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3 years ago

10

Simplify: (-2^3)^2 (there exponents signs btw :))

1 answer:

ollegr [7]3 years ago

3 0

Answer:

64

Step-by-step explanation:

Brainliest

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A 1,000 mL solution of acetic acid contains 26 mL of pure acetic acid. Find the percent concentration of this solution.

Mrac [35]

Given:

Total solution : 1000mL

Pure acetic acid = 26mL

To find:

The percent concentration of given solution.

Solution:

We know that,

$\text{Percent concentration}=\dfrac{\text{Pure acetic acid}}{\text{Total solution}}\times 100$

Putting the given values, we get

$\text{Percent concentration}=\dfrac{26}{1000}\times 100$

$\text{Percent concentration}=\dfrac{26}{10}$

$\text{Percent concentration}=2.6$

Therefore, the percent concentration of this solution is 2.6%.

6 0

2 years ago

An object is propelled upward from the top of a 300 foot building. The path that the object takes as it falls to the ground can

serg [7]

Answer:

As per the statement:

The path that the object takes as it falls to the ground can be modeled by:

h =-16t^2 + 80t + 300

where

h is the height of the objects and

t is the time (in seconds)

At t = 0 , h = 300 ft

When the objects hit the ground, h = 0

then;

-16t^2+80t+300=0

For a quadratic equation: ax^2+bx+c=0 ......[1]

the solution for the equation is given by:

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

On comparing the given equation with [1] we have;

a = -16 ,b = 80 and c = 300

then;

$t= \frac{-80\pm \sqrt{(80)^2-4(-16)(300)}}{2(-16)}$

$t= \frac{-80\pm \sqrt{6400+19200}}{-32}$

$t= \frac{-80\pm \sqrt{25600}}{-32}$

Simplify:

$t = -\frac{5}{2}$ = -2.5 sec and $t = \frac{15}{2}$ = 7.5 sec

Time can't be in negative;

therefore, the time it took the object to hit the ground is 7.5 sec

8 0

3 years ago

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

5 0

2 years ago

Find the value of y, rounded to the nearest tenth. Please help me, I'd appreciate it!!

Blababa [14]

If a secant<span> and a </span><span>tangent of a circle </span><span>are drawn from a point outside the circle, then the product of the lengths of the secant and its external segment equals the square of the length of the tangent segment.
</span><span>
y</span>² = 7(15+7)
<span>y</span>² = 7*22
<span>y</span>² = 154
<span>y = </span>√154
<span>y = 12.4 </span>← to the nearest tenth<span>
</span><span>
</span>

8 0

2 years ago

Help me w this pls pls pls

patriot [66]

Answer:

48

Step-by-step explanation:

2500 - 196 = $x^{2}$

x = 48

Please mark as brainliest for future answers :)

8 0

2 years ago