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3 years ago

If f(x)=2x+1 and g(x)=x^2+7, which of the following is equal to f(g(x))

SAT

2 answers:

topjm [15]3 years ago

7 0

Answer: $2x^{2} +15$

Explanation:

so basically f(g(x)) is telling you to plug the entire g(x) equation into x in the f(x) equation so it will look like this

2( $x^{2} +7$ ) + 1

and then simplify

$2x^{2} +14 + 1$

$2x^{2} +15$

user100 [1]3 years ago

6 0

2x exponent 2 • 15
Because

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A bat emits a sound at a frequency of 3. 00 × 104 hz as it approaches a wall. The bat detects beats such that the frequency of t

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Taking into account the Doppler efect, the speed of the bat is closest to 5.02 $\frac{m}{s}$ .

The Doppler effect is defined as the apparent frequency change of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion when the sender and receiver, or observer, move relative to each other.

The following expression is considered the general case of the Doppler effect:

$f1=f\frac{v+-vR}{v+-vE}$

Where:

f1, f: Frequency perceived by the receiver and frequency emitted by the issuer respectively. Its unit of measurement in the International System (S.I.) is the hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s⁻¹)
v: Velocity of propagation of the wave in the medium. It is constant and depends on the characteristics of the medium. In this case, the speed of sound in air is considered to be 340 m/s
vR, vE: Speed of the receiver and the sender respectively. Its unit of measure in the S.I. is the m/s
±, ∓:

We will use the + sign:

In the numerator if the receiver approaches the sender

In the denominator if the sender moves away from the receiver

We will use the sign -:

In the numerator if the receiver moves away from the emitter

In the denominator if the sender approaches the receiver

In this case, a bat emits a sound at a frequency of 3.00×10⁴ Hz= 30,000 Hz as it approaches a wall.

The bat detects beats such that the frequency of the echo is 900 hz higher than the frequency the bat is emitting.

So, you know:

f1= 30,000 Hz + 900 Hz= 30,900 Hz
f= 30,000 Hz
v= 340 m/s
vE=vR= vbat
Receiver approaches the sender

Replacing in the expression is considered the general case of the Doppler effect:

$30900 Hz=30000 Hz\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}$

Solving:

$\frac{30900 Hz}{30000 Hz} =\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}$

$1.03=\frac{340 \frac{m}{s} +vbat}{340\frac{m}{s} -vbat}$

1.03×( 340 $\frac{m}{s}$ -vbat)= 340 $\frac{m}{s}$ + vbat

1.03×340 $\frac{m}{s}$ - 1.03×vbat= 340 $\frac{m}{s}$ + vbat

350.2 $\frac{m}{s}$ - 1.03×vbat= 340 $\frac{m}{s}$ + vbat

350.2 $\frac{m}{s}$ - 340 $\frac{m}{s}$ = vbat + 1.03×vbat

10.2 $\frac{m}{s}$ = 2.03×vbat

10.2 $\frac{m}{s}$ ÷ 2.03 = vbat

5.02 $\frac{m}{s}$ = vbat

In summary, the speed of the bat is closest to 5.02 $\frac{m}{s}$ .

Learn more about the Doppler effect:

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