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3 years ago

5

Need help quickly! The equation A = 1300(1.02)^7 is being used to calculate the amount of money in a saving account. What does 1

.02 represent in this equation?

2 answers:

Mrrafil [7]3 years ago

7 0

Answer:

the answer should be 0.02 growth

Step-by-step explanation:

if the decimal was negative then the answer is decay.

HACTEHA [7]3 years ago

3 0

The answer is b
Your welcome

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Lostsunrise [7]

Answer:

B

Step-by-step explanation:

since 28 is the amount of time it takes for him to mow a lawn, we have to have an answer that involves time, which is B.

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3 years ago

A 220$ ring is on sale for 40% off. A sales tax of 7% will be applied to the sale price. What will be the final cost of the ring

ddd [48]

Answer:

Step-by-step explanation:

60% of 220 is 132. So thats the sale. The tax is 7% of that. So 7% of 132 is 9.24 So 132+9.24=141.24

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3 years ago

Thrice the sum of three fifths and two thirds less one half is what number?

Oduvanchick [21]

Answer:

Step-by-step explanation:

Thrice the sum of three fifths and two thirds less one half is expressed as 3(3/5 + 2/3)-1/2

Open the parenthesis

3(3/5) + 3(2/3) - 1/2

= 9/5 + 6/3 - 1/2

Find LCM

= 9(6)+10(6)-15/30

= 54+60-15/30

= 54+45/30

= 99/30

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Hence the number id 3 3/10

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3 years ago

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3 years ago

Read 2 more answers

It is known that the straight line l is tangent to the circle x^2+y^2=4 at a point on the x-axis, and intersects with the straig

Talja [164]

Step-by-step explanation:

The general equation of a circle is

$(x \ - \ h)^2 \ + \ (y \ - \ k)^2 \ = \ r^{2}$ ,

where <em>h</em> and <em>k</em> forms the coordinates of the centre of the circle.

When the circle has a centre at the origin, the equation reduces into

. $x^2 \ + \ y^2 \ = r^2$ .

Now, we are interested in solving for the <em>x</em>-intercepts (the <em>x</em>-coordinates when the circle intersects the <em>x</em>-axis), of the circle

$x^2 \ + \ y^2 \ = \ 4$ .

Thus,

$x^2 \ + \ (0)^2 \ = \ 4 \\ \\ \\ \-\hspace{1.3cm} x^{2} \ = \ 4 \\ \\ \\ \-\hspace{1.4cm} x \ = \ \pm \ 2$ .

Geometrically speaking, the tangent to the circle at the point defined by one of the <em>x</em>-intercepts of the circle is actually a vertical line, more specifically the lines $x \ = \ \pm \ 2$ .

First and foremost, for the vertical line $x \ = \ 2$ , it intersects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ , giving the y-coordinate for point P,

$y \ = \ \displaystyle\frac{1}{2}(2) \\ \\ \\ y \ = \ 1$ .

Hence, the coordinates of point P are $(1, \ 1)$ .

However, since there are no boundaries given in the question and a circle is symmetrical about its centre, thus, point P also exists when the vertical line $x \ = \ -2$ and interdects the straight line $y \ = \ \displaystyle\frac{1}{2}x$ .

$y \ = \ \displaystyle\frac{1}{2}(-2) \\ \\ \\ y \ = \ -1$ .

Therefore, the coordinates of point P are also $(1, \ -1)$ .

8 0

2 years ago