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Reil [10]
2 years ago
10

What can you add to 30 to make 50?

Mathematics
2 answers:
stepladder [879]2 years ago
7 0

Answer:well… 50 - 30 = 20 right? So we add 20 + 30 to get 50
30 + x = 50

30 + 20 = 50

X=20

Step-by-step explanation:

Archy [21]2 years ago
4 0

20, because 30+20 is 50.

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The length of a rectangle is increasing at a rate of 4 meters per day and the width is increasing at a rate of 1 meter per day.
puteri [66]

Answer:

\displaystyle \frac{dA}{dt} = 102 \ m^2/day

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Geometry</u>

Area of a Rectangle: A = lw

  • l is length
  • w is width

<u>Calculus</u>

Derivatives

Derivative Notation

Implicit Differentiation

Differentiation with respect to time

Derivative Rule [Product Rule]:                                                                              \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle l = 10 \ meters<u />

<u />\displaystyle \frac{dl}{dt} = 4 \ m/day<u />

<u />\displaystyle w = 23 \ meters<u />

<u />\displaystyle \frac{dw}{dt} = 1 \ m/day<u />

<u />

<u>Step 2: Differentiate</u>

  1. [Area of Rectangle] Product Rule:                                                                 \displaystyle \frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}

<u>Step 3: Solve</u>

  1. [Rate] Substitute in variables [Derivative]:                                                    \displaystyle \frac{dA}{dt} = (10 \ m)(1 \ m/day) + (23 \ m)(4 \ m/day)
  2. [Rate] Multiply:                                                                                                \displaystyle \frac{dA}{dt} = 10 \ m^2/day + 92 \ m^2/day
  3. [Rate] Add:                                                                                                      \displaystyle \frac{dA}{dt} = 102 \ m^2/day

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

Unit: Implicit Differentiation

Book: College Calculus 10e

8 0
3 years ago
Is this correct? pls be sure of yourself. if you arent, pls check
Elza [17]

Answer:

Yes It Is.

Step-by-step explanation:

8 0
2 years ago
(05.03 LC)
Paladinen [302]

f(x)=3x-23,\ g(x)=-4.5x+7\\\\f(x)=g(x)\iff3x-23=-4.5x+7\qquad\text{add 23 to both sides}\\\\3x=-4.5x+30\qquad\text{add 4.5x to both sides}\\\\7.5x=30\qquad\text{divide both sides by 7.5}\\\\\boxed{x=4}

6 0
3 years ago
Read 2 more answers
1 BRAINLIEST and 10 points!
erastovalidia [21]

Answer:

≈40.25 Feet

Step-by-step explanation:

The diagnal length of the orchard is the hypotenuse of the triangal with side lengths of 36 and 18.

Therefore, to calculate that, set the hypotenuse as x.

36²+18²=x²

√1620≈ 40.25yards

If I'm wrong I'm so sorry!  I'm wrong sorry

If I'm right Thank you

(lost)

3 0
3 years ago
Read 2 more answers
2v^2-12 =-12v <br> Would really appreciate it loves❤️
Black_prince [1.1K]

For this case we have the following equation:

2v ^ 2-12 = -12v

Rewriting we have:

2v ^ 2 + 12v-12 = 0

Dividing by 2 to both sides of the equation:

v ^ 2 + 6v-6 = 0

We apply the quadratic formula:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

We have to:

a = 1\\b = 6\\c = -6

Substituting:

x = \frac {-6 \pm \sqrt {6 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-6 \pm \sqrt {36 + 24}} {2}\\x = \frac {-6 \pm \sqrt {60}} {2}\\x = \frac {-6 \pm \sqrt {4 * 15}} {2}\\x = \frac {-6 \pm2 \sqrt {15}} {2}

Thus, we have two roots:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

ANswer:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

7 0
3 years ago
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