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antiseptic1488 [7]
3 years ago
11

I need help answering

Mathematics
1 answer:
Ghella [55]3 years ago
7 0
A is the right answer
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Two trains leave the same station at the same time, but are headed in opposite directions. One train travels at 70 mph and the o
uranmaximum [27]

Let

x--------> the time in hours

y-------> the distance in miles

we know that

The speed is equal to

Speed=\frac{distance}{time}

in this problem

Speed=\frac{y}{x}

y=(Speed)*x

<u>Equation First Train</u>

y1=70x

<u>Equation Second Train</u>

y2=80x

y1+y2=600\ miles ---------> equation 1

Substitute the values of y1 and y2 in the equation 1

70x+80x=600

Solve for x

150x=600

x=4\ hours

therefore

<u>the answer is</u>

4\ hours

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3 years ago
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You would pay £32.50
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4 years ago
¼ log b x – 2 log b 5 – 10log b y
melamori03 [73]

I will leave out the base, b since it's the same for all these logarithms.

\frac{1}{4} Log(x)-2Log(5)-10Log(y)  \\  \\ Power-Rule \\  \\ Log( x^{ \frac{1}{4} } )-Log( 5^{2} )-Log( y^{10} ) \\  \\ Quotient-Rule \\  \\ Log( x^{ \frac{1}{4} } ) -Log( \frac{25}{ y^{10} } ) \\  \\ Quotient-Rule \\  \\ Log( \frac{ x^{ \frac{1}{4} }  y^{10} }{25} )


7 0
3 years ago
What is the sound level for a noise that has an intensity of 9.9 × 10-5 watts/m2?
horsena [70]

Given: Sound Intensity ( I ) = 9.9 \times  10^{-5}\frac{watts}{m^2}

We know,  the approximate threshold of human hearing is at 1kHz. In watts/m^2 it's value is =\frac{9.9 \times  10^{-5}}{10^{-12}}⁻¹² W/m².

So, we can say,  Reference sound intensity(I_0) = 10⁻¹² W/m².

We have formula for "sound intensity level LI in dB" when entering sound intensity.

LI = 10×log (I / Io)  in dB

Plugging values of I and Io in formula.

LI = 10 × log (\frac{9.9 \times  10^{-5}}{10^{-12}})

LI = 10 × log (99000000)

= 10 × 7.99564

LI = 79.96 dB.

Therefore,  79.96 dB is the sound level for a noise that has an intensity of 9.9 × 10-5 watts/m2.

8 0
3 years ago
Which of the following is equivalent to log507 rounded to three decimal places?
Leya [2.2K]

Answer :     0.497 is what i think it is

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