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Elena L [17]
3 years ago
14

How do you write 8.45 × 10–5 in standard form

Mathematics
2 answers:
Ghella [55]3 years ago
5 0
We move the decimal point to the left by five figures to get 0.0000845
Elza [17]3 years ago
5 0

Answer:

845x10(-2) x .00010

Step-by-step explanation:

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21cm x by the 81 you will make a lot more than what you wish for
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Prove the following
DerKrebs [107]

\bf [cot(\theta )+csc(\theta )]^2=\cfrac{1+cos(\theta )}{1-cos(\theta )} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the left-hand side}}{[cot(\theta )+csc(\theta )]^2}\implies cot^2(\theta )+2cot(\theta )csc(\theta )+csc^2(\theta ) \\\\\\ \cfrac{cos^2(\theta )}{sin^2(\theta )}+2\cdot \cfrac{cos(\theta )}{sin(\theta )}\cdot \cfrac{1}{sin(\theta )}+\cfrac{1}{sin^2(\theta )}\implies \cfrac{cos^2(\theta )}{sin^2(\theta )}+\cfrac{2cos(\theta )}{sin^2(\theta )}+\cfrac{1}{sin^2(\theta )}

\bf \cfrac{\stackrel{\textit{perfect square trinomial}}{cos^2(\theta )+2cos(\theta )+1}}{sin^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{doing the right-hand-side}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}}\implies \stackrel{\textit{multiplying by the denominator's conjugate}}{\cfrac{1+cos(\theta )}{1-cos(\theta )}\cdot \cfrac{1+cos(\theta )}{1+cos(\theta )}}

\bf \cfrac{[1+cos(\theta )]^2}{\underset{\textit{difference of squares}}{[1-cos(\theta )][1+cos(\theta )]}}\implies \cfrac{[cos(\theta )+1]^2}{1^2-cos^2(\theta )} \\\\\\ \cfrac{[cos(\theta )+1]^2}{1-cos^2(\theta )}\implies \boxed{\cfrac{[cos(\theta )+1]^2}{sin^2(\theta )}}

recall that sin²(θ) + cos²(θ) = 1, thus sin²(θ) = 1 - cos²(θ).

7 0
3 years ago
POZ HELP RN PLZZZZZzzzz
soldier1979 [14.2K]

Answer: It’s the third answer.

Step-by-step explanation:

There are 2.54 cm in one inch and 12 inches in one foot, so 2.54x12 = 1 foot. Multiply this by 2 to get 2 feet.

7 0
3 years ago
Find the value of v and w to the nearest tenth.​
Anton [14]

Answer:

Trigonometry Ratio:

Sin65°=v/15

v=15 sin65°

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w=40.3°

Hope it helps! :)

6 0
2 years ago
J(1)²,
Natali [406]

The table when completed looks like this:

x   -2   -1   0   1   2

f(x) 4   2   1    2   4.

We plot these points on the graph: (-2, 4), (-1, 2), (0, 1), (1, 2), (2, 4).

<h3>What is a function?</h3>

A function exists as a relation between a dependent variable (f(x)) and an expression of an independent variable (x), utilized to define the value of a dependent variable from a provided value of an independent variable.

We have been given a function,

$f(x)=\left \{ {{(1/2)^{x},x\leq 0 } \atop {2^x > 0}} \right.

We have to estimate the value of f(x) when x = {-2, -1, 0, 1, 2}

If the values of x ≤ 0, take f(x) = (1/2)ˣ then

f(-2) = (1/2)⁻² = 2² = 4

f(-1) = (1/2)⁻¹ = 2¹ = 2

f(0) = (1/2)⁰ = 1

If the values of x > 0,  take f(x) = 2ˣ then

f(1) = 2¹ = 2

f(2) = 2² = 4

The table exists as follows

x   -2   -1   0   1   2

f(x) 4   2   1    2   4.

We plot these points on the graph: (-2, 4), (-1, 2), (0, 1), (1, 2), (2, 4).

To learn more about independent and dependent variable refer to:

brainly.com/question/11719274

#SPJ9

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