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2 years ago

What does AC=? Round your answer to the nearest hundredth. Please help

Mathematics

1 answer:

faltersainse [42]2 years ago

4 0

Answer: About 3.06

Step-by-step explanation:

We can use trigonometry functions to solve for AC. Let the ?, representing AC, be "x" in our mathematical work.

Since we have the hypotenuse and x is adjacent to the angle given, I am going to use cosine.

cos(θ) = $\frac{adjacent}{hypotenuse }$

cos(40) = $\frac{x}{4}$

0.766 ≈ $\frac{x}{4}$

3.06 ≈ x

x ≈ 3.06

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Perry can fill a sorter with oranges from the conveyer belt in 10 minutes. While Perry fills the sorter, Gillian takes oranges o

ladessa [460]

If x is the amount of time Gillian takes,
1/10 - 1/x = 1/25
Therefore, the answer is A

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3 years ago

Read 2 more answers

Need an answer fast!!!

trapecia [35]

Answer:

The length of the line segment is of 5.9 units.

Step-by-step explanation:

Distance between two points:

Suppose that we have two points, $(x_1,y_1)$ and $(x_2,y_2)$ . The distance between these two points is given by:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

How long is the line segment?

The distance between points P and Q. So

P(1,3), and Q(4,8).

$D = \sqrt{(4-1)^2+(8-3)^2} = \sqrt{3^2 + 5^2} = \sqrt{34} = 5.9$

The length of the line segment is of 5.9 units.

4 0

2 years ago

Apply The Remainder Theorem, Fundamental Theorem, Rational Root Theorem, Descartes Rule, and Factor Theorem to find the remainde

Over [174]

9514 1404 393

Answer:

possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}

actual roots: -1, (2 ±4i√2)/3

no turning points; no local extrema

end behavior is same-sign as x-value end-behavior

Step-by-step explanation:

The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.

The Rational Root Theorem tells us any rational roots will be of the form ...

±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}

= ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots

Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.

The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.

The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):

f(x) = (x +1)(3x^2 -4x +12)

The complex roots of the quadratic can be found using the quadratic formula:

x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6

x = (2 ± 4i√2)/3 . . . . complex roots

The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.

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3 years ago

A football team lost 9 yards on each of two consecutive plays. represent the total change as a product of sun numbers and then f

dexar [7]

2(-9)=-18
Do the total list is -18 yards

7 0

3 years ago

The focus of a parabola is (−5,−1) and the directrix is y=−3. What is an equation of the parabola? (x+5)2=8(y+5)

Anna35 [415]

Graph x2 = 4y and state the vertex, focus, axis of symmetry, and directrix.This is the same graphing that I've done in the past: y = (1/4)x2. So I'll do the graph as usual: The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:x2 = 4y Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved
(x – 0)2 = 4(y – 0)This rearrangement "shows" that the vertex is at (h, k) = (0, 0). The axis of symmetry is the vertical line right through the vertex: x = 0. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that 4p = 4, so p = 1. Then the focus is one unit above the vertex, at (0, 1), and the directrix is the horizontal line y = –1, one unit below the vertex.vertex: (0, 0); focus: (0, 1); axis of symmetry: x = 0; directrix: y = –1Graph y2 + 10y + x + 25 = 0, and state the vertex, focus, axis of symmetry, and directrix.Since the y is squared in this equation, rather than the x, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking y-values first and then finding the corresponding x-values for x = –y2 – 10y – 25:To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:y2 + 10y + 25 = –x 
(y + 5)2 = –1(x – 0)This tells me that 4p = –1, so p = –1/4. Since the parabola opens to the left, then the focus is 1/4 units to the left of the vertex. I can see from the equation above that the vertex is at (h, k) = (0, –5), so then the focus must be at (–1/4, –5). The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is 1/4 units to the right of the vertex. Putting this all together, I get:vertex: (0, –5); focus: (–1/4, –5); axis of symmetry: y = –5; directrix: x = 1/4Find the vertex and focus of y2 + 6y + 12x – 15 = 0The y part is squared, so this is a sideways parabola. I'll get the y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.y2 + 6y – 15 = –12x 
y2 + 6y + 9 – 15 = –12x + 9 
(y + 3)2 – 15 = –12x + 9 
(y + 3)2 = –12x + 9 + 15 = –12x + 24 
(y + 3)2 = –12(x – 2) 
(y – (–3))2 = 4(–3)(x – 2)Then the vertex is at (h, k) = (2, –3) and the value of p is –3. Since y is squared and p is negative, then this is a sideways parabola that opens to the left. This puts the focus 3 units to the left of the vertex.vertex: (2, –3); focus: (–1, –3)

8 0

3 years ago