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3 years ago

What is the volume of this object?

Mathematics

1 answer:

Anton [14]3 years ago

4 0

yo sup??

This figure can be resolved into two cuboids

one with l1,b1,h1=1,3,2

and other with l2,b2,h2=6,3,3

Volume of the two cuboids=l1*b1*h1+l2*b2*h2

=1*3*2+6*3*3

=6+54

=60 cm^3

Hope this helps.

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Ben is 19 Ishaan is 7

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Which of the is not a function

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2 years ago

Use Newton's method to find an approximate solution of ln(x)=10-x. Start with x_0 =9 and find x_2 .

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Answer:

x₂ = 7.9156

Step-by-step explanation:

Given the function ln(x)=10-x with initial value x₀ = 9, we are to find the second approximation value x₂ using the Newton's method. According to Newtons method xₙ₊₁ = xₙ - f(xₙ)/f'(xₙ)

If f(x) = ln(x)+x-10

f'(x) = 1/x + 1

f(9) = ln9+9-10

f(9) = ln9- 1

f(9) = 2.1972 - 1

f(9) = 1.1972

f'(9) = 1/9 + 1

f'(9) = 10/9

f'(9) = 1.1111

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x₁ = 9 - 1.1972/1.1111

x₁ = 9 - 1.0775

x₁ = 7.9225

x₂ = x₁ - f(x₁)/f'(x₁)

x₂ = 7.9225 - f(7.9225)/f'(7.9225)

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f(7.9225) = 2.0697 + 7.9225 -10

f(7.9225) = 0.0078

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x₂ = 7.9156

<em>Hence the approximate value of x₂ is 7.9156</em>

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3 years ago

The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

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3 years ago

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