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6 months ago

If f(x)=x²-6 x+2 and g(x)=-2x, find the following composition. (g ° f)(5) (g ° f)(5)=

Mathematics

1 answer:

Elan Coil [88]6 months ago

5 0

What the composite function in the question is actually trying to explain is;

$(gof)(x)=g(f(5))$

First of all, in the function $F(x)$ , we will find a result by writing $5$ where we see the letter $x$ .

$F(5)=5^2-6(5)+2=-3$

We will write this result in the $G(x)$ function where we see the letter $x$ and we will get the result.

$G(-3)=-2(-3)=6$

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A triangle-shaped machine part has sides of 2 1/2 inches, 3 1/4 inches, and 5 inches. The triangle's perimeter is

SOVA2 [1]

Answer:

Step-by-step explanation:

The perimeter of any shape is the distance around the edge of the shape. It is found by adding the side lengths of each side together.

Add each side length listed for the triangle part.

$P = 2\frac{1}{2} + 3\frac{1}{4} + 5 \\P = 2\frac{2}{4} + 3\frac{1}{4} + 5\\P = 10 \frac{3}{4}$

8 0

2 years ago

Built around 2600 BCE, the Great Pyramid of Giza in Egypt is 485 ft high (due to erosion, its current height is slightly less) a

Morgarella [4.7K]

Answer:1.69*10^12 J

Step-by-step explanation:

From figure above, using triangle ratio

485/755.5=y/l. Cross multiplying 485l=755.5y Divide via 485) hence l= 755.5y/485

Consider a slice volume Vslice= (755.5y/485)^2∆y; recall density =150lb/ft^3

Force slice = 150*755.5^2.y^2.∆y/485^2

From figure 2 in the attachment work done for elementary sclice

Wslice= 150.755.5^2.y^2.∆y.(485-y)/485^2

= (150*755.5^2*y^2)(485-y)∆y/485

To calculate the total work we integrate from y=0 to y= 485

Ie W=[ integral of 150*755.5^2 *y^2(485-y)dy/485] at y=0 and y= 485

Integrating the above

W= 150*755.5^2/485[485*y^3/3-y^4/4] at y= 0 and y=485

W= 150*755.5^2/485(485*485^3/3-484^4/4)-(485.0^3/3-0^4/4)

Work done 1.69*10^12joules

7 0

3 years ago

A circle has its center at (0,0) and passes through the point (0,9). What is the standard equation of the circle?

Alexxandr [17]

Answer:

x^2+y^2=9^2

Step-by-step explanation:

The standard equation of a circle is:

$(x-h)^2+(y-k)^2=r^2$

Where the center of a circle is (h,k) and r is the radius of the circle. In this case because a circle is equidistant from the center and we have a point where it passes through 9 that means that the radius is 9. However, since the standard equation states that we must write r in the form of $r^2$ this means that $r^2=9^2$ . Therefore by plugging in the values we have: