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RideAnS [48]
3 years ago
11

Which expression below has the same value as 9 to the power of 6

Mathematics
2 answers:
Kazeer [188]3 years ago
8 0
9^2 = 9 x 9

Therefore:
9^6 = 9 x 9 x 9 x 9 x 9 x 9
dimulka [17.4K]3 years ago
5 0
The answer is c. 9x9x9x9x9x9
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A kite Has four equal sides. If 5 inches is added to each side, the new perimeter will be 56 inches. Find a side length of the o
Zina [86]

Answer:

36

Step-by-step explanation:

7 0
3 years ago
Line w is perpendicular to the line y = −x − 5 and passes through the point (−2, 4). What is the y-intercept of line w?
Aleksandr [31]

The slope of a line perpendicular to another is the negative reciprocal of the original.  The negative reciprocal of -1 is 1.  The equation will read:

y=x+b

To solve for the y-intercept (b), plug in the given values of x and y:

4=-2+b

<em>*Add 2 to both sides*</em>

D. 6


Hope this helps!!

7 0
3 years ago
How many fixed-requirement constraints does a transportation problem with 5 factories and 6 customers have?a. 30.b. 11.c. 6.d. 5
bulgar [2K]

Answer:

Option B

Step-by-step explanation:

From the question we are told that:

Demand point m=5

Supply Point n=6

Generally the equation for fixed-requirement constraints is mathematically given by

 X=m+n

 X=5+6

 X=11

Therefore the correct option is

Option B

7 0
3 years ago
Solve (x + 4)2 – 3(x + 4) – 3 = 0 using substitution. u = Select the solution(s) of the original equation.
bezimeni [28]

Answer:

\dfrac{-5-\sqrt{21}}{2},\ \dfrac{-5+\sqrt{21}}{2}

Step-by-step explanation:

For the equation (x+4)^2-3(x+4)-3=0 use the substitution u=x+4. Then the equation will take look

u^2-3u-3=0.

Solve this quadratic equation:

D=(-3)^2-4\cdot 1\cdot (-3)=9+12=21,\\ \\u_{1,2}=\dfrac{-(-3)\pm \sqrt{21}}{2}=\dfrac{3\pm\sqrt{21}}{2}.

Thus,

x+4=\dfrac{3-\sqrt{21}}{2}\text{ or }x+4=\dfrac{3+\sqrt{21}}{2},\\ \\x_1=\dfrac{3-\sqrt{21}}{2}-4=\dfrac{-5-\sqrt{21}}{2}\text{ or }x_2=\dfrac{3+\sqrt{21}}{2}-4=\dfrac{-5+\sqrt{21}}{2}.

8 0
3 years ago
Read 2 more answers
Identify the monomial function(s) that have a maximum.
natima [27]

Answer:

Step-by-step explanation:

For a function f to have a maximum as per derivative rule we have to have

f'(x) =0, f"(x) <0

If second derivative =0 also then it is not maximum but point of inflections

Whenever f(x) = ax^n

we have

f'(x) = 0 gives x=0 and

f"(x) = n(n-1) ax ^(n-2)

So for n greater than or equal to there cannot be any maximum

And also for a straight line

y =-4x

y'=-4 and y"-0

No maximum

So only maximum can be for a funciton of the form y = ax^2

Here we do not have that all degrees are either 1 or greater than 1.

So no maximum for any funciton.

4 0
3 years ago
Read 2 more answers
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