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2 years ago

5

Seth found two new ways to get to school. Which route is faster for Seth, when school is 17 miles away?

1 answer:

nignag [31]2 years ago

6 0

Answer:answer is D relative house

Step-by-step explanation:

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Please need help thank you

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How many took 7.5 OR less. This means it includes the hikes that took 7.5 hours. Count all of the ones that start from 7.5 and lower

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3 years ago

Which strategy is the most appropriate strategy to solve (x−1)^2=9 ?

Leya [2.2K]

(x-1)^2=9 <=> x-1=3 <=> x=4

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3 years ago

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Draw three rectangles that 2 cm long and 1 cm wide. On each rectangle show a different way to divide the rectangle in half. Then

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3 years ago

How do I solve these problems?

lesya [120]

$Domain:\\D:x > 0\\\\\ln x=5.6+\ln(7.5)\ \ \ \ |-\ln(7.5)\\\\\ln x-\ln7.5=5.6\ \ \ |Use\ \log x-\log y=\log\dfrac{x}{y}\\\\\ln\dfrac{x}{7.5}=5.6\ \ \ \ |use\ \log_ab=c\iff a^c=b\\\\\dfrac{x}{7.5}=e^{5.6}\ \ \ \ |\cdot7.5\\\\\boxed{x=7.5e^{5.6}}\in D$

$\log x=5.6-\log7.5\ \ \ \ |+\log7.5\\\\\log x+\log7.5=5.6\ \ \ \ |use\ \log x+\log y=\log (xy)\\\\\log(7.5x)=5.6\ \ \ \ |use\ \log_ab=c\iff a^c=b\\\\7.5x=10^{5.6}\ \ \ \ |:7.5\\\\\boxed{x=\dfrac{10^{5.6}}{7.5}}\in D$

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3 years ago

A 10-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Alex777 [14]

Answer:

The area is changing at 15.75 square feet per second.

Step-by-step explanation:

The triangle between the wall, the ground, and the ladder has the following dimensions:

H: is the length of the ladder (hypotenuse) = 10 ft

B: is the distance between the wall and the ladder (base) = 6 ft

L: the length of the wall (height of the triangle) =?

dB/dt = is the variation of the base of the triangle = 9 ft/s

First, we need to find the other side of the triangle:

$H^{2} = B^{2} + L^{2}$

$L = \sqrt{H^{2} - B^{2}} = \sqrt{(10)^{2} - B^{2}} = \sqrt{100 - B^{2}}$

Now, the area (A) of the triangle is:

$A = \frac{BL}{2}$

Hence, the rate of change of the area is given by:

$\frac{dA}{dt} = \frac{1}{2}[L*\frac{dB}{dt} + B\frac{dL}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} + B\frac{d(\sqrt{100 - B^{2}})}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - B^{2}}*\frac{dB}{dt} - \frac{B^{2}}{(\sqrt{100 - B^{2}})}*\frac{dB}{dt}]$

$\frac{dA}{dt} = \frac{1}{2}[\sqrt{100 - 6^{2}}*9 - \frac{6^{2}}{\sqrt{100 - 6^{2}}}*9]$

$\frac{dA}{dt} = 15.75 ft^{2}/s$

Therefore, the area is changing at 15.75 square feet per second.

I hope it helps you!

5 0

3 years ago