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elixir [45]
3 years ago
9

A player earns 100 for each question answered correctly

Mathematics
2 answers:
garik1379 [7]3 years ago
7 0

Answer:

I don't understand what you're asking haha. Maybe you typed it wrong and didn't realise but your making a statement not a question .

Step-by-step explanation:

Readme [11.4K]3 years ago
6 0

We need more info to answer the question.

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Simplify using the distributive property. 6(y-8) =
Molodets [167]

6y- 48

hope that helps

6 0
3 years ago
Read 2 more answers
Find m A. 40°<br> B. 80°<br> C. 130°<br> D. 260°<br><br> Please hurry!!
PolarNik [594]

Answer:

C. 130

Step-by-step explanation:

Since the sum of the interior angles for a quadrilateral shape is 360 degrees, you take 55 and 45 away from 360 to get 260. You need to then divide that number by 2 to get Q and S, but you are looking for Q in this case, so the answer is 130.

3 0
4 years ago
Please helpp ASAP 2 5 pts + brainliest for best/right answer
dimulka [17.4K]

Answer:

The vertex is at (-3/2, 33/4)

x=-3/2

Step-by-step explanation:

y= -x^2 -3x+6

Factor out a negative from the first two terms

y = -(x^2 +3x)  +6

Complete the square

Take 3 /2  and square it  (3/2)^2  = 9/4

The - on the outside of the parentheses means we really are subtracting it so we need to add it on the outside

y = -(x^2 + 3x + 9/4) +9/4 +6

  = -(x+3/2)^2 + 9/4 +6

Get a common denominator of 4

  = -(x+3/2)^2 + 9/4 +6*4/4

y =- (x+3/2)^2 + 9/4 +24/4

y = -(x+3/2)^2 + 33/4

y = a(x-h)^2 +k  where (h,k) is the vertex

y = -(x - -3/2)^2 + 33/4

The vertex is at (-3/2, 33/4)

The axis of symmetry is at the x coordinate of the vertex

x=-3/2

7 0
3 years ago
Givea)Possible number of positive real rootsb)Possible number of negative real rootsc)Possible rational roolsd)Find the roots
CaHeK987 [17]

The function is given to be:

x^3-2x^2-3x+6

QUESTION A

We can use Descartes' Rule of Signs to check the positive real roots of a polynomial.

The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number.

If we have:

f(x)=x^3-2x^2-3x+6

The coefficients are: +1, -2, -3, +6.

We can see that there are only 2 sign changes; from the first to the second term, and from the third to the fourth term.

Therefore, there are 2 or 0 positive real roots.

QUESTION B

To find the number of negative real roots, evaluate f(-x) and check for sign changes:

\begin{gathered} f(-x)=(-x)^3-2(-x)^2-3(-x)+6 \\ f(-x)=-x-2x^2+3x+6 \end{gathered}

The coefficients are: -1, -2, +3, +6.

We can see that there is only one sign change; from the second term to the third term.

Therefore, there is 1 negative real root.

QUESTION C

To check the possible rational roots, we can use the Rational Root Theorem since all the coefficients are integers.

The Rational Root Theorem states that if the polynomial:

P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0

has any rational roots, they must be in the form:

\Rightarrow\pm\mleft\lbrace\frac{factors\text{ of }a_0}{factors\text{ of }a_n}\mright\rbrace

From the polynomial, the trailing coefficient is 6:

a_o=6

Factors of 6:

\Rightarrow\pm1,\pm2,\pm3,\pm6

The leading coefficient is 1:

a_n=1

Factors of 1:

\Rightarrow\pm1_{}

Write in the form

\Rightarrow\mleft\lbrace\frac{a_o}{a_n}\mright\rbrace

Therefore,

\Rightarrow\pm(\frac{1}{1}),\pm(\frac{2}{1}),\pm(\frac{3}{1}),\pm(\frac{6}{1})

Therefore, the possible rational roots are:

\Rightarrow\pm1,\pm2,\pm3,\pm6

QUESTION D

We can use a graph to check the roots of the polynomial. The graph is shown below:

The roots of the polynomial refer to the points when the graph intersects the x-axis.

Therefore, the roots of the polynomial are:

x=-1.732,x=1.732,x=2

7 0
2 years ago
The domain of y=cot x is given by x=npi
allsm [11]
True - apex just did the quiz
4 0
4 years ago
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