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Nookie1986 [14]

3 years ago

7

Simplify 5√2 +√3÷ 5√2-√3

2 answers:

eimsori [14]3 years ago

4 0

Answer:

<u>50√2+√6−10√3</u>

10

Step-by-step explanation:

Goshia [24]3 years ago

3 0

Answer:

13

Step-by-step explanation:

use this using the FOIL Method. 5⋅5+5(2√3)−2√3⋅5−2√3(2√3)Simplify

hope this helps

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Oranges cost 25 cents for 1/2 kilograms. How much would 8 kilograms of oranges cost?

eimsori [14]

$\dfrac{1}{2} \text{ kg} \longrightarrow 25 \text{ cents}$

Find 1 kg:
$1 \text{ kg} \longrightarrow 25 \times 2 \text{ cents}$

$1 \text{ kg} \longrightarrow 50 \text{ cents}$

Find 8 kg:
$8\text{ kg} \longrightarrow 50 \times 8 \text{ cents}$

$8\text{ kg} \longrightarrow 400 \text{ cents or } \$4.00$

Answer: 8 kg of oranges cost $4.00

6 0

3 years ago

Read 2 more answers

What is the solution to the inequality 15y-9<36

artcher [175]

Answer:

$\boxed{y$

Step-by-step explanation:

15y - 9 < 36 <em>add 9 to both sides</em>

15y < 45 <em>divide both sides by 15</em>

y < 3

6 0

3 years ago

The slope of (-2, -19) and (-12, 11)

ArbitrLikvidat [17]

I think the slope is m=-3 I hope it’s correct

6 0

3 years ago

The school want to purchase enough tarp material to cover their field during bad weather. They want to get exactly eleven percen

KonstantinChe [14]

Answer:

6660 yards^2

Step-by-step explanation:

Assumption: the field is 120 yards long and 50 yards wide.

First calculate the area of the field: 120*50 = 6000 yards^2

Then, calculate the 11% of that area: 6000 yards^2*11/100 = 660 yards^2

Finally, the area of the tarp material should be 6000 yards^2 + 660 yards^2 = 6660 yards^2

5 0

3 years ago

A large mixing tank initially contains 1000 gallons of water in which 30 pounds of salt have been dissolved. Another brine solut

serg [7]

Answer:

(B) $\dfrac{dA}{dt}=8-0.004A$

Step-by-step explanation:

Volume of fluid in the tank =1000 gallons

Initial Amount of Salt in the tank, A(0)= 30 pounds

Incoming brine solution of concentration 2 pounds of salt per gallon is pumped in at a rate of 4 gallons per minute.

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 4\frac{gal}{min})=8\frac{lbs}{min}$

The resulting mixture is pumped out at the same rate, therefore:

Rate Out =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{1000})( 4\frac{gal}{min})=\frac{A}{250}$

Therefore:

The rate of change of amount of salt in the tank,

$\dfrac{dA}{dt}=$Rate In-Rate out\\\dfrac{dA}{dt}=8-\dfrac{A}{250}\\\dfrac{dA}{dt}=8-0.004A$

5 0

4 years ago