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Fynjy0 [20]
3 years ago
8

Help Me Please.............(3)

Mathematics
1 answer:
Rom4ik [11]3 years ago
3 0

Answer:

0

Step-by-step explanation:

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Amy has $1,000 in a savings account at the beginning of the fall. She wants to have at least $500 in the account by the end of t
LuckyWell [14K]

Answer:

D. 1000 - 100w \geq 500; w \leq 5

Step-by-step explanation:

Given:

Initial amount in the bank = $1000

Money withdrawn each week = $100

Final amount should be at least $500.

Now, let the number of weeks the money is withdrawn be 'w'.

Therefore,

Money withdrawn in 'w' weeks = \textrm{Money withdrawn each week}\times w

Total Money withdrawn in 'w' weeks = 100w

Now, final amount after 'w' weeks is equal to the difference between initial amount and total withdrawal amount. Therefore,

Final amount = Initial amount - Total withdrawal amount

Final amount = 1000 - 100w

Now, final amount must be greater than or equal to $500. So,

\textrm{Final amount}\geq500\\\\1000-100w\geq500

Therefore, the  inequality that represents the inequality for the number of weeks Amy can withdraw money is:

1000-100w\geq500

Now, let us solve for 'w'.

Adding -500 and 100w both sides, we get:

1000-500-100w+100w\geq500-500+100w\\\\500\geq100w\\\\\textrm{The above inequality is reversed when taking 100w on the left side}\\\\100w\leq500\\\\w\leq\frac{500}{100}\\\\\therefore w\leq5

Therefore, the correct option is (D).

7 0
3 years ago
I think its C. but Im not sure how to determine the answer, however I know that using the two points I picked... (0, 2) & (-
9966 [12]
It is B.

You see that the line segment crosses the y-axis at 1? That is known as the y intercept.

Remember y = mx + b?

Well in this graph, b = 1
Therefore it is B
3 0
4 years ago
Read 2 more answers
HELLPP. ITS URGENT.50 PTS
ICE Princess25 [194]

Answer:

  • 4x² - 13x + 8 = 0
  • 4x² - 11x + 5 = 0
  • 16x² - 41x + 1 = 0
  • x² + 5x + 4 = 0
  • x² - 66x + 64 = 0

Step-by-step explanation:

<u>Given</u>

  • α and β are roots of 4x²-5x-1=0

<u>Then the sum and product of the roots are:</u>

  • α+b = -(-5)/4 = 5/4
  • αβ = -1/4

(i) <u>Roots are α + 1 and β + 1, then we have:</u>

  • (x - (α + 1))(x - (β + 1)) = 0
  • (x - α - 1)(x - β - 1) = 0
  • x² - (α+β+2)x + α+β+ αβ + 1 = 0
  • x² - (5/4+2)x +5/4 - 1/4 + 1 = 0
  • x² - 13/4x + 2= 0
  • 4x² - 13x + 8 = 0

(ii) <u>Roots are 2 - α and 2 - β, then we have:</u>

  • (x + α - 2)(x + β - 2) = 0
  • x² + (a + β - 4)x - 2(α + β) + αβ + 4 = 0
  • x² + (5/4 - 4)x - 2(5/4) - 1/4 + 4 = 0
  • x² - 11/4x - 10/4 - 1/4 + 16/4 = 0
  • x² - 11/4x + 5/4x = 0
  • 4x² - 11x + 5 = 0

(iii) <u>Roots are α² and β², then:</u>

  • (x - α²)(x-β²) = 0
  • x² -(α²+β²)x + (αβ)² = 0
  • x² - ((α+β)² - 2αβ)x + (-1/4)² = 0
  • x² - ((5/4)² -2(-1/4))x + 1/16 = 0
  • x² - ( 25/16 + 1/2)x + 1/16 = 0
  • x² - 33/16x + 1/16 = 0
  • 16x² - 33x + 1 = 0

(iv) <u>Roots are 1/α and 1/β, then:</u>

  • (x - 1/α)(x - 1/β) = 0
  • x² - (1/α+1/β)x + 1/αβ = 0
  • x² - ((α+β)/αβ)x + 1/αβ = 0
  • x² - (5/4)/(-1/4)x - 1/(-1/4) = 0
  • x² + 5x + 4 = 0

(v) <u>Roots are 2/α² and 2/β², then:</u>

  • (x - 2/α²)(x - 2/β²) = 0
  • x² - (2/α² + 2/β²)x + 4/(αβ)² = 0
  • x² - 2((α+β)² - 2αβ)/(αβ)²)x + 4/(αβ)² = 0
  • x² - 2((5/4)² - 2(-1/4))/(-1/4)²x + 4/(-1/4)² = 0
  • x² - 2(25/16 + 8/16)/(1/16)x + 4(16) = 0
  • x² - 2(33)x + 64 = 0
  • x² - 66x + 64 = 0

3 0
3 years ago
Read 2 more answers
Solve for<br> 9(x + 1) = 25+x
den301095 [7]

Answer:

2

Step-by-step explanation:

9(x+1)=25+x

9x+9=25+x

9x-x=25-9

8x=16

x=16/8

x=2

4 0
3 years ago
Read 2 more answers
Which statements describe a residual plot for a line of best fit that is a good model for a scatterplot? Check all that apply. T
lidiya [134]

Answer:

The points are randomly scattered with no clear pattern

The number of points is equal to those in the scatterplot.

Step-by-step explanation:

The points in the residual plot of the line of best fit that is a good model for a scatterplot are randomly scattered with no clear pattern (like a line or a curve).

The number of points in the residual plot is always equal to those in the scatterplot.

It doesn't matter if there are about the same number of points above the x-axis as below it, in the residual plot.

The y-coordinates of the points are not the same as the points in the scatterplot.

8 0
3 years ago
Read 2 more answers
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