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3 years ago

Please do number 6 only

Mathematics

1 answer:

Troyanec [42]3 years ago

7 0

-$67 because she lost or spent 67 dollars which is larger than 34 dollars.

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The midpoint of AB is M(3,-1).If the coordinates of A are (8,4) what are the coordinates of B

Alchen [17]

Answer:

B(-2,-6)

Step-by-step explanation:

The midpoint of AB is M, then A(8,4), M(3,-1)

So we can find:

$x_{B} =2x_{M} -x_{A}= 2*3-8= -2\\y_{B}= 2y_{M} -y_{A} = 2*(-1)- 4= -6\\$

then B(-2,-6), hope that useful.

6 0

4 years ago

A square was altered so that one side is increased by 9 inches in the other side is decreased by 2 inches. The area of the resul

nevsk [136]

Answer:

area of the original square = 36 in²

Step-by-step explanation:

given data

one side is increased = 9 inches

other side is decreased = 2 inches

area of the resulting rectangle = 60 in²

solution

we consider here x as the length of any one side

and

longer side of the resulting rectangle = x + 9

and

the shorter side x - 2

so that area of this rectangle is (s+9) × (s-2) = 60 in²

standard form of equation is x² + 7x - 18 = 60 .....................1

it will be

x² + 7x - 78 = 0

(s+13) (s-6) = 0

so that here x = -13

and x = 6

here we Discard x = -13 because the side length cannot be negative.

so x = 6

and area of the original square = 36 in²

5 0

3 years ago

Convert to a fraction in simplest terms: .45

photoshop1234 [79]

Answer:

9/20

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Help I will make you a brainlist!!

enyata [817]

You answer would be Oy= 7x-1. I hope this helps!

4 0

3 years ago

Let R be the region in the first quadrant bounded by the graphs of y =x^2 and y=2x, as shown in the figure above. The region R i

Kobotan [32]

Answer:

The area between the two functions is approximately 1.333 units.

Step-by-step explanation:

If I understand your question correctly, you're looking for the area surrounded by the the line y = 2x and the parabola y = x², (as shown in the attached image).

To do this, we just need to take the integral of y = x², and subtract that from the area under y = 2x, within that range.

First we need to find where they intersect:

2x = x²

2 = x

So they intersect at (2, 4) and (0, 0)

Now we simply need to take the integrals of each, subtracting the parabola from the line (as the parabola will have lower values in that range):

$a = \int\limits^2_0 {2x} \, dx - \int\limits^2_0 {x^2} \, dx\\\\a = x^2\left \{ {{x=2} \atop {x=0}} \right - \frac{x^3}{3}\left \{ {{x=2} \atop {x=0}} \right\\\\a = (2^2 - 0^2) - (\frac{2^3}{3} - \frac{0^3}{3})\\\\a = 2^2 - \frac{2^3}{3}\\a = 4 - 8/3\\a \approx 1.333$

So the correct answer is C, the area between the two functions is 4/3 units.

3 0

3 years ago