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2 years ago

Un triángulo rectángulo ABC recto en B , en la region relativa al lado AB se ubica el punto P, tal que la m

Mathematics

1 answer:

Murrr4er [49]2 years ago

8 0

this queation is tricky give me a sec

Step-by-step explanation:

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G The swimmer's legs appear shorter while in teh pool because of the of lights

goblinko [34]

Answer: c. refraction

Step-by-step explanation:

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3 years ago

Find the product. Write fractions in the simplest form. 5/6(-8/15) *

aliina [53]

Answer:

-75/48

Step-by-step explanation:

multiply 5 by 15

and 6 by 8

5x15 is 75 , put it in numerator

6x8 is 48 , denominator

then the answer should have the negative sign

4 0

3 years ago

Read 2 more answers

For which value of x is the equation 2(1 + x) = x+3 true? a.1 b.2 c.3 d.4

dexar [7]

Answer:

a. 1

Step-by-step explanation:

Replace 1 with x

2 ( 1 + 1 ) = 1 + 3

2 ( 2 ) = 1 + 3

4 = 1 + 3

4 = 4

5 0

3 years ago

Read 2 more answers

Hello, may I please have some help on these 4 questions? Thanks very much! {Also, if you could answer them quick, that would be

wlad13 [49]

1.

Parallelogram with base a = 20 in. and height h = 16 in.
Triangle with base a = 20 in. and height b = 8 in.

so:

$A = A_P+A_T=a\cdot h+\frac{1}{2}\cdot a\cdot b=20\cdot16+\frac{1}{2}\cdot20\cdot8=320+80=\\\\=\boxed{400\text{ in}^2}$

2.

Trapezoid with base b₁ = 14 cm, base b₂ = 4 cm and height h = 10 cm
Triangle with base b₁ = 14 cm and height x = 18 cm - 10 cm = 8 cm

so:

$A=A_T+A_\Delta=\frac{1}{2}\cdot(b_1+b_2)\cdot h+\frac{1}{2}\cdot b_1\cdot x=\\\\=
\frac{1}{2}\cdot(14+4)\cdot 10+\frac{1}{2}\cdot 14\cdot8=\frac{1}{2}\cdot18\cdot 10+\frac{1}{2}\cdot 14\cdot8=90+56=\\\\=\boxed{146\text{ cm}^2}$

3.

We have three rectangles:

$A=A_1+A_2+A_3=1\cdot 4+6\cdot5+18\cdot6=4+30+108=\boxed{142\text{ ft}^2}$

4.

Area of a circle:

$A_\circ=\pi\cdot r^2=3.14\cdot5^2=3.14\cdot25=78.5 \text{ cm}^2$

Area of a rectangle:

$A_R=8\cdot 6=48\text{ cm}^2$

Area of the shaded region:

$A=A_\circ-A_R=78.5-48=\boxed{30.5\text{ cm}^2}$

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3 years ago

Zeke deposited $300 into a savings account. The interest rate on the account is 3%, compounded continuously. To determine how lo

Vedmedyk [2.9K]

$\bf \begin{array}{llll} \textit{logarithm of factors} \\\\ \log_a(xy)\implies \log_a(x)+\log_a(y) \end{array} ~\hspace{4em} \begin{array}{llll} \textit{Logarithm of rationals} \\\\ \log_a\left( \frac{x}{y}\right)\implies \log_a(x)-\log_a(y) \end{array}$

$\bf \begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array} ~\hspace{7em} \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\textit{let's use this one}}{log_a a^x = x}\qquad \quad a^{log_a x}=x \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 900=300e^{0.03t}\implies \log_e(900)=\log_e(300e^{0.03t})$

$\bf \log_e(900)=\log_e(300)+\log_e(e^{0.03t})\implies \ln(900)=\ln(300)+0.03t\cdot \ln(e) \\\\\\ \ln(900)-\ln(300)=0.03t\implies \ln\left( \cfrac{900}{300} \right)=0.03t\implies \ln(3)=0.03t$

3 0

3 years ago