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3 years ago

7 × 10^7 in standard form

Mathematics

1 answer:

svp [43]3 years ago

4 0

Answer:

70000000

Step-by-step explanation:

move the dot 7 times to the right

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P=21+2w; if p=18 when l=5, find w

jolli1 [7]

Answer:

if im not mistaken.. i would say its -3.5 or -7/2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Jina wanted to study how the area of a rectangle changes with the length if it’s width is fixed. She computed the areas of sever

olganol [36]

Answer:

The domain and the range of the function are, respectively:

$Dom\{f\} = [0\,m,5\,m]$

$Ran\{f\} = [0\,m^{2}, 10\,m^{2}]$

Step-by-step explanation:

Jina represented a function by a graphic approach, where the length, measured in meters, is the domain of the function, whereas the area, measured in square meters, is its range.

$Dom\{f\} = [0\,m,5\,m]$

$Ran\{f\} = [0\,m^{2}, 10\,m^{2}]$

8 0

3 years ago

What is the answer for. 2x+16y-36y-9+1-x+3x

poizon [28]

After solving $2x+16y-36y-9+1-x+3x$ we get $4x-20y-8$

Step-by-step explanation:

We need to solve $2x+16y-36y-9+1-x+3x$

Combining the like terms:

$2x+16y-36y-9+1-x+3x\\=2x-x+3x+16y-36y-9+1\\=4x-20y-8$

So, After solving $2x+16y-36y-9+1-x+3x$ we get $4x-20y-8$

Keywords: Polynomials

Learn more about Polynomials at:

#learnwithBrainly

3 0

4 years ago

One ten-thousandth scientific notation

Nataliya [291]

1 times 10^1=10
1 times 10^-1=1/10
1 times 10^-2=1/100
1 ten thousandth is 1/10,000, that is 1 times 10^-4

answer is
1*10⁻⁴

8 0

4 years ago

Solve the given initial-value problem. y'' 4y' 5y = 35e−4x, y(0) = −5, y'(0) = 1

meriva

The solution for the initial value problem is $y_{g} = e^{-2x} (-12cos(x) + 5sin(x)) + 7e^{-4x}$

Given,

y" + 4y' + 5y = 35 $e^{-4x}$

y(0) = -5

y'(0) = 1

Solve this homogenous equation to get $y_{h}$

According to differential operator theorem,

$y_{h}$ = $e^{ax}$ ( A cos (bx) + B sin (bx)), where A and B are constants.

Therefore,

y" + 4y' + 5y = 0

( $D^{2}$ + 4D + 5)y = 0

D = -2± i

$y_{h} = e^{-2x} ( A cos (x) + B sin (x))$

Now, solve for $y_{p}$

A function of the kind $ce^{-4x}$ is the function on the right, we are trying a solution of the form $y_{p} =ce^{-4x}$ , here c is a constant.

$y_{p} " + 4y_{p} ' + 5y_{p} = 35e^{-4x} \\=16ce^{-4x} -16ce^{-4x} +5ce^{-4x} = 35e^{-4x} \\= 5ce^{-4x} =35e^{-4x} \\c=\frac{35}{5} =7\\y_{p} =7e^{-4x}$