All categories

2 years ago

Solve the inequality. -1/3w ≤ -5

Mathematics

1 answer:

omeli [17]2 years ago

3 0

Answer:

$\frac{ - 1}{3} w \leqslant - 5$

$w \leqslant - 5 \div \frac{ - 1}{3}$

$w \leqslant - 5 \times - 3$

$w \leqslant 15$

You might be interested in

First you to find the worksheet and download it plase I need help

Goshia [24]

Answer:

a) The horizontal asymptote is y = 0

The y-intercept is (0, 9)

b) The horizontal asymptote is y = 0

The y-intercept is (0, 5)

c) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

d) The horizontal asymptote is y = 3

The y-intercept is (0, 4)

e) The horizontal asymptote is y = -1

The y-intercept is (0, 7)

The x-intercept is (-3, 0)

f) The asymptote is y = 2

The y-intercept is (0, 6)

Step-by-step explanation:

a) f(x) = $3^{x + 2}$

The asymptote is given as x → -∞, f(x) = $3^{x + 2}$ → 0

∴ The horizontal asymptote is f(x) = y = 0

The y-intercept is given when x = 0, we get;

f(x) = $3^{0 + 2}$ = 9

The y-intercept is f(x) = (0, 9)

b) f(x) = $5^{1 - x}$

The asymptote is fx) = 0 as x → ∞

The asymptote is y = 0

Similar to question (1) above, the y-intercept is f(x) = $5^{1 - 0}$ = 5

The y-intercept is (0, 5)

c) f(x) = 3ˣ + 3

The asymptote is 3ˣ → 0 and f(x) → 3 as x → ∞

The asymptote is y = 3

The y-intercept is f(x) = 3⁰ + 3= 4

The y-intercept is (0, 4)

d) f(x) = 6⁻ˣ + 3

The asymptote is 6⁻ˣ → 0 and f(x) → 3 as x → ∞

The horizontal asymptote is y = 3

The y-intercept is f(x) = 6⁻⁰ + 3 = 4

The y-intercept is (0, 4)

e) f(x) = $2^{x + 3}$ - 1

The asymptote is $2^{x + 3}$ → 0 and f(x) → -1 as x → -∞

The horizontal asymptote is y = -1

The y-intercept is f(x) = $2^{0 + 3}$ - 1 = 7

The y-intercept is (0, 7)

When f(x) = 0, $2^{x + 3}$ - 1 = 0

$2^{x + 3}$ = 1

x + 3 = 0, x = -3

The x-intercept is (-3, 0)

f) $f(x) = \left (\dfrac{1}{2} \right)^{x - 2} + 2$

The asymptote is $\left (\dfrac{1}{2} \right)^{x - 2}$ → 0 and f(x) → 2 as x → ∞

The asymptote is y = 2

The y-intercept is f(x) = $f(0) = \left (\dfrac{1}{2} \right)^{0 - 2} + 2 = 6$

The y-intercept is (0, 6)

7 0

2 years ago

Indicate the answer choice that best completes the statemento Dr What is the value of |-6|?

sleet_krkn [62]

6 absolute value is 6

4 0

2 years ago

NEED HELP PLEASE!

USPshnik [31]

Answer:

$(x + (3+5i))^2 = x^2 + 6x + (10x+30)i - 16$

Step-by-step explanation:

Complex numbers identity:

The complex numbers identity is:

$i^2 = -1$

Square of the sum:

$(x + y)^2 = x^2 + 2xy + y^2$

In this question:

$(x + (3+5i))^2$

$x^2 + 2x(3+5i) + (3+5i)^2$

$x^2 + 6x + 10xi + 9 + 30i + 25i^2$

$x^2 + 6x + 9 + (10x+30)i - 25$

$x^2 + 6x + (10x+30)i - 16$

$(x + (3+5i))^2 = x^2 + 6x + (10x+30)i - 16$

5 0

2 years ago

Solve for t: t + 1 = 24 A. t = 23 B. t = 25 C. t = 31 D. t = 20

Flauer [41]

Answer:

t = 25

Step-by-step explanation:

T is 25 because you take 1 away from the t+1 side and add the 1 to the 24 side.

3 0

2 years ago

Read 2 more answers

PLEASE HELP 100 POINTS!!!!!!

horrorfan [7]

Answer:

A) See attached for graph.

B) (-3, 0) (0, 0) (18, 0)

C) (-3, 0) ∪ [3, 18)

Step-by-step explanation:

Piecewise functions have multiple pieces of curves/lines where each piece corresponds to its definition over an interval.

Given piecewise function:

$g(x)=\begin{cases}x^3-9x \quad \quad \quad \quad \quad \textsf{if }x < 3\\-\log_4(x-2)+2 \quad \textsf{if }x\geq 3\end{cases}$

Therefore, the function has two definitions:

$g(x)=x^3-9x \quad \textsf{when x is less than 3}$

$g(x)=-\log_4(x-2)+2 \quad \textsf{when x is more than or equal to 3}$

When graphing piecewise functions:

Use an open circle where the value of x is not included in the interval.
Use a closed circle where the value of x is included in the interval.
Use an arrow to show that the function continues indefinitely.

First piece of function

Substitute the endpoint of the interval into the corresponding function:

$\implies g(3)=(3)^3-9(3)=0 \implies (3,0)$

Place an open circle at point (3, 0).

Graph the cubic curve, adding an arrow at the other endpoint to show it continues indefinitely as x → -∞.

Second piece of function

Substitute the endpoint of the interval into the corresponding function:

$\implies g(3)=-\log_4(3-2)+2=2 \implies (3,2)$

Place an closed circle at point (3, 2).

Graph the curve, adding an arrow at the other endpoint to show it continues indefinitely as x → ∞.

See attached for graph.

The x-intercepts are where the curve crosses the x-axis, so when y = 0.

Set the first piece of the function to zero and solve for x:

$\begin{aligned}g(x) & = 0\\\implies x^3-9x & = 0\\x(x^2-9) & = 0\\\\\implies x^2-9 & = 0 \quad \quad \quad \implies x=0\\x^2 & = 9\\\ x & = \pm 3\end{aligned}$

Therefore, as x < 3, the x-intercepts are (-3, 0) and (0, 0) for the first piece.

Set the second piece to zero and solve for x:

$\begin{aligned}\implies g(x) & =0\\-\log_4(x-2)+2 & =0\\\log_4(x-2) & =2\end{aligned}$

$\textsf{Apply log law}: \quad \log_ab=c \iff a^c=b$

$\begin{aligned}\implies 4^2&=x-2\\x & = 16+2\\x & = 18 \end{aligned}$

Therefore, the x-intercept for the second piece is (18, 0).

So the x-intercepts for the piecewise function are (-3, 0), (0, 0) and (18, 0).

From the graph from part A, and the calculated x-intercepts from part B, the function g(x) is positive between the intervals -3 < x < 0 and 3 ≤ x < 18.

Interval notation: (-3, 0) ∪ [3, 18)

Learn more about piecewise functions here:

brainly.com/question/11562909

3 0

1 year ago

Solve the inequality. -1/3w ≤ -5​

Solve the inequality. -1/3w ≤ -5