Answer:
x = 12 cm
Step-by-step explanation:
The product of lengths from the secant intersection point to the "near" and "far" circle intersection points is the same for both secants. When one "secant" is a tangent, the lengths to the circle intersection points are the same (so their product is the square of the tangent segment length).
8^2 = 4·(4 +x) . . . . . . measures in centimeters
16 = 4 +x . . . . . . divide by 4
12 = x . . . . . . . . . subtract 4
4 x 1,151.65 = 4,606.60
<span>This is invested in two accounts. Twenty-five percent is invested in savings (4,606.60 x 25% = 1,151.65). The remainder is invested in a CD (4,606.60 - 1,151.65 = 3,454.95) </span>
<span>Interest accrued for savings. *Note: The interest rates given are for a full year, but you want to calculate interest for only 60 days. </span>
<span>1,151.65 x 0.033 x 60/360 = $6.33 </span>
<span>Interest accrued for CD </span>
<span>3,454,95 x 0.043 x 60/360 = $24.76 </span>
<span>Total interest accrued </span>
<span>6.33 + 24.76 = $31.09 </span>
<span>Note: 0.033 and 0.043 = 3.3% and 4.3%. Also, the 360 day year is normally used when doing these types of problems.</span>
For the area of the deck to be doubled, he should increase each dimension by 3.
<h3>How to find the dimension increase to double the area?</h3>
The deck is 4 feet by 21 feet.
She wants to increase each dimension by equal lengths so that its area is doubled.
Therefore,
initial area = 4 × 21 = 84 ft²
Hence,
The increase by equal length
width = x + 4
length = x + 21
area = 2(84) = 168 ft²
Therefore,
(x + 4)(x + 21) = 168
x² + 21x + 4x + 84 = 168
x² + 25x + 84 = 168
x² + 25x + 84 - 168 = 0
x² + 25x - 84 = 0
(x + 28) • (x - 3) = 0
x = -28 or 3
It can only be positive.
Therefore, she should increase each dimension by 3.
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Answer: 9
x
3
(
5
x
+
8
)
Step-by-step explanation:
Factor 9
x
3 out of 45
x
4
+
72x
3
.
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).
The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).
b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.
c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.