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2 years ago

In 1 year earth travels about how many miles in its orbit

Mathematics

2 answers:

VashaNatasha [74]2 years ago

5 0

940 million kilometers (584 million miles).

gogolik [260]2 years ago

5 0

the earth travels 584 miles in one year

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Solve for f. 5 = f + 1 f=

4vir4ik [10]

Answer:

f=4

Step-by-step explanation:

1) first we have to get the "f" by itself. to do this we subtract the 1 from the right side of the equation.

2) and what you do to one side you ALWAYS do to the other side!!! so subtract 1 from the 5 also

3) this leaves you with 4=f

hope this helps!

8 0

3 years ago

Read 2 more answers

4 (X+7) = 38 WHAT IS X AND EXPLAIN HOW YOU KNOW

xxMikexx [17]

Answer:

X=2.5

Step-by-step explanation:

38÷4 = 9.5

9.5 - 7=2.5

Divide each side by four to get rid of that numbr from the elft and subtract seven from both sides to get rid of the seven so your left with x=2.5

7 0

2 years ago

Read 2 more answers

Plzzzzzzzzzzzzzzzzaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

erastovalidia [21]

Answer:

Option A

Step-by-step explanation:

Number of employees exceeded their sales quota = 17

Number of employees met their sales quota = 13

Number of employees didn't exceed their sales quota = 3

Now, we need to find the ratio of the number employees who exceeded their sales quota to the number of employees who didn't exceed their sales quota,

$17:13 + 3\\=17:16$

So, Option 'A' is correct.

Please mark as Brainliest!!

3 0

3 years ago

Jill is thinking of a number. Twenty less than one forth of the number is -45. Find Jill's number.

Bezzdna [24]

Answer:

The number is -100, Jill can think of something else, now.

Step-by-step explanation:

x = the number

x/4 - 20 = -45 {twenty less than one-fourth of the number is -45}

x/4 = -25 {added 20 to each side}

x = -100 {multiplied each side by 4}

5 0

2 years ago

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If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

4 0

2 years ago