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yawa3891 [41]
3 years ago
5

Evaluate this expression for x= 3.1.7x-5The value of the expression is​

Mathematics
2 answers:
Butoxors [25]3 years ago
6 0

Evaluate this expression for

x= 3.1 [Given]

Now,

7x - 5

<em>Putting</em><em> </em><em>the</em><em> </em><em>value</em><em> </em><em>of</em><em> </em><em>x</em><em> </em>

⇒<em> </em>7 × 3.1 - 5

⇒ 21.7 - 5

⇒ 16.7

The value of the expression is 16.7

lozanna [386]3 years ago
5 0

Answer:

Answer is 16.7

Step-by-step explanation:

Given \: expression \: is \: \:  7x - 5 \\ tofind \: the \: value \: of \: the \: expression \:  \\ when \: x = 3.1 \\ on \: substituting \: x = 3.1 \: in \: the \: given \: expression \\ 7x - 5 = 7(3.1) - 5 \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 21.7 - 5 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \: = 16.7

HAVE A NICE DAY!

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION.

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The question is incomplete. The complete question is :

The volume of a right circular cone with radius r and height h is V = pir^2h/3. a. Approximate the change in the volume of the cone when the radius changes from r = 5.9 to r = 6.8 and the height changes from h = 4.00 to h = 3.96.

b. Approximate the change in the volume of the cone when the radius changes from r = 6.47 to r = 6.45 and the height changes from h = 10.0 to h = 9.92.

a. The approximate change in volume is dV = _______. (Type an integer or decimal rounded to two decimal places as needed.)

b. The approximate change in volume is dV = ___________ (Type an integer or decimal rounded to two decimal places as needed.)

Solution :

Given :

The volume of the right circular cone with a radius r and height h is

$V=\frac{1}{3} \pi r^2 h$

$dV = d\left(\frac{1}{3} \pi r^2 h\right)$

$dV = \frac{1}{3} \pi h \times d(r^2)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

a). The radius is changed from r = 5.9 to r = 6.8 and the height is changed from h = 4 to h = 3.96

So, r = 5.9  and dr = 6.8 - 5.9 = 0.9

     h = 4  and dh = 3.96 - 4 = -0.04

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (5.9)(4)(0.9)+\frac{1}{3} \pi (5.9)^2 (-0.04)$

$dV=44.484951 - 1.458117$

$dV=43.03$

Therefore, the approximate change in volume is dV = 43.03 cubic units.

b).  The radius is changed from r = 6.47 to r = 6.45 and the height is changed from h = 10 to h = 9.92

So, r = 6.47  and dr = 6.45 - 6.47 = -0.02

     h = 10  and dh = 9.92 - 10 = -0.08

Now, $dV = \frac{2}{3} \pi r h (dr)+\frac{1}{3} \pi r^2 dh$

$dV = \frac{2}{3} \pi (6.47)(10)(-0.02)+\frac{1}{3} \pi (6.47)^2 (-0.08)$

$dV=-2.710147-3.506930$

$dV= -6.22$

Hence, the approximate change in volume is dV = -6.22 cubic units

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