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3 years ago

WHICH IS WARMER -9 OR -12 DEGRESS

Mathematics

2 answers:

sertanlavr [38]3 years ago

4 0

Answer:

-9 Degrees is warmer because that number is bigger than -12.

Step-by-step explanation:

frosja888 [35]3 years ago

4 0

Answer:

-9 is warmer

Step-by-step explanation:

Since this negative we want to get closer to zero to be warmer

-12 means you owe 12

-9 means you owe 9

The more you owe means is further away from zero

-12 < -9 <0

You might be interested in

If the radius is 14cmand The perimeter of the sector is 57.32cm. what is the size of the angle?

Degger [83]

Answer:

120°

Step-by-step explanation:

A sector of a circle is the portion or region of a circle enclosed by two radii and an arc. The perimeter of a the sector of a circle is given by the formula:

Perimeter of sector = $\frac{\theta}{360} *2\pi r$ + 2r

Where θ is the angle which forms the sector and r is the radius of the circle.

Given that Perimeter of sector = 57.32 cm, radius (r) = 14 cm, we can find the angle θ using:

Perimeter of sector = $\frac{\theta}{360} *2\pi r$ + 2r

$57.32=\frac{\theta}{360} *2\pi *14+2(14)\\\\57.32-28=\frac{\theta}{360} *2\pi *14\\\\29.32=\frac{\theta}{360} *2\pi *14\\\\ \frac{\theta}{360}=\frac{29.32}{2\pi *14} \\\\\frac{\theta}{360}=0.33\\\\\theta=120^o$

7 0

3 years ago

Prove that:

Lorico [155]

A.)

$\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)$

Use the identities $\csc x = 1 / \sin x$ and $\tan x = \sin x / \cos x$ on the left-hand side

$\begin{aligned}
\text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\
&= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\
&= \frac{1}{\cos^2 (x)} - 1
\end{aligned}$

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1 - \cos^2 x}{\cos^2 (x)}
\end{aligned}$

Use Pythagorean identity for the numerator.

If $\sin^2 (x) + \cos^2(x) = 1$ then subtracting both sides by $\cos^2 (x)$ yields $\sin^2(x) = 1 - \cos^2(x)$ . We can substitute that into the numerator

$\begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\
&= \frac{\sin^2 (x)}{\cos^2 (x)} \\
&= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\
&= \text{RHS}
\end{aligned}$

======

b.)

$\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1$

For the left-hand side:
By definition, $\sec(x) = 1/\cos(x)$ and $\tan (x) = 1/\cot (x)$

$\begin{aligned}
\text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} \\
&= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} 
\end{aligned}$

Since $\cot (x) = \cos (x) / \sin (x)$

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}$

Using Pythagorean identity, $\cos^2(x) = 1 - \sin^2(x)$ so

$\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}$

6 0

3 years ago

8. A number is of another number. The difference of the numbers is 18. (Assume that you are subtracting the

bezimeni [28]

Answer:

Step-by-step explanation:

"A number is [WHAT?] of another number." It appears the WHAT?? is missing. If it is a word such as "twice," it can be solved in this fashion.

Let the numbers be X and Y, with Y the larger number.

A number is twice another number: Y = 2X

The difference of the numbers is 18: Y - X = 18

Use the first equation (Y = 2X) in the second by replacing "Y" with (2X):

Y - X = 18

2X - X = 18

X = 18

Y = 2X, so Y = 36

5 0

3 years ago

Which best describes a standard thermometer...

Katen [24]

Answer:

inteager

Step-by-step explanation:

temperature can express both positive and negative

6 0

3 years ago

The perimeter of an equilateral triangle is 12\/3 cm. Find the radius, apothem and area of the equilateral triangle.

Irina-Kira [14]

Answer:

Radius - 4

Apothem - 2

Area - $12\sqrt{3}$ or 20.8

Step-by-step explanation:

Try to draw out my explanation so you know what this looks like.

~~~~

This is an equilateral triangle, so all sides are the same length. The perimeter can be divided by 3 to get each side length...

$12\sqrt{3} \div3 = 4\sqrt{3}$

Now that we know the side lengths, we can get this started!

~~~~

Now, an equilateral triangle has angles that all equal 60 degrees. We can bisect this into TWO special case triangles of measures 90-60-30.

4 times the square root of 3 will be the hypotenuse, and the smallest leg is always half of that. The larger leg is represented as the small leg times the square root of 3.

$Small=2\sqrt{3} \\Hypotenuse = 4\sqrt{3} \\Long =6$

By the way, the formula for the area can be either of the two:

$A= \frac{1}{2} bh\\or\\A = \frac{1}{2} NAS$

we can easily find the area using the first formula.

$A= \frac{1}{2} (4\sqrt{3} )(6)\\A=12\sqrt{3}$

The radius is the distance from the center to the corners & the apothem is the distance from the center to a side...

So we can divide the big triangle into a mini triangle at the bottom left/right

The height (small leg) of that triangle would be the apothem

The hypotenuse of that triangle would be the radius.

I did the math really quick because this is getting long. Anyhow, the small leg is 2 so is the apothem, and the hypotenus is 4, so is the radius

Happy April Fool's btw, lol.

4 0

3 years ago