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sineoko [7]
3 years ago
8

Bryce is solving this problem.

Mathematics
1 answer:
zavuch27 [327]3 years ago
6 0

Answer:

The answer is C

2.640 and 10,560

Step-by-step explanation:

264 * 410 = (264 * 400) + (264 * 10) = 105,600 + 2640 = 2.640 and 10,560

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What is the answer of this<br><br> pls say, I will mark as brilliant
Diano4ka-milaya [45]

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3 years ago
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Alejandro has gone to school StartFraction 5 Over 7 EndFraction of the last 35 days. Which expression can be used to determine t
Katarina [22]

Answer:

35 times StartFraction 5 Over 7 EndFraction

Step-by-step explanation:

Alejandro has gone to school (5/7) of the last 35 days.

We need to find the expression that can be used to determine the number of days Alejandro has gone to school.

x=\dfrac{5}{7}\times 35\\\\=5\times 5\\\\=25

So, the correct option is (b) "35 times StartFraction 5 Over 7 EndFraction".

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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
A firework is launched at the rate of 10 feet per second from a point on the ground 50 feet from an observer. to 2 decimal place
Kazeer [188]

The rate of change of the angle of elevation when the firework is 40 feet above the ground is 0.12 radians/second.

First we will draw a right angle triangle ΔABC, where ∠B = 90°

Lets, assume the height(AB) = h and base(BC)= x

If the angle of elevation, ∠ACB = α, then

tan(α) = \frac{AB}{BC} = \frac{h}{x}

Taking inverse trigonometric function, α = tan⁻¹ (\frac{h}{x}) .............(1)

As we need to find the rate of change of the angle of elevation, so we will differentiate both sides of equation (1) with respect to time (t) :

\frac{d\alpha}{dt}=[\frac{1}{1+ \frac{h^2}{x^2}}]*(\frac{1}{x})\frac{dh}{dt}

Here, the firework is launched from point B at the rate of 10 feet/second and when it is 40 feet above the ground it reaches point A,

that means h = 40 feet and \frac{dh}{dt} = 10 feet/second.

C is the observer's position which is 50 feet away from the point B, so x = 50 feet.

\frac{d\alpha}{dt}= [\frac{1}{1+ \frac{40^2}{50^2}}] *\frac{1}{50} *10\\ \\ \frac{d\alpha}{dt} = [\frac{1}{1+\frac{16}{25}}] *\frac{1}{5}\\ \\ \frac{d\alpha}{dt} = [\frac{25}{41}] *\frac{1}{5}\\   \\ \frac{d\alpha}{dt}= \frac{5}{41} =0.1219512

= 0.12 (Rounding up to two decimal places)

So, the rate of change of the angle of elevation is 0.12 radians/second.

5 0
3 years ago
From her window 5m above the ground, Sherri spots a turtle on the ground at a 34 degree angle of depression. To the nearest tent
Thepotemich [5.8K]
Let y be the distance from the turtle to the base of the building.

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-----------------------------------------------------------------
Answer: The distance is 7.4 meters (Answer A)
-----------------------------------------------------------------
3 0
4 years ago
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