All categories

3 years ago

What is the square rout of 56

Mathematics

1 answer:

Mars2501 [29]3 years ago

6 0

$\sqrt{56}=\sqrt{4\cdot14}=\sqrt4\cdot\sqrt{14}=2\sqrt{14}$

You might be interested in

A box contains 3 plain pencils and 9 pens. a second box contains 7 color pencils and 3 crayons. one item from each box is chosen

SpyIntel [72]

Out of the first box the possibility would be 3:12 or 3 out of 12. The second box would be 7:10 or 7 out of 10

3 0

3 years ago

Read 2 more answers

What is the value of x?

vitfil [10]

Answer:

5 cm

Step-by-step explanation:

We know these are similar triangles, so we can use ratios of side lengths

x/40 = 4/32

Using cross products

32x = 4*40

32x = 160

Divide each side by 32

32x/32 = 160/32

x =5

3 0

3 years ago

Someone please help!

Natalija [7]

The box plot shows the range and quartiles of the data. It does not show numbers or the mean. The only appropriate choice is

... The lower quartile for the data

3 0

3 years ago

Celia is scheduled to work a total of 20 hours this week. So far she has worked

Fudgin [204]

Answer:

15 hours.

Step-by-step explanation:

20 can be divided into quarters, 3 quarters is 15/20 hours.

6 0

3 years ago

I don't know how to do this or what i'm doing plz help

mote1985 [20]

recalling that d = rt, distance = rate * time.

we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?

$\bf \begin{array}{ccll} miles&hours\\ \cline{1-2} 12&1\\ 18&x \end{array}\implies \cfrac{12}{18}=\cfrac{1}{x}\implies 12x=18\implies x=\cfrac{18}{12}\implies x=\cfrac{3}{2}$

so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.

then Wanda kicks in, rolling like a lightning at 16mph.

let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.

the distance both have travelled is the same "d" miles, reason why they "meet", same distance.

$\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}$

$\bf \stackrel{\textit{substituting \underline{d} in the 2nd equation}}{\boxed{(12)\left( t+\frac{3}{2} \right)}=16t}\implies 12t+18=16t \\\\\\ 18=4t\implies \cfrac{18}{4}=t\implies \cfrac{9}{2}=t\implies \stackrel{\textit{four and a half hours}}{4\frac{1}{2}=t}$

7 0

3 years ago