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Alisiya [41]
3 years ago
8

The value of Peter's house has risen 20% a year for the last 3 years. If the value of Peter's home 3 years ago was $90,000.00, w

hat is the value now?
Mathematics
1 answer:
natima [27]3 years ago
4 0

Answer:

$155,520

Step-by-step explanation:

When the value rises 20%, it is 1.2 times what it was. This has occurred 3 times, so the value is now 1.2³ = 1.728 times what it was.

... $90,000 × 1.2³ = $155,520

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Answer:

D.

Step-by-step explanation:

(f - g) (x) ➡ f(x) - g(x)

5x - 2 - 2x - 1 = 3x - 3

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3 years ago
If you have 63 books and you donate 1/3 of you’re books how many will you have left
ryzh [129]

Answer: 42

Step-by-step explanation: you would do 63 divided by 3 then you would get 21 and do 21×2 and get 42

6 0
3 years ago
FIND THE THIRD,FIFTH AND TENTH TERM SEQUENCE DESCRIBED BY EACH RULE
Karolina [17]
In order to solve for a nth term in an arithmetic sequence, we use the formula written as:<span>

an = a1 + (n-1)d

where an is the nth term, a1 is the first value in the sequence, n is the term position and d is the common difference.

</span><span>THIRD
</span><span>A3=4+(3-1)(-5)
A3 = -6

A(3)=-2(3-1)(-5)
A3 = 20
 </span><span>
FIFTH
</span>A5=4+(5-1)(-5)
A5 = -16

A(5)=-2(5-1)(-5)
A5 = 40<span>

TENTH
</span>A10=4+(10-1)(-5)
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A(10)=-2(10-1)(-5)
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8 0
3 years ago
Let f(x)=x^3-x-1
alexira [117]
f(x) = x^3 - x - 1

To find the gradient of the tangent, we must first differentiate the function.

f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1

The gradient at x = 0 is given by evaluating f'(0).

f'(0) = 3(0)^2 - 1 = -1

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

y = -x + c

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

f(0) = (0)^3 - (0) - 1 = -1

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}
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3 years ago
Please help me out ​
Tema [17]

Answer:

UwU thank you

please remember to smile, because when you do the whole world gets brighter

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Step-by-step explanation:

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2 years ago
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