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3 years ago

.000000759 scientific notation

Mathematics

2 answers:

lara31 [8.8K]3 years ago

8 0

Answer

7.59 x 10⁻⁷

Step-by-step explanation:

navik [9.2K]3 years ago

8 0

7.59*10^-7 since the number is a decimal the exponent in the equation is negative then you count the zeros from the first real number (7)

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weeeeeb [17]

Answer:

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solution 27

Step-by-step explanation:

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3 years ago

The parallelogram has a base of 9 cm and a high of 21 cm what is the area of the parallelogram

Assoli18 [71]

$a = b \times h$

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3 years ago

Find equation of line y=mx+b

vesna_86 [32]

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3 years ago

Determine the slope for the line that meets each condition stated. Show your work. Passes through the points ( 3 , 5 ) and ( − 1

Kamila [148]

Answer:

The slope of the line that passes through (-1, -7) and (3, 5) is 3

Explanation:

The slope of the line is calculated using the following rule:

$Slope = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

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(3 , 5) representing (x₁ , y₁)

(-1 , -7) representing (x₂ , y₂)

To get the slope, we will simply substitute with the givens in the above equation

This is done as follows:

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You haven't attached the table for the second part. However, follow the same procedure. Pick two points from the table, substitute with them in the above equation and you'll get the slope

Hope this helps :)

4 0

3 years ago

I will give brainliest to anyone who gives me a correct answer

Pachacha [2.7K]

Answer:

Step-by-step explanation:

Let's check the first condition [y=110, t=0] for all the choices and eliminate:

$y = 110e^{7t} - 80\\y = 110e^{0} - 80\\y=30$

$y = 110 - 80e^{-5t}\\y = 110 - 80e^{0}\\y=30$

$y = 30 + 80e^{-2t}\\y = 30 + 80e^{0}\\y=110$

YES

$y = 110 - 80e^{2t}\\y = 110 - 80e^{0}\\y=30$

$y = 30 - 80e^{-4t}\\y = 30 - 80e^{0}\\y=-50$

Note: e^0 = 1

Only C is satisfied. But let's make sure the 2nd condition applies as well.

$y = 30 + 80e^{-2t}\\y = 30 + 80e^{-\infty}\\y=30+\frac{80}{e^{\infty}}\\y>>30$

Hence, y approaches 30 as t goes to infinity. 2nd condition is satisfied as well.

Answer is C

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4 years ago