All categories

3 years ago

15. Which expression can go in the blank to make the equation true?

Mathematics

1 answer:

topjm [15]3 years ago

6 0

The answer is A
-6.9+6.8 is equal to -0.1, and -5.4+5.5 is 0.1. If you add those two together, you get zero.

You might be interested in

Please help I'm stuck on this

Digiron [165]

8-13= -5 //hope this helped :)

3 0

4 years ago

Read 2 more answers

Can someone help me with this question

bija089 [108]

Answer:

what is the lesson? abot?

Step-by-step explanation:

8 0

2 years ago

Identify the range of the functions shown in the graph

miss Akunina [59]

Answer:

A. y >= 0

Step-by-step explanation:

The range of a function is the set of all values that the y-coordinate can have.

Look at the graph. Look at the red curve. The lowest y-coordinate is 0. As the curve goes up to the right, the y-coordinates are all real numbers greater than or equal to 0.

Answer: A. y >= 0

7 0

4 years ago

Read 2 more answers

Juan read 15 books in 9 months. How many books did he read in 24 months? And how many months did it take him to read 5 books?

larisa86 [58]

0.8 months stan loona

8 0

3 years ago

Read 2 more answers

Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B

telo118 [61]

In order to prove

$\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)$

Let's write both sides in terms of $\sin(x),\ \sin^2(x),\ \cos(x),\ \cos^2(x)$ only.

Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

$\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)$

and

$\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)$

So, their multiplication is

$\sin(x+y)\sin(x-y)=(\cos(y)\sin(x))^2-(\cos(x)\sin(y))^2\\=\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)$

So, the left hand side simplifies to

$\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now, on with the right hand side. We have

$1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

Now simply make this expression one fraction:

$1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}$

And as you can see, the two sides are equal.

6 0

4 years ago