xy D) -3x + 11xy

Vedmedyk [2.9K]

3xy + 10xy - 5x + 2x - 2yx

Add 3xy to 10xy

13xy - 5x + 2x - 2yx

Add -5x to 2x

Final Answer: B.) 13xy -3x - 2yx

3 0

2 years ago

Read 2 more answers

Simplify these 12 radicals:

Alexandra [31]

1)
$4 \sqrt{5}$
2)
$8$
3)
$3 \sqrt{5}$
4)
$3 \sqrt{6}$
5)
$4 \sqrt{2}$
6)
$6$
7)
$5 \sqrt{6}$

8)
$2 \sqrt{5}$
9)
$6 \sqrt{5}$
10)
$6 \sqrt{2}$
11)
$2 \sqrt{2}$
12)
$5 \sqrt{2}$

6 0

3 years ago

What is<br> |-3+45-(-14)|+(-|-87+12/46|<br> simplified

USPshnik [31]

|x|=absoulte value of x which means whatever x is, make it positive
so first solve any division/multiplication

the only one is 12/46=6/23
so we do the absoulte value signs
-3+45-(-14)
-(-14)=+14
-3+45+14=56
absoulute value of 56=56

56+(-|-87+6/23|)
-87+6/23=-86 and 17/23
absoulte value of -86 and 17/23=86 and 17/23
there is a negative sign in front of the absoulte vaulue so make it negative
56-86 and 17/23=-30 and 17/23 or about -30.73913

4 0

3 years ago

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

2 years ago