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3 years ago

12

You are given a graph of a linear equation, where the x-axis is time in months and the y-axis is money (in thousands) in your ba

nk account. The equation of the line goes through the points (0, 5) and (10, 0). Explain what these points represent in the context of the situation.

2 answers:

zzz [600]3 years ago

5 0

Answer:The y-intercept, at the point (0, 5), indicates that you started with $5,000 in your bank account. The x-intercept, at the point (10, 0), indicates that, after 10 months, there was no money in your bank account.

Step-by-step explanation:

LuckyWell [14K]3 years ago

4 0

0,5).....is saying that in 0 months, u will have $5000.
(10,0)....is saying that in 10 months, u will have $ 0.

u just went thru $5000 in 10 months...which really isn't much...about 500 per month

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X- (5-3x)<=2x-1 solve

kkurt [141]

Answer:

x<=2

Step-by-step explanation:

7 0

3 years ago

Pweeze help and I will give brainliest

Brums [2.3K]

Answer:

Part A) plot 1 and -2.5.

B) The opposite of 2 is -2. The only number that has its own opposite is 0. The first statement is false.

The opposite of 1.5 is -1.5, so the opposite of the opposite of 1.5 is equal to the opposite of -1.5, which is 1.5. The second statement is true.

Hope this helps sorry for not answering earlier

3 0

3 years ago

The length of a rectangle is 5m less than three times the width, and the area of the rectangle is 50m^(2). Find the dimensions o

Ilia_Sergeevich [38]

Answer:

The width is 5m and the length is 10m.

Step-by-step explanation:

Rectangle:

Has two dimensions: Width(w) and length(l).

It's area is:

$A = w*l$

The length of a rectangle is 5m less than three times the width

This means that $l = 3w - 5$

The area of the rectangle is 50m^(2)

This means that $A = 50$ . So

$A = w*l$

$50 = w*(3w - 5)$

$3w^{2} - 5w - 50 = 0$

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$ .

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$ , given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$3w^{2} - 5w - 50 = 0$

So

$a = 3, b = -5, c = -50$

$\bigtriangleup = (-5)^{2} - 4*3*(-50) = 625$

$w_{1} = \frac{-(-5) + \sqrt{625}}{2*3} = 5$

$w_{2} = \frac{-(-5) - \sqrt{625}}{2*3} = -3.33$

Dimension must be positive result, so

The width is 5m(in meters because the area is in square meters).

Length:

$l = 3w - 5 = 3*5 - 5 = 10$

The length is 10 meters

6 0

3 years ago

1. Quadrilateral ABCD has vertices A(-1, 1), B(2, 3), C(6, 0) and D(3, -2). Determine using coordinate geometry whether or not t

deff fn [24]

Answer:

The Conclusion is

Diagonals AC and BD,

a. Bisect each other

b. Not Congruent

c. Not Perpendicular

Step-by-step explanation:

Given:

[]ABCD is Quadrilateral having Vertices as

A(-1, 1),

B(2, 3),

C(6, 0) and

D(3, -2).

So the Diagonal are AC and BD

To Check

The diagonals AC and BD

a. Bisect each other. B. Are congruent. C. Are perpendicular.

Solution:

For a. Bisect each other

We will use Mid Point Formula,

If The mid point of diagonals AC and BD are Same Then

Diagonal, Bisect each other,

For mid point of AC

$Mid\ point(AC)=(\dfrac{x_{1}+x_{2} }{2},\dfrac{y_{1}+y_{2} }{2})$

Substituting the coordinates of A and C we get

$Mid\ point(AC)=(\dfrac{-1+6}{2},\dfrac{1+0}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Similarly, For mid point of BD

Substituting the coordinates of B and D we get

$Mid\ point(BD)=(\dfrac{2+3}{2},\dfrac{3-2}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Therefore The Mid point of diagonals AC and BD are Same

Hence Diagonals,

a. Bisect each other

B. Are congruent

For Diagonals to be Congruent We use Distance Formula

For Diagonal AC

$l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting A and C we get

$l(AC) = \sqrt{((6-(-1))^{2}+(0-1)^{2} )}=\sqrt{(49+1)}=\sqrt{50}$

Similarly ,For Diagonal BD

Substituting Band D we get

$l(BD) = \sqrt{((3-2))^{2}+(-2-3)^{2} )}=\sqrt{(1+25)}=\sqrt{26}$

Therefore Diagonals Not Congruent

For C. Are perpendicular.

For Diagonals to be perpendicular we need to have the Product of slopes must be - 1

For Slope we have

$Slope(AC)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Substituting A and C we get

$Slope(AC)=\dfrac{0-1}{6--1}\\\\Slope(AC)=\dfrac{-1}{7}$

Similarly, for BD we have

$Slope(BD)=\dfrac{-2-3}{3-2}\\\\Slope(BD)=\dfrac{-5}{1}$

The Product of slope is not -1

Hence Diagonals are Not Perpendicular.

6 0

4 years ago

Round 129.392 to the nearest ten,one,tenth,and hundredth. I need help plz

Soloha48 [4]

Nearest ten=130
nearest one=129
nearest tenth= 129.4
nearest hundredth= 129.39

8 0

4 years ago