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3 years ago

Given: 3x + 1 = -14; Prove: x = -5

Mathematics

2 answers:

MrRissso [65]3 years ago

7 0

Answer:

x = -5

Step-by-step explanation:

Solve for x:

3 x + 1 = -14

Subtract 1 from both sides:

3 x + (1 - 1) = -14 - 1

1 - 1 = 0:

3 x = -14 - 1

-14 - 1 = -15:

3 x = -15

Divide both sides of 3 x = -15 by 3:

(3 x)/3 = (-15)/3

3/3 = 1:

x = (-15)/3

The gcd of -15 and 3 is 3, so (-15)/3 = (3 (-5))/(3×1) = 3/3×-5 = -5:

Answer: x = -5

kozerog [31]3 years ago

6 0

Answer:

Step-by-step explanation:

3x + 1 = -14;

substr 1 : 3x+1-1 = - 14 -1

3x = -15

divid by 3

x = -5

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Y=-4.3x-1.3<br> Y=1.7x+4.7

Contact [7]

Answer:

0.566666666

Step-by-step explanation:

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3 years ago

let t : r2 →r2 be the linear transformation that reflects vectors over the y−axis. a) geometrically (that is without computing a

tangare [24]

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

See the figure for the graph:

(a) for any (x, y) ∈ R² the reflection of (x, y) over the y - axis is ( -x, y )

∴ x → -x hence '-1' is the eigen value.

∴ y → y hence '1' is the eigen value.

also, ( 1, 0 ) → -1 ( 1, 0 ) so ( 1, 0 ) is the eigen vector for '-1'.

( 0, 1 ) → 1 ( 0, 1 ) so ( 0, 1 ) is the eigen vector for '1'.

(b) ∵ T(x, y) = (-x, y)

T(x) = -x = (-1)(x) + 0(y)

T(y) = y = 0(x) + 1(y)

Matrix Representation of $T = \left[\begin{array}{cc}-1&0\\0&1\end{array}\right]$

now, eigen value of 'T'

$T - kI = \left[\begin{array}{cc}-1-k&0\\0&1-k\end{array}\right]$

after solving the determinant,

we get two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Hence,

(a) ( 1, 0 ) is the eigen vector for '-1' and ( 0, 1 ) is the eigen vector for '1'.

(b) two eigen values of 'k' = 1, -1

for k = 1, eigen vector is $\left[\begin{array}{c}0\\1\end{array}\right]$

for k = -1 eigen vector is $\left[\begin{array}{c}1\\0\end{array}\right]$

Learn more about " Matrix and Eigen Values, Vector " from here: brainly.com/question/13050052

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1 year ago

Alvarez has $650 deposited into his bank account. The account earns simple interest of 5. 5% per year. No other money is added o

Elena-2011 [213]

Answer:

$793

Step-by-step explanation:

The amount of money in an account earning simple interest is given by the formula ...

A = P(1 +rt)

where P is the principal invested at rate r for t years.

Using the values given in the problem statement, the account balance can be found to be ...

A = $650(1 +0.055×4) = $650×1.22 = $793

There will be $793 in the account after 4 years.

4 0

2 years ago

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT

fredd [130]

Here you go!! It's always best to graph these questions, if you have graph paper near.
If not there are online graphing websites to use!! :))

ANSWER: Choice D (4th answer)

7 0

3 years ago

Which graph best represents the solution to this system of inequalities?

Agata [3.3K]

Answer:

Step-by-step explanation:

D is the only graph where the shaded region is on 6 and less.

6 0

2 years ago