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3 years ago

Which shape does the intersection of the horizontal plane with a prism look like

Mathematics

1 answer:

slega [8]3 years ago

3 0

Answer:

Triangle

Step-by-step explanation:

Given: A prism

To find: the intersection of the horizontal plane with a prism

Solution:

A prism is a polyhedron having two parallel faces The other faces are always parallelograms. A prism can have rectangular or other polygon shapes.

The intersection of the horizontal plane with a prism looks like a triangle.

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Three more than the product of two and a number x

Nonamiya [84]

$2x+3$

You are taking two times a number, which is defined as x. That is equivalent to 2x. Then, you are adding three more to that product.

3 0

3 years ago

Read 2 more answers

Graph the functions and approximate an x-value in which the exponential function exceeds the polynomial function. y = 4x y = 7x2

nordsb [41]

Answer:

The Exponential function is

$y=4^x$

And the polynomial function is

$y=7 x^2 +4 x -2$

And, we have to find the value of x for which, exponential function exceeds the polynomial function which can be written as

$4^x> 7 x^2 +4 x -2$

1. When , x= -1

LHS

$4^{-1}=\frac{1}{4}=0.25$

RHS

$=7*(-1)^2 +4 *(-1)-2\\\\=7-4-2=1$

2. When , x=0

L HS

$4^0=1$

RHS

$7*0+4*0-2= -2$

3. When ,x= 0.5

L HS

$4^{0.5}=2$

RHS

$=7*0.25 +4*0.5 -2\\\\= 1.75+2-2\\\\=1.75$

4. When , x=2

LHS

$4^{2}=16$

RHS

$=7*4^2+4*4-2\\\\=112+16-2\\\\=126$

The Minimum value for which exponential function exceeds the polynomial function is , x= 0.5

But,there is other value for which exponential function exceeds the polynomial function is , x=2.

5 0

3 years ago

Principal: $15,000

zhannawk [14.2K]

I like eating poop that’s the answer

7 0

3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

inysia [295]

Answer:

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

Step-by-step explanation:

We are given that

$f(x)=7tan^{-1}(x)$

a=1

$T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}$

Substitute n=3 and a=1

$t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}$

$f(x)=7tan^{-1}(x)$

$f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}$

Where $tan^{-1}(1)=\frac{\pi}{4}$

$f'(x)=\frac{7}{1+x^2}$

Using the formula

$\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}$

$f'(1)=\frac{7}{2}$

$f''(x)=\frac{-14x}{(1+x^2)^2}$

$f''(1)=-\frac{7}{2}$

$f''(x)=-14x(x^2+1)^{-2}$

$f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})$

By using the formula

$(uv)'=u'v+v'u$

$f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}$

$f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}$

$f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}$

Substitute the values

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3$

$t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3$

7 0

3 years ago

If r || s and M<5 = 128, what’s is M<1?? A. 128 B. 59 C. 62 D. 64

Fudgin [204]

We have been given that r is parallel to s. Now, if we look at angle 5 and angle 1, we can see that they are corresponding angles on parallel lines. Corresponding angles are ALWAYS equal, so therefore, your answer is 128 degrees; option A! :)

Hope this helped!

7 0

4 years ago