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damaskus [11]
3 years ago
5

Ollie has $400 balance on her credit card.She pays $60 a month on the card.Does this situation have a positive or negative rate

of change?explain
Mathematics
1 answer:
Maurinko [17]3 years ago
3 0

Answer:

It has a negative rate of change.

Step-by-step explanation:

She is losing money so the rate of change is negative

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F^-1= positive or negative x square root
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Ostrovityanka [42]
Maybe 40 degrees
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6 0
3 years ago
Please please 1,2,3 please help!!
zhannawk [14.2K]
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6 0
4 years ago
I need help ASAP! Plz ( Thank You )
Assoli18 [71]

Answer:

6a) x = -6

Let's solve your equation step-by-step.


14= 2x + 26


Step 1: Flip the equation.


2x + 26 = 14


Step 2: Subtract 26 from both sides.


2x + 26 − 26 = 14 − 26


2x = −12


Step 3: Divide both sides by 2.


2x/2 = −12/2


x=−6

--------------------

7a) x = 4.25

Let's solve your equation step-by-step.


−30 =4 − 8x


Step 1: Simplify both sides of the equation.


−30 = 4 − 8x


−30 = 4 + −8x


−30 = −8x + 4


Step 2: Flip the equation.


−8x + 4 = −30


Step 3: Subtract 4 from both sides.


−8x + 4 − 4 = −30 − 4


−8x = −34

x=17/4 (Decimal: 4.25)

---------------------------------

8a) x = 8

Let's solve your equation step-by-step.


3(x − 4) = 12


Step 1: Simplify both sides of the equation.


3(x − 4) = 12


(3)(x) + (3)(−4)= 12(Distribute)


3x + −12 = 12


3x − 12= 12


Step 2: Add 12 to both sides.


3x − 12 +1 2 = 12 + 12


3x = 24


Step 3: Divide both sides by 3.

3x/3=24/3

x=8


3 0
4 years ago
1. Find the vertices and locate the foci for the hyperbola whose equation is given.
Irina18 [472]
\bf \cfrac{(x-{{ h}})^2}{{{ a}}^2}-\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1
\qquad center\ ({{ h}},{{ k}})\qquad
 vertices\ ({{ h}}\pm a, {{ k}})\\\\
-----------------------------\\\\
\textit{now let's take a look at yours}
\\\\\\
49x2 - 16y2 = 784\implies \cfrac{49x^2}{784}-\cfrac{16y^2}{784}=1
\\\\\\
\cfrac{x^2}{16}-\cfrac{y^2}{49}=1\implies \cfrac{(x-0)^2}{4^2}-\cfrac{(y-0)^2}{7^2}=1
\\\\\\
recall\implies center\ ({{ h}},{{ k}})\qquad vertices\ ({{ h}}\pm a, {{ k}})
\\\\\\


\bf \textit{now, for the foci, the foci are "c" distance from the center point}\\\\\
whereas\qquad c=\sqrt{a^2+b^2}\qquad \textit{ that is }\qquad  h\pm \sqrt{a^2+b^2}

notice your "a" and "b" components, to get the distance "c" from the center to either foci and the vertices, of course, are h + a, k and h - a, k
5 0
3 years ago
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