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3 years ago

Which statement is true?

Mathematics

2 answers:

Basile [38]3 years ago

7 0

Answer:

The mode is equal to the minimum

Step-by-step explanation:

Find out the mean, median, mode and minimum

minimum = 48

mean = 50.4

mode = 48

median = 50

patriot [66]3 years ago

6 0

Answer:The mode is equal to the minimum

Step-by-step explanation:

Find out the mean, median, mode and minimum

minimum = 48

mean = 50.4

mode = 48

median = 50

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(35 points) GEOMETRY PROOF

Lisa [10]

Please give me brainlest

click this link.

https://artofproblemsolving.com/wiki/index.php/2019_AMC_8_Problems/Problem_24

or copy it then paste it in search.

5 0

3 years ago

Solve.<br><br> 3 = –(–y + 6)

o-na [289]

3=−(−y+6)

Simplify:

3=−(−y+6)

3=y+−6(Distribute)

3=y−6

Flip the equation.

y−6=3

Add 6 to both sides.

y−6+6=3+6

y=9

6 0

2 years ago

Will mark brainliest for whoever answers

Ostrovityanka [42]

Answer:

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

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

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

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

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

2 years ago

Help please!!!!!!!!!!!!!!!!!!!!!!!!!!!!

MatroZZZ [7]

9514 1404 393

Answer:

second step is wrong; x = 5

Step-by-step explanation:

The correct solution is ...

3(x -3) +9 = 15

3x -9 +9 = 15 . . . . . . . not 3x -3

3x = 15

x = 5

_____

Jeremy did what a lot of students do -- failed to apply the outside factor to <em>all</em> of the terms in parentheses.

6 0

2 years ago