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elena55 [62]
3 years ago
5

Anita has a cellphone contract that costs her R100 per month plus 85 cents per peak time SMS,and 25 cents per off-peak time SMS.

if she sends 45 SMSs during peak time and 105 SMSs during off-peak time In a month, what will her monthly bill come to?
Mathematics
1 answer:
algol [13]3 years ago
6 0

Answer:

Anita's monthly bill will be $164.25.

Step-by-step explanation:

Since Anita has a cellphone contract that costs her R100 per month plus 85 cents per peak time SMS, and 25 cents per off-peak time SMS, if she sends 45 SMSs during peak time and 105 SMSs during off-peak time In a month To determine what will her monthly bill come to, the following calculation must be performed:

100 + (0.85 x 45) + (0.25 x 105) = X

100 + 38.25 + 26.25 = X

138.25 + 26.25 = X

164.25 = X

Therefore, her monthly bill will be $ 164.25.

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Pls help with this question I don't know the answer. Don't put wrong answers pls.
Lapatulllka [165]

Answer:

-2 I believe. because the 1st exponent would make it 9, and then you'd remove 2

Step-by-step explanation:

I hope this is correct and helpful, I tried my best! <3

7 0
3 years ago
A rectangle has vertices at these coordinates.(3,6), (3,4), (6,4)What are the coordinates of the fourth vertex of the rectangle?
omeli [17]
Hi, i hope this helps! The coordinates are (6,6). I figured this out by plotting the given points and remembering that a rectangle has parallel sides and right angles :))

5 0
3 years ago
Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction
Oliga [24]

The base case is the claim that

\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}

which reduces to

\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}

By the induction hypothesis,

\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}

Now compare this to the upper bound we seek:

\dfrac{2k+1}{k+1}  > \dfrac{2k+2}{k+2}

because

(2k+1)(k+2) > (2k+2)(k+1)

in turn because

2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0

6 0
2 years ago
Read 2 more answers
Solve each equation 3x+2(5x-3)=7
Sloan [31]

Answer:

1

Step-by-step explanation:

Step 1:

3x + 2 ( 5x - 3 ) = 7

Step 2:

3x + 10x - 6 = 7

Step 3:

13x - 6 = 7

Step 4:

13x = 13

Answer:

x = 1

Hope This Helps :)

7 0
3 years ago
Based on the graphs of the equations y = –2x + 3 and y = x2 – x + 1, the solutions are located at points:
Ray Of Light [21]

The solutions of the equations are the point of intersection of the equations

The solutions are located at points (-2,7) and (1,1)

<h3>How to determine the solutions?</h3>

The equations are given as:

y = –2x + 3 and y = x2 – x + 1

Next, we plot the graph of both equations

From the graph of the equations (see attachment), we have the solution to be:

(x,y) = (-2,7) and (1,1)

Hence, the solutions are located at points (-2,7) and (1,1)

Read more about system of equations at:

brainly.com/question/14323743

8 0
2 years ago
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