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Aleksandr-060686 [28]
2 years ago
11

Budgeting for home maintenance early can save money in the long run. Why save early compared to later, especially if the home is

new?
a.
Home maintenance is unpredictable, but it is important to do it promptly, or it will result in home damage that requires repair.
b.
Homebuilders are often untrustworthy and do low quality work, requiring home owners to fix or finish a lot of things on their own.
c.
Homeowners often run into problems soon after purchasing a brand new house.
d.
Homeowners always have a lot of money after moving into a house they purchased, so they save it for the future.
Mathematics
2 answers:
Kruka [31]2 years ago
4 0
Its A. all the other answers are unreasonable

Korolek [52]2 years ago
3 0

Answer:

A. Home maintenance is unpredictable, but it is important to do it promptly, or it will result in home damage that requires repair.  

Step-by-step explanation:

We are given that, 'Budgeting for home maintenance early can save money in the long run'.

Now, it is necessary as home maintenance is unpredictable and requires repair during any damage. So, it is important to budget home maintenance at the start.

So, option A is correct.

Looking at the other options, we see that, all other options does not provide strong reasoning in order to budget the home maintenance early.

Since, not all home builders do low quality work which may need repair further in life.

Also, the statement that home owners run into problems soon after purchasing a new house, is not universally true.

Further, not all home owners are rich in order to save it for the future.

Hence, option A gives the correct reasoning as to why early budgeting is necessary.

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Step-by-step explanation:

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Which statements about the local maximums and minimums for the given function are true? Choose three options.
Nostrana [21]

Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.  

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the  local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum  is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

8 0
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I don't understand what integer is being asked for. The question is poorly worded.

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calculeaza lungimea segmentului ab in fiecare dintre cazuri:A(1,5);B(4,5);A(2,-5),B(2,7);A(3,1)B(-1,4);A(-2,-5)B(3,7);A(5,4);B(-
Tatiana [17]

Answer:

1. 3; 2. 12; 3. 5; 4. 13; 5. 10; 6. 10

Step-by-step explanation:

We can use the distance formula to calculate the lengths of the line segments.

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

1. A (1,5), B (4,5) (red)

d = \sqrt{(x_{2} - x_{1}^{2}) + (y_{2} - y_{1})^{2}} = \sqrt{(4 - 1)^{2} + (5 - 5)^{2}}\\= \sqrt{3^{2} + 0^{2}} = \sqrt{9 + 0} = \sqrt{9} = \mathbf{3}

2. A (2,-5), B (2,7) (blue)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(2 - 2)^{2} + (7 - (-5))^{2}}\\= \sqrt{0^{2} + 12^{2}} = \sqrt{0 + 144} = \sqrt{144} = \mathbf{12}

3. A (3,1), B (-1,4 ) (green)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-1 - 3)^{2} + (4 - 1)^{2}}\\= \sqrt{(-4)^{2} + 3^{2}} = \sqrt{16 + 9} = \sqrt{25} = \mathbf{5}

4. A (-2,-5), B (3,7) (orange)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(3 - (-2))^{2} + (7 - (-5))^{2}}\\= \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13}

5. A (5,4), B (-3,-2) (purple)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-3 - 5)^{2} + (-2 - 4)^{2}}\\= \sqrt{(-8)^{2} + (-6)^{2}} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10}

6. A (1,-8), B (-5,0) (black)

d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} = \sqrt{(-5 - 1)^{2} + (0 - (-8))^{2}}\\-= \sqrt{(-6)^{2} + (-8)^{2}} = \sqrt{36 + 64} = \sqrt{100} = \mathbf{10}

6 0
2 years ago
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