All categories

2 years ago

Solve for x −9x+5≤17 OR 13x+25≤−1 correct answer gets: 100 points

Mathematics

2 answers:

victus00 [196]2 years ago

7 0

Answer:

The answer is x≤−2 OR x≥−4/3

dimaraw [331]2 years ago

6 0

-9x+5<17 answer x >-4/3 how I do to it
subtract -5 from both side and you get 12. so it be -9x<12
than you divide -9 from both side because you want x by it self you get -4/3 but you have to flip the sign since is negavite and you will be left over with ×>-4/3

You might be interested in

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

#SPJ4

6 0

10 months ago

How to write y-1=-x-6 In slope intercept form

DiKsa [7]

y=-x-5. you just need to add 1 to another side to get rid of 1 from the left side

4 0

3 years ago

Read 2 more answers

PLESE HELPPP im struggling please.

Gekata [30.6K]

Answer:

ab - b^2

Step-by-step explanation:

The area of the large rectangle is

A = l*w

=a*b

The area of the unshaded region is

A = b*b = b^2

The area of the shaded region is the area of the rectangle minus the area of the unshaded region

ab - b^2

8 0

2 years ago

What is the equation in slope-intercept form of a line that has a slope of -- and a y-intercept of 6?

Elza [17]

Answer:

Step-by-step explanation:

I fits the graph that is described

5 0

2 years ago

Solve the equation : <img src="https://tex.z-dn.net/?f=%20%5Ccos%28x%29%20-%20%20%5Csin%28x%29%20%20%3D%20%20%5Csqrt%7B2%20%

Debora [2.8K]

Answer:

General solution is

$x = n \pi + \frac{\pi }{8}$

Step-by-step explanation:

Step(i):-

Given cos x - sin x = √2 cos (3 x)

Dividing '√2' on both sides , we get

$\frac{1}{\sqrt{2} } cos (x) - \frac{1}{\sqrt{2} } sin (x) = \frac{\sqrt{2} cos (3 x)}{\sqrt{2} }$

we will use trigonometry formulas

a) Cos ( A + B) = Cos A Cos B - sin A sin B

b) $cos \frac{\pi }{4} = \frac{1}{\sqrt{2} }$

Step(ii):-

$\frac{1}{\sqrt{2} } cos (x) - \frac{1}{\sqrt{2} } sin (x) = \frac{\sqrt{2} cos (3 x)}{\sqrt{2} }$

$cos (\frac{\pi }{4} ) cos x - sin(\frac{\pi }{4} ) sin x = cos 3x$

$cos (\frac{\pi }{4}+x ) = cos 3 x$

Step(iii):-

General solution of cos x = cos ∝ is x = 2 nπ+∝

we have $cos (\frac{\pi }{4}+x ) = cos 3 x$

The general solution of $cos (\frac{\pi }{4}+x ) = cos 3 x$ is

⇒ $3 x = 2 n \pi + (\frac{\pi }{4}+x )$

⇒ $3 x- x = 2 n \pi + \frac{\pi }{4}$

$2x = 2 n \pi + \frac{\pi }{4}$

final answer:-

General solution is

$x = n \pi + \frac{\pi }{8}$

8 0

2 years ago