All categories

2 years ago

Enter the number using calculator notation. 8.5x 103 The number in calculator notation is

Mathematics

1 answer:

BabaBlast [244]2 years ago

4 0

Answer: 875.5x

Step-by-step explanation:

You might be interested in

adam has come to town to attend his friend's wedding. He is staying in a hotel. To which person should he offer a tip

Varvara68 [4.7K]

Answer: The hotel doorman for opening the door for him

Step-by-step explanation: that one was tricky

8 0

3 years ago

Maggie volunteered to bake a cake she decided to double the recipe if the original recipe calls for 3/4 cup brown sugar how much

zlopas [31]

Answer:

3/2

Step-by-step explanation:

3/4 * 2 = 6/4 = 3/2

7 0

3 years ago

Solve the equation using the zero-product property. (2x − 8)(7x + 5) = 0 x = –2 or x = 7 x = 4 or x = x = 4 or x = x = –4 or x =

aliya0001 [1]

$(2x-8)(7x+5)=0\iff2x-8=0\ \vee\ 7x+5=0\\\\2x-8=0\qquad\text{add 8 to both sides}\\2x=8\qquad\text{divide both sides by 2}\\\boxed{x=4}\\\\7x+5=0\qquad\text{subtract 5 from both sides}\\7x=-5\qquad\text{divide both sides by 7}\\\boxed{x=-\dfrac{5}{7}}$

5 0

3 years ago

A.756<br> b.936<br> c.1,008<br> d.1,080<br> HELLLPPPPP

luda_lava [24]

Answer:

936 is the correct option

6 0

3 years ago

Read 2 more answers

Find a particular solution to y" - y + y = 2 sin(3x)

leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$

8 0

2 years ago