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3 years ago

15

What is the percentage change of 50 to 35 pounds

1 answer:

Artist 52 [7]3 years ago

4 0

30%
hope I helped you with this

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What is the product of 2.5 × 10−15 and 3.9 × 1026? 6.4 × 101 9.75 × 1011 6.4 × 1041 9.75 × 1041

Doss [256]

$(2.5*10^{-15})(3.9*10^{26})$ . We begin by multiplying the "regular" numbers first. 2.5 × 3.9 = 9.75. That's good; it's already in the proper format for scientific notation. Now we multiply the bases of 10 with their exponents. Since the bases are both 10, they are like. When we multiply like bases, we add their exponents. $10^{-15+26}=10^{11}$ . Our answer then is $9.75*10^{11}$ . There you go!

5 0

3 years ago

Read 2 more answers

What is the answer to (14/5)/4

Misha Larkins [42]

14 / 5 = 2.8 / 4 = 0.7

Always divide what's in the bracket first.

8 0

3 years ago

Helphelphelphelphelp

victus00 [196]

5/16x24/15

Reduce the number with greatest common divisor 5

1/16x24/3

Reduce the numbers with greatest common divisor 8

1/2x3/3

Reduce the numbers with greatest common divisor 3

1/2

Answer
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7 0

4 years ago

3x + 2x – x + 2x² = <br> Please help !!

baherus [9]

Answer:

3 + 2 = 5

add x to 5

5x + 2x to the 2nd power

2 times 2 = 4x = 5 and 4x

5x + 4x

5 0

3 years ago

First-order linear differential equations

kkurt [141]

Answer:

$(1)\ logy\ =\ -sint\ +\ c$

$(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Step-by-step explanation:

1. Given differential equation is

$\dfrac{dy}{dt}+ycost = 0$

$=>\ \dfrac{dy}{dt}\ =\ -ycost$

$=>\ \dfrac{dy}{y}\ =\ -cost dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}$

$=>\ logy\ =\ -sint\ +\ c$

Hence, the solution of given differential equation can be given by

$logy\ =\ -sint\ +\ c.$

2. Given differential equation,

$\dfrac{dy}{dt}\ -\ 2ty\ =\ t$

$=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty$

$=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})$

$=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}$

$=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c$

$=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Hence, the solution of given differential equation is

$log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

8 0

4 years ago