I really need some help. I need to find the inverse of this equation in the picture below. Thanks!

Mathematics

1 answer:

serg [7]3 years ago

4 0

Answer:

y=81x^4+648x^3+1944x^2+2592x+1289

Step-by-step explanation:

1. Interchange x and y.

y=\frac{\sqrt[4]{x+7}}{3}-2

x=\frac{\sqrt[4]{y+7}}{3}-2

2. Then solve x=\frac{\sqrt[4]{y+7}}{3}-2 for y.

3. y=81x^4+648x^3+1944x^2+2592x+1289

I hope this helps.

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Sveta_85 [38]

If you're asking how to simplify, then this is how you do it.

when a power meets another power but in brackets, then you have to multiply the powers. ¼ × 4 would be 1. i simplified 81 to 3⁴.

7 0

3 years ago

(High points) please solve with explanation

AysviL [449]

Answer:

The area and the perimeter of the picture are:

Area = 160 cm^2
Perimeter = 67.31 cm

Step-by-step explanation:

To find the area of that figure, you can find the area how if it was a rectangle and next subtract the area of the triangle in the upper part. The area of a rectangle could be found by the next formula:

Area of a rectangle = base * height

As you can see in the picture, the base is 16 cm and the height is 12 cm, then we replace in the formula:

Area of a rectangle = 16 cm * 12 cm
Area of a rectangle = 192 cm^2

Now, we calculate the area of the triangle to subtract to the area we found and obtain the real area, the formula to obtain the area of a triangle is:

Area of a triangle = (base * height) / 2

The height of the triangle is 8 cm, and the base is 8 cm too, because you subtract to the base of the rectangle (16 cm) the measurements in the upper part (16 - 4 - 4 = 8), Now, we replace in the formula:

Area of a triangle = (8 cm * 8 cm) / 2
Area of a triangle = (64 cm^2) / 2
Area of a triangle = 32 cm^2

We subtract to the found area:

Area of the picture = 192 cm^2 - 32 cm^2
Area of the picture = 160 cm^2

To find the perimeter, you must add all the sides of the picture, but, as you can see, there is a side that doesn't have the measurent, this is the hypotenuse of the triangle used before, but how we know the other sides, we can use Pythagorean theorem: