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3 years ago

T= m/x + p for m. Find m

Mathematics

1 answer:

Usimov [2.4K]3 years ago

4 0

You only have variables, so this problem is unsolvable

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The unit circle has a radius of 1 unit and is centered at the origin. It is dilated so that it passes through the point . What i

yuradex [85]

Answer:

Step-by-step explanation:

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3 years ago

Find the difference of 6/7 minus 1/14

VMariaS [17]

Answer:

11/14

Step-by-step explanation:

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3 years ago

Read 2 more answers

Barbara drives between Miami, Florida, and West Palm Beach, Florida. She drives 50 mi in clear weather and then encounters a thu

cluponka [151]

Answer:

Her speed driving in nice weather is 50 mph and in thunderstorm is 32 mph.

Step-by-step explanation:

Barbara drives 50 miles in clear weather and then encounters a thunderstorm for the last 16 miles.

Suppose, her speed in nice weather is $x$ mph.

As she drives 18 mph slower through the thunderstorm than she does in clear weather, so her speed in thunderstorm will be: $(x-18) mph$

We know that, $Time = \frac{Distance}{Speed}$

So, the time of driving in clear weather $=\frac{50}{x}$ hours

and the time of driving in thunderstorm $=\frac{16}{x-18}$ hours.

Given that, the total time for the trip is 1.5 hours. So, the equation will be......

$\frac{50}{x}+ \frac{16}{x-18}=1.5 \\ \\ \frac{50x-900+16x}{x(x-18)}=1.5\\ \\ \frac{66x-900}{x(x-18)}=1.5 \\ \\ 1.5x(x-18)=66x-900\\ \\ 1.5x^2-27x=66x-900\\ \\ 1.5x^2-93x+900=0\\ \\ 1.5(x^2 -62x+600)=0\\ \\ x^2 -62x+600=0\\ \\ (x-50)(x-12)=0$

Using zero-product property.........

$x-50=0\\ x=50\\ \\ and\\ \\ x-12=0\\ x=12$

We need to ignore $x=12$ here, otherwise the speed in thunderstorm will become negative.

So, her speed driving in nice weather is 50 mph and her speed driving in thunderstorm is (50-18) = 32 mph

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3 years ago

A recursive rule for a geometric sequence is a1=2; an= 4/9 an-1

Oxana [17]

A recursive rule for a geometric sequence:

$a_1\\\\a_n=r\cdot a_{n-1}$

---------------------------------------------------

$a_1=2;\ a_n=\dfrac{4}{9}a_{n-1}\to r=\dfrac{4}{9}$

---------------------------------------------------

The exciplit rule:

$a_n=a_1r^{n-1}$

Substitute:

$a_n=2\left(\dfrac{4}{9}\right)^{n-1}=2\cdot\left(\dfrac{4}{9}\right)^n\cdot\left(\dfrac{4}{9}\right)^{-1}=2\cdot\left(\dfrac{4}{9}\right)^n\cdot\dfrac{9}{4}\\\\\boxed{a_n=\dfrac{9}{2}\cdot\left(\dfrac{4}{9}\right)^n}$