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3 years ago

Should hypothesis the part of scientific inquiry

Mathematics

1 answer:

Vadim26 [7]3 years ago

6 0

It is an assumption of something not scientific.

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Can you please help me and also please be sure to show your work.

adoni [48]

Use Pythagorean theorem
/17 ^ 2 + x ^ 2 = 20 ^ 2
17 + x^2 = 400
x^2 = 383
x = 19.57

8 0

3 years ago

This is my last question

WARRIOR [948]

Answer:

the first one

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the surface area of the cone? Express your answer in terms of π.

Licemer1 [7]

Answer:

The surface area is about 408.47

Step-by-step explanation:

We must use Pythagorean Theorem in order to find the the height of the cone in order to use the surface area of a cone formula. The formula that you use would be A=πr(r+h2+r2). You already have the radius which is 5.

Pythagorean Theorem

a^2 + b^2 = c^2

You have two leg measurements of the triangle, one of which is the hypotenuse. The hypotenuse is 21.

a^2 + (5)^2 = (21)^2
a^2 +25 = 441

Subtract 25

a^2 = 416
a = 20.4

Take the square root of 416 and you get about 20.4. So the height of the cone would be about 20.4.

πr(r+h2+r2) ----> π(5)(5+(20.4)2+(5)2) and you get about 408.47

7 0

4 years ago

Consider the radical equation (3√6-x)+4=-8. Explain why the calculation in Problem 1 does not produce a solution to the equation

murzikaleks [220]

Answer:

See explanation below

Step-by-step explanation:

<u>First we will solve the radical equation</u> (which I guess was problem 1),

Let's start by simplifying it:

$3\sqrt{6-x}+4=-8\\ 3\sqrt{6-x}=-8-4\\ 3\sqrt{6-x}=-12\\\sqrt{6-x}=-4$

Now we will solve the equation by squaring both sides of the equation:

$\sqrt{6-x} =-4\\6-x=-4^{2} \\6-x=16\\-x=16-6\\-x=10\\x=-10$

So the calculation for x was that x = -10

However, this does not produce a solution to the equation: When we plug this value into the radical equation we get:

$3\sqrt{6-x} +4= -8\\3\sqrt{6-(-10)} +4=-8\\3\sqrt{16}+4 = -8\\ 3(4)+4 = -8\\12 + 4 = -8$

This happens because <u>when we first squared both sides of the equation in the first part of the problem we missed one value for x </u>(remember that all roots have 2 answers, a positive one and a negative one) while squares are always positive.

When we squared the root, we missed one value for x and that is why the calculation does not produce a solution to the equation.

6 0

3 years ago

Help please. I need assistance

madam [21]

The first translation is 3 units to the right giving in new equation
y = sqrt(x - 3)

then translation of 2 units vertically upwards makes
h(x) = sqrt(x - 3) + 2

6 0

3 years ago