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3 years ago

Are any of the figures B, C, or D scaled coples of figure A? Explain how you know.

Mathematics

1 answer:

belka [17]3 years ago

3 0

Answer:

Step-by-step explanation:

C is the only one that is an enlarged version of A.

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What is a-5/245 Don't hate but be truthful and don't cap do this teen boy look good ra-te him plz

Inessa [10]

No if its you sorry I don’t really mean it

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2 years ago

I WILL GIVE YOU BRAINLIEST!!

ollegr [7]

Answer:

32 goldfish

Step-by-step explanation:

Given:

Number of fish tanks = 8
Volume of water per fish tank = 5,544 in³
Water required by 1 goldfish = min 1,386 in³

Number of fish per tank = water per tank ÷ water per goldfish

= 5544 ÷ 1386

= 4

Therefore, each tank can hold a maximum of 4 goldfish.

Greatest number of goldfish = number of tanks × fish per tank

= 8 × 4

= 32

3 0

2 years ago

For the figure above: Assume that <ABD has a mesure of 105° ANSWER ASAP WILL GIVE 15 POINTS AND BRAINLIEST

kykrilka [37]

Answer:

DBC = 75

Step-by-step explanation:

ABD and DBC are supplementary angles. They add to 180 degrees.

ABD + DBC = 180

105 + DBC = 180

Subtract 105 from each side

105-105 + DBC = 180-105

DBC = 75

4 0

2 years ago

Read 2 more answers

The rate of change of the number of mountain lions N(t) in a population is directly proportional to 725 - N(t), where t is the t

wolverine [178]

Answer:

a) The implied differential equation is $\frac{dN}{725 - N} = kdt$

b) The general equation is $N = 725 - C e^{-kt}$

c) The particular equation is $N = 725 - 325 e^{-0.49t}$

d) The population when t = 5, N(5) = 697 = 700( to the nearest 50)

Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, $\frac{dN}{dt} \alpha (725 - N(t))$

$\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt$

Integrating both sides of the equation

$\int {\frac{1 }{725 - N}} \, dN = \int {k} \, dt\\- ln (725 - N) = kt + C\\ ln (725 - N) = -(kt + C)\\725 - N = e^{-(kt + C)} \\725 - N = e^{-kt} * e^{-C} \\725 - N = C e^{-kt}\\N = 725 - C e^{-kt}$

When t = 0, N = 400

$400 = 725 - C e^{-k*0}\\400 = 725 - C\\C = 725 - 400\\C = 325$

When t = 3, N = 650

$650 = 725 - (325 * e^{-3k})\\325 * e^{-3k} = 75\\e^{-3k} = 75/325\\e^{-3k} = 0.23\\-3k = ln 0.23\\-3k = -1.47\\k = 1.47/3\\k = 0.49$

The equation for the population becomes:

$N = 725 - 325 e^{-0.49t}$

At t = 5, the population becomes:

$N = 725 - 325 e^{-0.49*5}\\N = 725 - 325 e^{-2.45}\\N = 696.95\\N(5) = 697$

N(5) = 700 ( to the nearest 50)

6 0

3 years ago

Find the value of x. Then find the measure of each label length.

sweet-ann [11.9K]

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3 years ago