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kogti [31]
3 years ago
8

Question

Mathematics
2 answers:
Volgvan3 years ago
7 0

Answer:

C.- x = 80

Step-by-step explanation:

vaieri [72.5K]3 years ago
4 0

Answer:

C

Step-by-step explanation:

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What is 66 2/3% as a decimal to the nearest tenth
KonstantinChe [14]

We first know that 66 as a decimal is well 66. Now, 2/3 as a decimal would be 66.666666 and so on on. Therefore, 66 and 2/3 as a decimal is 66.66666 and so on.
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I really need some help with this question (Image attached)
ella [17]

Answer:

KL = 23

KM = 29

Step-by-step explanation:

By applying cosine rule in the given triangle KLM,

cos(∠M) = \frac{\text{Adjacent side}}{\text{Hypotenuse}}

cos(52°) = \frac{LM}{KM}

cos(52°) = \frac{18}{KM}

KM = \frac{18}{\text{cos}52}

KM = 29.24

KM ≈ 29

By applying sine rule,

sin(52°) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

sin(52°) = \frac{KL}{29.24}

KL = 29.24[sin(52°)]

    = 23.03

    ≈ 23

KL = 23

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3 years ago
Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the
ArbitrLikvidat [17]

Answer:

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because \frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi } the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let r be the radius, d the diameter, and V the volume of the spherical balloon.

We know \frac{dV}{dt}=20 \:{in^3/s} and we want to find \frac{dr}{dt}

The volume of a spherical balloon is given by

V=\frac{4}{3} \pi r^3

Taking the derivative with respect of time of both sides gives

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}

We now substitute the values we know and we solve for \frac{dr}{dt}:

d=2r\\\\r=\frac{d}{2}

r=\frac{12}{2}=6

\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and \frac{dr}{dt} is:

\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because \frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi } the radius is changing more rapidly when the diameter is 12 inches.

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Where is the pictures?
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