All categories

3 years ago

Someone PLEASE HELP I’m struggling rn

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

8 0

Answer:

The value of the constant of proportionality is k=5.

You might be interested in

Marie and fe work in a flower shop 6 hours on weekend to add their daily allowance in school. They spends 2 1/3 hours for making

Shkiper50 [21]

Answer:

2 1/6 hours

Step-by-step explanation:

2 1/3 + 1 1/2 + x = 6

convert all mixed numbers to improper fractions

7/3 + 3/2 + x = 6/1

change all numbers to have a common denominator

14/6 + 9/6 + x = 36/6

combine like terms

23/6 + x = 36/6

subtract 23/6 from both sides

x = 13/6

convert improper fraction to mixed number

2 1/6

7 0

4 years ago

if you have a box of candy that is 8 inches long 5 inches wide and 2 inches tall how much space do you have for candy

stiv31 [10]

8 x 5 x 2 = 80 square cubits

4 0

3 years ago

Read 2 more answers

Ms. Palmer has 24 crystals and 18 pearls. She wants all of the necklaces to look the same. She wants to use all of the crystals

marysya [2.9K]

she can make six neckleces.

3 0

3 years ago

The Laplace Transform of a function f(t), which is defined for all t > 0, is denoted by L{f(t)} and is defined by the imprope

lesya692 [45]

(1) D

$L_s\left\{t\right\} = \displaystyle\int_0^\infty te^{-st}\,\mathrm dt$

Integrate by parts, taking

$u = t \implies \mathrm du=\mathrm dt$

$\mathrm dv = e^{-st}\,\mathrm dt \implies v=-\dfrac1se^{-st}$

Then

$L_s\left\{t\right\} = \displaystyle \left[-\frac1ste^{-st}\right]\bigg|_{t=0}^{t\to\infty}+\frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle \frac1s\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{t\right\} = \displaystyle -\frac1{s^2}e^{-st}\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{t\right\} = \displaystyle \boxed{\frac1{s^2}}$

(2) A

$L_s\left\{1\right\} = \displaystyle\int_0^\infty e^{-st}\,\mathrm dt$

$L_s\left\{1\right\} = \displaystyle\left[-\frac1se^{-st}\right]\bigg|_{t=0}^{t\to\infty}$

$L_s\left\{1\right\} = \displaystyle\boxed{\frac1s}$

7 0

3 years ago

3x^2 - 4x - 5x^2 + 8x <br> simplify it please

Levart [38]

$3x^2 - 4x - 5x^2 + 8x$

Set up

$-2x^2 + 4x$

Combine like terms

$2x(-x + 2)$

Take out a common factor

$2x(2 - x)$

Rearrange to remove a starting negative coefficient inside the parentheses

The simplification is $\boxed{2x(2 - x)}$ .

3 0

3 years ago