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3 years ago

A student dissolves some solid sodium chloride in a beaker of water.

Mathematics

1 answer:

Margarita [4]3 years ago

3 0

Solute = what is dissolved

solvent = that which dissolves or can dissolve

You take it from here.

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Which expression is equivalent to (5^-2)(5^-1) A. -1/125 B. -1/5 C. 1/125 D. 1/5

n200080 [17]

Answer:

1/125

Step-by-step explanation:

(5^-2)(5^-1) = 5^-3=1/5^3=1/125

3 0

3 years ago

6.find zeroes by factorisation 4x2 + 5√2 x - 3

dangina [55]

-13.213203440000000000000

7 0

3 years ago

The radius of a sphere-shaped balloon increases at a rate of 2 centimeters (cm) per second. If the surface area of

djverab [1.8K]

Answer:

7 seconds

Step-by-step explanation:

The radius of a sphere-shaped balloon increases at a rate of 2 centimeters (cm) per second.

surface area of the completely inflated balloon is 784 cm²

SA=4r²

784= 4r²

784/4= r²

196 = r²

14cm = r

Yhe radius of the complete inflated balloon is 14cm

If the ball inflate at the rate of 2 cm per seconds

Then it took 14/2 = 7 seconds to inflate fully

5 0

3 years ago

How do I solve ?? It’s cylinders

UkoKoshka [18]

V= 3.14r^2h
10/2= r
5^2= 25
(3.14)(25)(6)
V=471

3 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 43 ft/s. Its height

Lina20 [59]

Answer:

Instantaneous Velocity at t = 1 is 20 feet per second

Step-by-step explanation:

We are given he following information in the question:

$y(t)=43t-23t^{2}$

B) Instantaneous Velocity at t = 1

$y(1) = 43(1)-23(1)^2 = 20$

A) Formula:

Average velocity =

$\displaystyle\frac{\text{Displacement}}{\text{Time}}$

1) 0.01

$y(1 + 0.01)=y(1.01) = 43(1.01)-23(1.01)^{2} = 19.9677\\\\\text{Average Velocity} = \dfrac{y(1.01)-y(1)}{1.01-1} =\dfrac{19.9677-20}{1.01-1}= \dfrac{-0.0323}{0.01} = -3.230000 \text{feet per second}$

2) 0.005 s

$y(1 + 0.005)=y(1.005) = 43(1.005)-23(1.005)^{2} = 19.984425\\\\\text{Average Velocity} = \dfrac{y(1.005)-y(1)}{1.005-1} =\dfrac{19.984425-20}{1.005-1} = -3.1150000 \text{feet per second}$

3) 0.002 s

$y(1 + 0.002)=y(1.002) = 43(1.002)-23(1.002)^{2} = 19.993908\\\\\text{Average Velocity} = \dfrac{y(1.002)-y(1)}{1.002-1} =\dfrac{19.993908-20}{1.002-1} = -3.0460000 \text{feet per second}$

4) 0.001 s

$y(1 + 0.001)=y(1.001) = 43(1.001)-23(1.001)^{2} = 19.996977\\\\\text{Average Velocity} = \dfrac{y(1.001)-y(1)}{1.001-1} =\frac{19.996977-20}{1.001-1} = -3.0230000 \text{feet per second}$

3 0

2 years ago