The speed of sound in air is 340 m/s. How far , in meters, would a sound of wave travel on 7 seconds?

Mathematics

1 answer:

Rasek [7]2 years ago

4 0

Answer:

2 380m

Step-by-step explanation:

very simple just multiply 7×340

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Please help me on this will give you brainliest!!

vovangra [49]

Answer:

∠3 would be 134° because it is alternate interior to ∠5

∠2 would be 46° because it is supplementary to ∠1

Hope this helped

4 0

3 years ago

Question 15 of 43

Andrews [41]

Answer:

Option A

Step-by-step explanation:

$\frac{12}{25} \times \frac{11}{24} \times \frac{10}{23} \times \frac{9}{22}$

we start with 12 red marbles out of 25 marbles, that's the probability for the first.

in the second step we have one red marble less, wich also means one marble less overall

and so on

$\frac{12 \times 11 \times 10 \times 9}{25 \times 24 \times 23 \times 22}$

$= \frac{11880}{303600}$

can be reduced by 1320 (step by step) to

$\frac{9}{230}$

6 0

3 years ago

Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.

vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$