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Artyom0805 [142]
3 years ago
5

Find the square root of: (i) 729(ii) 1296if anyone knows the pls help me in this ​

Mathematics
1 answer:
Sveta_85 [38]3 years ago
6 0

\sqrt{729} =27\\ \sqrt{1296} =36

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How much money at 12 weeks?<br><br>1). 330<br>2). 290<br>3). 360
zloy xaker [14]

To answer this question, you need to write a linear equation (because the graph is a visibly straight line) that models this situation, then plug in 12.

The general format for a linear equation is y=mx+b, where m is the slope and b is the y intercept.  The y intercept of a line is where it crosses the y axis; for this line, it crosses the y axis at (0, 0) so the y intercept is 0.

To find the slope, use the slope equation: (y2-y1)/(x2-x1).  For y1, y2, x1, and x2, just pick any two points on the line.  We will use (1, 30) and (2, 60):

(60-30)/(2-1)

30/1

30

The line is y=30x.

Plug in 12 for x:

y=30(12)

y=360

3). 360

Hope this helps!!

3 0
3 years ago
The price of a 7-minute phone call is $1.05. What is the price of a 19-minute phone call
Leni [432]

Answer:

19.95

Step-by-step explanation:

if you take 1.05 times it by 19 you get your answer

7 0
3 years ago
-2x+8x=20 ans can you please include the steps. Thank you
sladkih [1.3K]

Hope this helps :).    

3 0
3 years ago
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Use 3.14 for pi. ROUND YOUR FINAL ANSWER TO THE NEAREST TENTHS PLACE!
Natasha2012 [34]

The area of the given circle is 95 square ft.

Calculation for the Area of the Circle:

It is given in the diagram that,

The diameter of the circle, d = 11 ft.

Now, we know that the radius of a circle is equal to half of the diameter.

Therefore, radius of the given circle, r = 11/2 ft.

The formula for the area of the circle is given as follows,

A = π × r²

Substituting the values, π = 3.1, and r = 11/2, we get,

A = (3.14) × (11/2)²

A =  (3.14) × 30.25

A = 94.985

A ≈ 95

Hence, the area of the given circle with diameter 11 ft. comes out to be 95 ft².

Learn more about a circle here:

brainly.com/question/11833983

#SPJ1

7 0
2 years ago
Read 2 more answers
A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per
Tems11 [23]

Answer:

The pressure is changing at \frac{dP}{dt}=3.68

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} and we want to find at what rate is the pressure changing.

The equation that model this situation is

PV^{1.4}=k

Differentiate both sides with respect to time t.

\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)

Apply this rule to our expression we get

V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0

Solve for \frac{dP}{dt}

V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}

when P = 23 kg/cm2, V = 35 cm3, and \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} this becomes

\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68

The pressure is changing at \frac{dP}{dt}=3.68.

7 0
3 years ago
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