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xxTIMURxx [149]
3 years ago
14

Connor just switched to a new phone plan. His old phone plan was $10 plus 10 cents per gigabyte of data overage. His new plan is

$12 plus 8 cents per gigabyte of data overage. Connor claims that he is going to pay less on his new plan because he always goes over his data limit. His data overage was 105 gigabytes. Is Connor's claim true?
A- Yes, because 0.10(105) > 0.08(105).
B- Yes, because 10 + 0.10(105) > 12 + 0.08(105).
C- No, because 10 + 0.10(105) < 12 + 0.08(105).
D- No, because 0.10(105) < 0.08(105).
Mathematics
1 answer:
Archy [21]3 years ago
3 0
First, let's write an expression for both plans. We will use x for the data overage in gigabytes.

Old plan: 10 + 0.10x

New plan: 12 + 0.08x

Now let's plug in 105 to the expressions to determine the costs of the plans.

Old plan: 10 + 0.10(105) = $20.50

New plan: 12 + 0.08(105) = $20.40

So yes, Connor was correct because when we compare his plans, the new one costs 10 cents less.

The answer is B.


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melomori [17]

Answer:

The quadratic polynomial with integer coefficients is y = 81\cdot x^{2}-36\cdot x -19.

Step-by-step explanation:

Statement is incorrectly written. Correct form is described below:

<em>Find a quadratic polynomial with integer coefficients which has the following real zeros: </em>x = \frac{2}{9}\pm \frac{\sqrt{23}}{9}<em>. </em>

Let be r_{1} = \frac{2}{9}+\frac{\sqrt{23}}{9} and r_{2} = \frac{2}{9}-\frac{\sqrt{23}}{9} roots of the quadratic function. By Algebra we know that:

y = (x-r_{1})\cdot (x-r_{2}) = x^{2}-(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} (1)

Then, the quadratic polynomial is:

y = x^{2}-\frac{4}{9}\cdot x -\frac{19}{81}

y = 81\cdot x^{2}-36\cdot x -19

The quadratic polynomial with integer coefficients is y = 81\cdot x^{2}-36\cdot x -19.

5 0
3 years ago
Find the values for c and d
baherus [9]

Answer:

c=-5

d=1

Step-by-step explanation:

(cy^2)(4y^d)=-20y^3

I'm going to reorder the left-hand side.  Multiplication is commutative.

(4c)(y^2y^d)=-20y^3

Since the bases are the same in y^2y^d and the operation is multiplication, I'm going to add the exponents giving me:

4cy^{2+d}=-20y^3

So this implies we have two equations to solve:

4c=-20 and 2+d=3

So the first equation can be solved by dividing both sides by 4 giving you c=-5.

The second equation can be solved by subtracting 2 on both sides giving you d=1.

6 0
3 years ago
What is the solution to 3/4b=8
soldier1979 [14.2K]

Answer:

b=\frac{32}{3}

Step-by-step explanation:

\frac{3}{4}b=8

Multiply both sides by 4:

4\times \frac{3}{4}b=8\times4

3b=32

Divide both sides by 3:

\frac{3b}{3}=\frac{32}{3}

b=\frac{32}{3}

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3 years ago
Help<br><br><br><br><br><br><br><br><br><br> Sjsjjjsjajajahjajaja
motikmotik

Answer:

hi

Step-by-step explanation:

{5}^{2}  +  {x}^{2}  =  {7}^{2}  \\ 25 +  {x}^{2}  = 49 \\  {x}^{2}  = 49 - 25 \\  {x}^{2}  = 24 \\ x =  \sqrt{24}

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eduard

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Step-by-step explanation:

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