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babymother [125]
3 years ago
11

PLEASE PLEASE HELP WITH THIS

Mathematics
1 answer:
Phoenix [80]3 years ago
4 0
Since triangles ABC and BCD are similar, we can establish a proportion between the lengths of corresponding sides:
\frac{AC}{BC} = \frac{BC}{CD}

We can infer from our picture that BC=m. Also we can find the length of AC by adding AD and CD:
AC=AD+CD
AC=11+7
AC=18

Lets replace the values in our proportion:
\frac{AC}{BC} = \frac{BC}{CD}
\frac{18}{m} = \frac{m}{7}
Solving for m:
m^2=(18)(7)
m^2=126
m= \sqrt{126}

We can conclude that the correct answer is: D \sqrt{126}
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Step-by-step explanation:

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Option (4)

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From the graph attached,

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y-coordinate of A → -4

y-coordinate of B → 0

y-coordinate of C → -4

Ends of the curve are moving towards infinity, so the absolute maximum is not defined.

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Therefore, point representing local maximum is B(0, 0).

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Option (4) will be the correct option.

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3 years ago
Jake tosses a coin up in the air and lets it fall to the ground. The equation that models the height (in feet) and time (in seco
adoni [48]

Answer:

The maximum height of coin Jake tosses is 15 feet.

Step-by-step explanation:

Given Jake tosses a coin up in the air

and the height of coin is model by h(t)=(-16)t^{2} +24t+6

To find height of the coin when Jake tosses it:

When a coin is in the hand of Jake, time t=0

Height of coin is h(t)=(-16)t^{2} +24t+6

h(0)=(-16)(0)^{2} +24(0)+6

h(0)=6 feet.

Therefore, Height of coin at time t=0 is 6.

For maximum height of the coin,

h(t)=(-16)t^{2} +24t+6

Differentiating both side,

\frac{d}{dt}h(t)=\frac{d}{dt}[(-16)t^{2} +24t+6]

\frac{d}{dt}h(t)=\frac{d}{dt}[(-32)t+24]

\frac{d}{dt}h(t)=0

\frac{d}{dt}[(-32)t+24]=0

t=\frac{24}{32}

t=\frac{3}{4}

t=0.75

Now,

h(t)=(-16)t^{2} +24t+6

h(0.75)=(-16)(0.75)^{2} +24(0.75)+6

h(0.75)=15 feet.

The maximum height of coin jake tosses is 15 feet.

7 0
4 years ago
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