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wariber [46]
3 years ago
12

Plssssssssssss help asap !

Mathematics
2 answers:
7nadin3 [17]3 years ago
6 0

Answer:

B one solution to the problem

AveGali [126]3 years ago
6 0

Answer:

B: one solution

Step-by-step explanation:

yas!

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I need help on this please SHOW WORK
elena-s [515]

\boxed{\large{\bold{\blue{ANSWER~:) }}}}

See this attachment

7 0
4 years ago
Write g(x) = 4x2 + 88x in vertex form. The function written in vertex form is g(x) = ___ (x +11)2 +____.
wel

The\ vertex\ form:\\\\f(x)=a(x-h)^2+k

\text{Use}\ (a+b)^2=a^2+2ab+b^2\qquad(*)

g(x)=4x^2+88x=4(x^2+22x)=4(x^2+2\cdot x\cdot11)\\\\=4(\underbrace{x^2+2\cdot x\cdot11+11^2}_{(*)}-11^2)=4[(x+11)^2-121]=4(x+11)^2+(4)(-121)=4(x+11)^2-484

5 0
3 years ago
Read 2 more answers
The sum of the first number cubed and the second number is 500 and the product is a maximum.
soldier1979 [14.2K]

We are given the following information:

the sum of the first number cubed and the second number is 500

their product is a maximum

We are looking for the 2 missing numbers.

To answer this, let's represent the two numbers as x and w.

From the given, we can form the following equation:

x^3+w=500

We can then express y as:

w=500-x^3

We can express their product as:

x(500-x^3)=500x-x^4

To find the maximum value of x, let's solve for the derivative of -x^4 + 500 x.

\begin{gathered} f(x)=-x^4+500x \\ f^{\prime}(x)=-4x^3+500 \end{gathered}

Then we solve for the value of x where f'(x) = 0.

\begin{gathered} -4x^3+500=0 \\ -4x^3=-500 \\ x^3=125 \\ x=5 \end{gathered}

Then we use x = 5 to solve for the second number, w.

\begin{gathered} w=500-x^3 \\ w=500-5^3 \\ w=500-125 \\ w=375 \end{gathered}

Therefore, the two numbers are 5 and 375.

3 0
1 year ago
HELP PLS, IT’S FOR A TEST!
solniwko [45]

Answer:

Its scale/actual, good luck on the test

4 0
4 years ago
PLS HELP ME WILL GIVE BRAIN THING
inysia [295]

Answer:

quadrant III (3)

Step-by-step explanation:

If you graph the point it is located in quadrant III

5 0
3 years ago
Read 2 more answers
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