Compute the area of triangle, if x equals 4 more than 6. A) 10 B) 50 C) 100 D) 200 E) 400

Mathematics

1 answer:

harkovskaia [24]3 years ago

5 0

Answer:

76cm²

Step-by-step explanation:

The triangle from all indication is an isosceles triangle:

Let x represent the side

∴ x = 4 - 6 = -2

x = -2

x + 2 = 0

Using the fomular, as area of triangle, that is:

Area = √s(s - a)(s - b)(s - c)

s = a + b+ c/2, where a = x + 2; b = x+ 2; c = x

∴ s = x + 2 + x + 2 + x/2 = 3x + 4/2

Area = √3x + 4/2( 3x + 4/2 - x - 2)(3x + 4/2 - x - 2)(3x + 4/2 - x)

= √3x + 4/2( x/2)(x/2)(x + 4/2)

= √3x + 4/2(x²/4)(x + 4)/2

Let's assume x = 12

∴ Area = √5760 = 76cm²

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The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$