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2 years ago

3 Which expression is equivalent - 3/8 / - 1/4

Mathematics

1 answer:

FrozenT [24]2 years ago

4 0

Answer:

3/8 - 1/4 = 1/

= 0.125

Step-by-step explanation:

brainliest please.

if your not sure with my answer don't go with it.

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Which one is the better buy $3 and 2 refills $4 and 3 refills

Brums [2.3K]

Answer:

Step-by-step explanation:

3 and 2

6 0

3 years ago

Read 2 more answers

A school holds classes from 8:00 a.m -2:00 p.m. For what percent of a 24- hour day does this school hold classes?

kati45 [8]

8:00 a.m all the way until 2:00 p.m.

8:00 a.m. to 12:00 p.m is 4 hours.
12:00 p.m. to 2:00 p.m is 2 hours

8:00 a.m. to 2:00 p.m. is 6 hours.

6 hours out of a 24 hour day.

$\frac{6}{24} =\frac{x}{100}\\\\\frac{1}{4} =\frac{x}{100}\\\\\sf{Cross~multiplying}\\\\100 = 4x\\\\\sf{Divide~4~on~both~sides$

$\boxed{\bf{x =25}}$

The school holds classes for 25% of the 24-hour day.

3 0

3 years ago

Read 2 more answers

Change the fraction 7/12 to a decimal. Round your answer to the nearest thousandth. A. 1.714 B. 1.700 C. .600 D. .583

mrs_skeptik [129]

Divide

7/12 = <span>.583

We get answer: </span><span>D. .583</span>

4 0

3 years ago

An airliner maintaining a constant elevation of 2 miles passes over an airport at noon traveling 500 mi/hr due west. At 1:00 PM,

butalik [34]

Answer:

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

Step-by-step explanation:

Let suppose that airliners travel at constant speed. The equations for travelled distance of each airplane with respect to origin are respectively:

First airplane

$r_{A} = 500\,\frac{mi}{h}\cdot t\\r_{B} = 550\,\frac{mi}{h}\cdot t$

Where t is the time measured in hours.

Since north and west are perpendicular to each other, the staight distance between airliners can modelled by means of the Pythagorean Theorem:

$s=\sqrt{r_{A}^{2}+r_{B}^{2}}$

Rate of change of such distance can be found by the deriving the expression in terms of time:

$\frac{ds}{dt}=\frac{r_{A}\cdot \frac{dr_{A}}{dt}+r_{B}\cdot \frac{dr_{B}}{dt}}{\sqrt{r_{A}^{2}+r_{B}^{2}} }$

Where $\frac{dr_{A}}{dt} = 500\,\frac{mi}{h}$ and $\frac{dr_{B}}{dt} = 550\,\frac{mi}{h}$ , respectively. Distances of each airliner at 2:30 PM are:

$r_{A}= (500\,\frac{mi}{h})\cdot (1.5\,h)\\r_{A} = 750\,mi$

$r_{B}=(550\,\frac{mi}{h} )\cdot (1.5\,h)\\r_{B} = 825\,mi$

The rate of change is:

$\frac{ds}{dt}=\frac{(750\,mi)\cdot (500\,\frac{mi}{h} )+(825\,mi)\cdot(550\,\frac{mi}{h})}{\sqrt{(750\,mi)^{2}+(825\,mi)^{2}} }$

$\frac{ds}{dt}\approx 743.303\,\frac{mi}{h}$

6 0

3 years ago

What is the sum of the infinite geometric series 9−6+4−83+...9-6+4-83+...??

vivado [14]

Successive terms have a common factor of $-\dfrac23$ , which means the $n$ th partial sum of the series can be written as

$S_n=9-6+4-\dfrac83+\cdots+9\left(-\dfrac23\right)^{n-1}$
$S_n=9\left(1+\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\cdots+\left(-\dfrac23\right)^{n-1}$

$\implies-\dfrac23S_n=9\left(\left(-\dfrac23\right)^1+\left(-\dfrac23\right)^2+\left(-\dfrac23\right)^3+\left(-\dfrac23\right)^4+\cdots+\left(-\dfrac23\right)^n\right)$

$\implies S_n-\left(-\dfrac23\right)S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$\dfrac53S_n=9\left(1-\left(-\dfrac23\right)^n\right)$
$S_n=\dfrac{27}5\left(1-\left(-\dfrac23\right)^n\right)$

As $n\to\infty$ , the geometric term vanishes, leaving you with

$S=\displaystyle\lim_{n\to\infty}S_n=\dfrac{27}5$

7 0

3 years ago