All categories

4 years ago

5x+20=70 solve for x

2 answers:

7 0

1st minus from 20. that cancels out. Then put it on the other side so it looks like 5x=70-20. simplify. 5x=50. next, divide by 5. 50/5=10. so x=10

Anarel [89]4 years ago

5 0

5x + 20 = 70
5x = 70 - 20
5x = 50
x = 50/5
x = 10

5(10) + 20 = 70
50 + 20 = 70
70 = 70

hope this helps :)

You might be interested in

38.2834. V1435. 36- 5637. v40. V4141. 214265 43. 12 445 8 -3ja 47. V3248. V4749. V99 50

aliya0001 [1]

No, your answer are wrong (I assume that the question asks to simplify the roots). You can only remove (their square is placed in from of the root and multiplied by it) the numbers from the root that are squares, the lest must stay, like this:

$\sqrt{14} = \sqrt{2*7} ,$ no roots are among the factors, nothing can be removed
$\sqrt{14} = \sqrt{2*7} ,$ nothing can be removed
$-\sqrt{39} = -\sqrt{3*13}$ nothing can be removed
$-\sqrt{56} = -\sqrt{2*2*2*7}=-2\sqrt{14} ,$
$-\sqrt{77} = -\sqrt{11*7} ,$ nothing can be removed
$\sqrt{41} = \sqrt{41} ,$ nothing can be removed
$\sqrt{21} = \sqrt{3*7} ,$ nothing can be removed
$- \sqrt{65} = -\sqrt{5*13} ,$ nothing can be removed
$-\sqrt{12} = -\sqrt{2*3*4} =-2\sqrt{4},$
$\sqrt{13} = \sqrt{13} ,$ nothing can be removed
$\sqrt{32} = \sqrt{4*4*2} =4\sqrt{2},$
$\sqrt{47} = \sqrt{47} ,$ (47 is prime) nothing can be removed
$-\sqrt{99} = -\sqrt{9*11}= -3\sqrt{11} ,$

4 0

3 years ago

7. Da'Quan is reading a 532-page book for English. His teacher says that each student must have read at least half of the book b

Radda [10]

Answer: x > 38

Step-by-step explanation:

Easy

7 0

3 years ago

Money from an allowance or job is known as what

sineoko [7]

Answer:

salary my guesss don't know right or wrong

3 0

3 years ago

Read 2 more answers

Consider the following two ordered bases of R3:

grigory [225]

Answer:

Let $A = (a_1, ..., a_n)$ and $B = (b_1, ..., b_n)$ bases of V. The matrix of change from A to B is the matrix n×n whose columns are vectors columns of the coordinates of vectors $b_1, ..., b_n$ at base A.

The, we case correspond to find the coordinates of vectors of C,

$\{\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right], \left[\begin{array}{ccc}2\\0\\-1\end{array}\right], \left[\begin{array}{ccc}-3\\1\\2\end{array}\right] \}$

at base B.

1. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&-1\\0&-1&1&-1\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&1\end{array}\right]$

now we use backward substitution

$c=1\\-b+c=-1,\; b=2\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\-1\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\2\\1\end{array}\right]$

2. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$

Then we find these values solving the linear system

$\left[\begin{array}{cccc}1&-2&2&2\\-1&2&-1&0\\0&-1&1&-1\end{array}\right]$

Using rows operation we obtain the echelon form of the matrix

$\left[\begin{array}{cccc}1&-2&2&2\\0&-1&1&-1\\0&0&1&2\end{array}\right]$

now we use backward substitution $c=2\\-b+c=-1,\; b=3\\a-2b+2c=2,\; a=4$

Then the coordinate vector of $\left[\begin{array}{ccc}2\\0\\-1\end{array}\right]$ is $\left[\begin{array}{ccc}4\\3\\2\end{array}\right]$

3. We need to find $a,b,c\in\mathbb{R}$ such that

$\left[\begin{array}{ccc}-3\\1\\2\end{array}\right]=a\left[\begin{array}{ccc}1\\-1\\0\end{array}\right]+b\left[\begin{array}{ccc}-2\\2\\-1\end{array}\right]+c\left[\begin{array}{ccc}2\\-1\\1\end{array}\right]$