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4 years ago

Which bank costs less for the feature or service?

Mathematics

1 answer:

mario62 [17]4 years ago

3 0

I think its oakley bank which costs less oakley bank, but it has a higher overdraft fee

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Name the congruent sides.

olga nikolaevna [1]

The answer is B.
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3 years ago

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Simora [160]

Answer: 10

Step-by-step explanation:

add 2.75 to 0

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3 years ago

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Please help me put these in the correct order. *No wrong answers* List the following in order Can you list the answers in order

tankabanditka [31]

Answer:

Add or subtract. Show all work.

1) 9.069

2) 43.95

3) 20.64

4) 32.145

5) 7.16

6) 12.2

7) 25.95

8) 15.815

Increasing order (smallest to largest): 5, 1, 6, 8, 3, 7, 4, 2

Decreasing order (largest to smallest): 2, 4, 7, 3, 8, 6, 1, 5

Multiply or divide. Show all work.

1) 44.95

2) 225.9

3) 20.271

4) 11.557

5) 6.78

6) 41.3

7) 120

8) 515

Increasing order (smallest to largest): 5, 4, 3, 6, 1, 7, 2, 8

Decreasing order (largest to smallest): 8, 2, 7, 1, 6, 3, 4, 5

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3 0

3 years ago

TIME REMAINING

mars1129 [50]

Answer:

A pint of veggies is approximately $1.75

Step-by-step explanation:

(7 × 2.85) + 5v = 28.70. 19.95 + 5v = 28.70. 5v = 28.70 - 19.95. 5v = 8.75. v = 8.75/5. v = 1.75

Hope this helps!!

4 0

3 years ago

A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a

Rashid [163]

Answer:

Length of original rectangle: 11 units.

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

$\text{Perimeter of new rectangle}=20$

Step-by-step explanation:

Let x represent width of the original rectangle.

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be $x+6$ .

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be $x-2$ .

The length of new rectangle would be $x+6-4=x+2$ .

The area of new rectangle would be $(x+2)(x-2)$ .

Now we will equate area of new rectangle with 21 and solve for x as:

$(x+2)(x-2)=21$

Applying difference of squares, we will get:

$x^2-2^2=21$

$x^2-4=21$

$x^2-4+4=21+4$

$x^2=25$

Since width cannot be negative, so we will take positive square root of both sides.

$\sqrt{x^2}=\sqrt{25}$

$x=5$

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be $x+6\Rightarrow x+5=11$ .

Therefore, the length of original rectangle is 11 units.

$\text{Area of original rectangle}=5\times 11$

$\text{Area of original rectangle}=55$

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

$\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}$

We know that perimeter of rectangle is two times the sum of length and width.

$\text{Perimeter of new rectangle}=2((x+2)+(x-2))$

$\text{Perimeter of new rectangle}=2((5+2)+(5-2))$

$\text{Perimeter of new rectangle}=2(7+3)$

$\text{Perimeter of new rectangle}=2(10)$

$\text{Perimeter of new rectangle}=20$

Therefore, the perimeter of the new rectangle is 20 units.

7 0

4 years ago