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2 years ago

15

Order the following from greatest to least -0.4,0.8

1 answer:

JulijaS [17]2 years ago

3 0

Certainly negative number is smaller than a positive number

So the order is 0.8 , -0.4
Here 0.8 is the greatest while -0.4 is least

Hope This Helps You!

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The ratio of teachers to students in a school is 1:18.

sineoko [7]

Answer:

if there are 360 students, the number of teacher will be 6,480...

360:6,480

6 0

2 years ago

Read 2 more answers

How do you slove 3x+4y=8 in steps

gavmur [86]

<span><span><span>3x</span>+<span>4y</span></span>=8
</span>Add -4y to both sides
<span><span><span><span>3x</span>+<span>4y</span></span>+<span>−<span>4y</span></span></span>=<span>8+<span>−<span>4y</span></span></span></span><span><span>3x</span>=<span><span>−<span>4y</span></span>+8
</span></span>Then you divide both sides by 3
<span><span><span>3x/</span>3</span>=<span><span><span>−<span>4y</span></span>+8/</span>3</span></span><span>x=<span><span><span><span>−4/</span>3</span>y</span>+<span>8/3
</span></span></span>And your answer is ...
<span>x=<span><span><span><span>−4/</span>3</span>y</span>+<span>8/<span>3</span></span></span></span>

7 0

2 years ago

(cot^2x - 1)/(csc^2x) = cos2x

Alex787 [66]

Answer:

Step-by-step explanation:

The idea here is to get the left side simplified down so it is the same as the right side. Consequently, there are 3 identities for cos(2x):

$cos(2x)=cos^2x-sin^2x$ ,

$cos(2x)=1-2sin^2x$ , and

$cos(2x)=2cos^2x-1$

We begin by rewriting the left side in terms of sin and cos, since all the identities deal with sines and cosines and no cotangents or cosecants. Rewriting gives you:

$\frac{\frac{cos^2x}{sin^2x} -\frac{sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

Notice I also wrote the 1 in terms of sin^2(x).

Now we will put the numerator of the bigger fraction over the common denominator:

$\frac{\frac{cos^2x-sin^2x}{sin^2x} }{\frac{1}{sin^2x} }$

The rule is bring up the lower fraction and flip it to multiply, so that will give us:

$\frac{cos^2x-sin^2x}{sin^2x} *\frac{sin^2x}{1}$

And canceling out the sin^2 x leaves us with just

$cos^2x-sin^2x$ which is one of our identities.

5 0

2 years ago

Can someone help me, please ?? im stuck

Tresset [83]

Part A

5x - 5 >= 10 or-3x + 1 > 13

1) This is a disjunction (OR)

5x - 5 >= 10

5x >= 15

x >= 3

or

-3x + 1 > 13

-3x > 12

x < -4

Answer

x <- 4 or x >= 3

Part C:

1) This is a conjunction (And)

5x+ 3 <= 18 and 4 - x < 6

5x <= 15 and - x < 2

x <= 3 and x > -2

Answer

-2 < x <= 3

8 0

3 years ago

Find the area of △LKJ.

nirvana33 [79]

Answer:

$A=600\: u^{2}$

<u>The correct answer is letter c.</u>

Step-by-step explanation:

The area equation of a triangle is given by:

$A=\frac{bh}{2}$ (1)

Where:

b is the base

h is the height

Let's find first the angle MJK (α), using the tangent definition.

$tan(\alpha)=\frac{24}{32}$

$\alpha=tan^{-1}(\frac{24}{32})$

$\alpha=36.9^{\circ}$

Now, if α = 37° then β (angle JLK) must be:

$37^{\circ} + \beta + 90^{\circ}=180^{\circ}$

$\beta=53^{\circ}$

Now, we can find the distance ML using again the tangent definition:

$tan(\beta)=\frac{24}{ML}$

$ML=\frac{24}{tan(53)}$

$ML=18$

The base of the triangle LKJ will be:

$b=32+ML=32+18=50$

Therefore, using equation (1) the area of △LKJ will be:

$A=\frac{50*24}{2}$

$A=600\: u^{2}$

The correct answer is c.

I hope it helps you!

5 0

2 years ago