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svetlana [45]
3 years ago
5

Write 720,080 in expanded form with exponents

Mathematics
1 answer:
Kamila [148]3 years ago
5 0
7 x 10^5
+ 2 x 10^4
+ 0 x 10^3
+ 0 x 10^2
+ 8 x 10^1
+ 0 x 10^0
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PLEASEEEE HELP URGENT
kotegsom [21]

Answer:

7.1 cm

Step-by-step explanation:

The diagonal of a cube connects two opposite corners.

You can use the diagonal of a cube in a right triangle in which it is the hypotenuse. The legs of the triangle are a diagonal of a face and an edge.

Let the length of the edge be s.

a^2 + b^2 = c^2

s^2 + 10^2 = (\sqrt{150})^2

s^2 + 100 = 150

s^2 = 50

s = \sqrt{50}

s = 5\sqrt{2}

s = 7.1

Answer: 7.1 cm

7 0
3 years ago
Pree point sa inyong lahat​
zimovet [89]

Answer:

Thanksss

Step-by-step explanation:

:)

4 0
2 years ago
Read 2 more answers
(6+3)+21=6+(3+21) name each property of addition or multiplication
Anna007 [38]
(6 + 3) + 21 = 6 + (3 + 21)

It's an ASSOCIATIVE PROPERTY
(a + b) + c = a + (b + c)
7 0
3 years ago
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

3 0
3 years ago
Show that (x+1) is a factor of f(x)=19x^42+18x-1
Dmitry_Shevchenko [17]

Answer:

see explanation

Step-by-step explanation:

If (x + 1) is a factor then f(- 1) = 0

f(- 1) = 19(-1)^{42} + 18(- 1) - 1 = 19 - 18 - 1 = 0

Thus (x + 1) is a factor of f(x)

6 0
3 years ago
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