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antiseptic1488 [7]
4 years ago
13

Which inequality is true? Use the number line to help. A number line going from negative 2.5 to positive 2.5 in increments of 0.

5. Negative 1.5 greater-than 0.5 Negative 0.5 greater-than 0 Negative 1.5 less-than negative 0.5 One-half greater-than 0.5
Mathematics
2 answers:
Ede4ka [16]4 years ago
7 0

Answer:

C

Step-by-step explanation:

I got it right on edge

alexgriva [62]4 years ago
3 0

Given:

A number line going from negative 2.5 to positive 2.5 in increments of 0.5.

To find:

The inequality which is true.

Solution:

Consider the number line as shown in the below figure.

We know that, on a line as we go to the left direction, the number become smaller and as we go right, the numbers become larger.

From the number line it is clear that -1.5 is on the left side of 0.5. So,

-1.5

Hence, option A is incorrect.

-0.5 is on the left side of 0. So,

-0.5

Hence, option B is incorrect.

-1.5 is on the left side of -0.5. So,

-1.5

Hence, option C is correct.

One half is equal to 0.5.

\dfrac{1}{2}=0.5

Hence, option D is incorrect.

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Prove Sin3theta = 3sintheta - 4sin^3theta
trapecia [35]

Express the left hand side as

sin3theta=sin(theta+2theta)

now expand the right side of this equation using color(blue)"Addition formula"

color(red)(|bar(ul(color(white)(a/a)color(black)(sin(A±B)=sinAcosB±cosAsinB)color(white)(a/a)|)))

rArrsin(theta+2theta)=sinthetacos2theta+costhetasin2theta.......(A)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos2theta=cos^2theta-sin^2theta=2cos^2theta-1=1-2sin^2theta)color(white)(a/a)|)))

The right hand side is expressed only in terms of sintheta's

so we use cos2theta=1-2sin^2theta........(1)

color(red)(|bar(ul(color(white)(a/a)color(black)(sin2theta=2sinthetacostheta)color(white)(a/a)|)))........(2)

Replace cos2theta" and " sin2theta by the expansions (1) and (2)
into (A)

sin(theta+2theta)=sinthetacolor(red)((1-2sin^2theta))+costhetacolor(red)((2sinthetacostheta)

and expanding brackets gives.

sin(theta+2theta)=sintheta-2sin^3theta+2sinthetacos^2theta....(B)

color(red)(|bar(ul(color(white)(a/a)color(black)(cos^2theta+sin^2theta=1rArrcos^2theta=1-sin^2theta)color(white)(a/a)|)))

Replace cos^2theta=1-sin^2theta" into (B)"

rArrsin(theta+2theta)=sintheta-2sin^3theta+2sintheta(1-sin^2theta)

and expanding 2nd bracket gives.

sin(theta+2sintheta)=sintheta-2sin^3theta+2sintheta-2sin^3theta

Finally, collecting like terms.

sin3theta=3sintheta-4sin^3theta="R.H.S hence proven"
3 0
3 years ago
In the figure below, KL=9 and JK=4. Find JL.
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Answer:

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Explanation:

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JL → 4 + 9

JL → 13

8 0
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