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3 years ago

Find the value of x.

Mathematics

1 answer:

Leona [35]3 years ago

5 0

Answer:

$\boxed{\text{\sf \Large 3}}$

Step-by-step explanation:

Angles in a triangle add up to 180°

$180-(24x+3+25x)$

$-49x+177$

Angles on a straight line add up to 180°

$-49x+177+150=180$

$x=3$

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Determine the common difference for each arithmetic sequence 10.5, 13, 15.5, 18

andrew11 [14]

Answer:

+2.5

Step-by-step explanation:

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2 years ago

The perimeter of a square is 12p +4. One side of the square equals? Explain your answer.

Hoochie [10]

We know that a square, by definition, has four congruent sides.

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3 years ago

the diameter d in inches of a rope required to lift a weight of w tons is given by the function d = 10w π . How much weight can

faltersainse [42]

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Step-by-step explanation:

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3 years ago

Where would you insert parentheses to make the statement true?

9966 [12]

Just around the 15 and 3. So it would be

(15-3)x4+9=57

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8 0

3 years ago

Read 2 more answers

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:

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

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

4 0

3 years ago

Find the value of x.​

Find the value of x.