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4 years ago

14

Please help!! Show work please!!

2 answers:

wel4 years ago

3 0

Answer:

x^3

-------------

16 y^5 z

Step-by-step explanation:

4^-2 x^3 y^-5 z^-1

negative exponents flip

x^3 x^3

------------- = -------------

4^2 y^5 z^1 16 y^5 z

Nonamiya [84]4 years ago

3 0

$4^{-2}x^3y^{-5}z^{-1}=\dfrac{1}{4^2}\cdot x^3\cdot\dfrac{1}{y^5}\cdot\dfrac{1}{z^1}=\boxed{\dfrac{x^3}{16y^5z}}\\\\Used:\\\\a^{-n}=\dfrac{1}{a^n}$

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Preform the indicated operation 15 - (-11)

xenn [34]

Answer:It would be: 15 - (-11)

= 15 + 11

= 26

Step-by-step explanation:

5 0

4 years ago

PLEASE HELP ILL MARK BRAINLIEST !!!

Alexus [3.1K]

Answer:

a) the common difference is 20

b) $x_8=115 , x_{12}=195$

c) the common difference is -13

d) $a_{12}=52, a_{15}=13$

Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

36-16 =20

So, the common difference is 20

b) what is x_8? what is x_12

The formula used is: $x_n=x_1+(n-1)d$

We know common difference d= 20, we need to find $x_1$

Using $x_3=16$ we can find $x_1$

$x_n=x_1+(n-1)d\\x_3=x_1+(3-1)d\\15=x_1+2(20)\\15=x_1+40\\x_1=15-40\\x_1=-25$

So, We have $x_1 = -25$

Now finding $x_8$

$x_n=x_1+(n-1)d\\x_8=x_1+(8-1)d\\x_8=-25+7(20)\\x_8=-25+140\\x_8=115$

So, $\mathbf{x_8=115}$

Now finding $x_{12}$

$x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195$

So, $\mathbf{x_{12}=195}$

c) what is the common difference of the sequence $a_m$

Looking at the table, we get a_7=104, a_8=91 and a_9= 78

Deterring the common difference by subtracting a_7 from a_8 we get

91-104 =-13

So, the common difference is -13

d) what is a_12? what is a_15?

The formula used is: $a_n=a_1+(n-1)d$

We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

Now finding $a_{12}$ , put n=12

$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

So, $\mathbf{a_{12}=52}$

Now finding $a_{15}$ , put n=15

$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

So, $\mathbf{a_{15}=13}$

5 0

3 years ago

Factor completely 3x - 12. (1 point)

yarga [219]

Answer:

3(x - 4)

Step-by-step explanation:

Given

3x - 12 ← factor out 3 from each term

= 3(x - 4)

3 0

3 years ago

75/12=□/4=160/□what is the answer?<br><br><br><br><br>

Vlada [557]

Answer:

75/12= 6.25/4=160/102.4

4 0

3 years ago

Triangle TRE has vertices T(3,6), R(-3,10), and E(-9,4). Find the coordinates of point M if line TM is a median of triangle TRE

Sergeeva-Olga [200]

Answer:

The coordinate of point M = (-6,7)

Explanation:

The Median of a triangle is a line segment from a vertex to the midpoint of the opposite side of a triangle.

Given: $\triangle TRE$ has vertices T(3,6) , R(-3,10) and E(-9,4).

Here, line TM is a median of triangle TRE where M is the midpoint of RE.

The midpoint of M of the line segment from R(-3,10) to E(-9,4) is;

M = $(\frac{-3+(-9)}{2}, \frac{10+4}{2}) = ( \frac{-12}{2}, \frac{14}{2} )=(-6,7)$

Therefore, the coordinate of point M is, (-6,7).

7 0

3 years ago