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kondaur [170]
3 years ago
12

Please Help! 15 points!

Mathematics
1 answer:
elena-s [515]3 years ago
8 0

height/shadow constant

h/30 = 6/4.5

height of building = h = 30•6/4.5 = 40 feet long

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PLS help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Darina [25.2K]

9514 1404 393

Answer:

  y = -1

Step-by-step explanation:

Sides marked with the same marking are congruent, so we have two equations:

  • 3x -y = 16
  • 2x +y = 9

Adding the equations gives ...

  5x = 25

  x = 5 . . . . . . divide by 5

Then the second equation tells us ...

  y = 9 -2x = 9 -2(5) = -1

The value of y is -1.

5 0
3 years ago
3175÷66 partial quotient
Serga [27]
20.83333333333333333
3 0
3 years ago
Plz help me, thank you
Kruka [31]

Answer:

P=40(1.03526)^{t}

Step-by-step explanation:

<u>Exponential Growth </u>

The natural growth of some magnitudes can be modeled by the equation:

P=P_o(1+r)^{t}

Where P is the actual amount of the magnitude, Po is its initial amount, r is the growth rate and t is the time.

The initial number of bacteria is Po=40 and it doubles (P=2Po) at t=20 min. With that point we can find the value of r:

2P_o=P_o(1+r)^{20}

Simplifying:

(1+r)^{20}=2

Solving for 1+r:

1+r=\sqrt[20]{2}

1+r=1.03526

The exponential function that models the situation is:

\mathbf{P=40(1.03526)^{t}}

4 0
3 years ago
How To Solve these? ​
Inessa [10]

Answer:

a. 15/23

b. 13/27

c. 400g

Step-by-step explanation:

a. When the denominators are the same, you can just sum the numerators.

Which becomes, 13+2=15--> 15/23

b. Same, when the denominator is the same, you can just minus the numerators. Which becomes, 25-12=13--> 13/27

c. 1kg=1000g. 1000/5=200✖️2=400

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%281-x%5E%7B2%7D%20%29%5E%7B3%2F2%7D%20%7D%20%5C%2C%20dx" id="TexFo
Ludmilka [50]

Answer:\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}General Formulas and Concepts:

<u>Pre-Calculus</u>

  • Trigonometric Identities

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Integration

  • Integrals
  • Definite/Indefinite Integrals
  • Integration Constant C

Integration Rule [Reverse Power Rule]:                                                               \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

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

U-Substitution

  • Trigonometric Substitution

Reduction Formula:                                                                                               \displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx

<u>Step 2: Integrate Pt. 1</u>

<em>Identify variables for u-substitution (trigonometric substitution).</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle x = sin(u)
  2. [<em>u</em>] Differentiate [Trigonometric Differentiation]:                                         \displaystyle dx = cos(u) \ du
  3. Rewrite <em>u</em>:                                                                                                       \displaystyle u = arcsin(x)

<u>Step 3: Integrate Pt. 2</u>

  1. [Integral] Trigonometric Substitution:                                                           \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{\frac{3}{2}} \, du
  2. [Integrand] Rewrite:                                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{\frac{3}{2}} \, du
  3. [Integrand] Simplify:                                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \int\limits^a_b {cos^4(u)} \, du
  4. [Integral] Reduction Formula:                                                                       \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{4 - 1}{4}\int \limits^a_b {cos^{4 - 2}(x)} \, dx + \frac{cos^{4 - 1}(u)sin(u)}{4} \bigg| \limits^a_b
  5. [Integral] Simplify:                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4}\int\limits^a_b {cos^2(u)} \, du
  6. [Integral] Reduction Formula:                                                                          \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg|\limits^a_b + \frac{3}{4} \bigg[ \frac{2 - 1}{2}\int\limits^a_b {cos^{2 - 2}(u)} \, du + \frac{cos^{2 - 1}(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  7. [Integral] Simplify:                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}\int\limits^a_b {} \, du + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  8. [Integral] Reverse Power Rule:                                                                     \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3}{4} \bigg[ \frac{1}{2}(u) \bigg| \limits^a_b + \frac{cos(u)sin(u)}{2} \bigg| \limits^a_b \bigg]
  9. Simplify:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(u)sin(u)}{4} \bigg| \limits^a_b + \frac{3cos(u)sin(u)}{8} \bigg| \limits^a_b + \frac{3}{8}(u) \bigg| \limits^a_b
  10. Back-Substitute:                                                                                               \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{cos^3(arcsin(x))sin(arcsin(x))}{4} \bigg| \limits^a_b + \frac{3cos(arcsin(x))sin(arcsin(x))}{8} \bigg| \limits^a_b + \frac{3}{8}(arcsin(x)) \bigg| \limits^a_b
  11. Simplify:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x)}{8} \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{\frac{3}{2}}}{4} \bigg| \limits^a_b + \frac{3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b
  12. Rewrite:                                                                                                         \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{\frac{3}{2}} + 3x\sqrt{1 - x^2}}{8} \bigg| \limits^a_b
  13. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:              \displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}

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

Unit: Integration

Book: College Calculus 10e

8 0
3 years ago
Read 2 more answers
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