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Rudik [331]
3 years ago
15

Caroline is doing a traffic survey. On the first day, the ratio of

Mathematics
1 answer:
Goryan [66]3 years ago
4 0

Answer:

1st day = 13/20×100 = 65%

2nd day = 8/13×100=61.53%

1st day answer

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What do the graphs of sine and cosine have in common with the swinging you see?
lilavasa [31]

The relationship between the cosine and sine graphs is that the cosine is the same as the sine — only it’s shifted to the left by 90 degrees, or π/2. The trigonometry equation that represents this relationship is: cosx= sin (x+π/2)

The graphs of the sine and cosine functions illustrate a property that exists for several pairings of the different trig functions. The property represented here is based on the right triangle and the two acute or complementary angles in a right triangle. The identities that arise from the triangle are called the cofunctionidentities.

5 0
4 years ago
Suppose that you have three consecutive positive integers where the product of the smaller and larger number is equal to 15 more
EleoNora [17]

Answer:

The numbers are either {-3, -2, -1} or {7, 8, 9}.

Step-by-step explanation:

Let the smallest integer be represented by x. The middle number is x+1, and the largest number is x+2.

The product of the smaller and larger number is (x)(x+2).

15 more than 6 times the middle number is 6(x+1)+15.

(x)(x+2)=6(x+1)+15

x^2+2x=6x+6+15

x^2+2x=6x+21

x^2-4x-21=0

(x-7)(x+3)=0

x=-3 or x=7

The numbers are either {-3, -2, -1} or {7, 8, 9}.

3 0
3 years ago
The height of a ball​ (in feet) after t seconds is given by the quadratic function h=-16(t-5)^2+116
Dmitriy789 [7]

Check the picture below, so pretty much reaches its maximum height at the vertex, now let's take a peek at the equation above hmmmm

~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h(t)=-16(t ~~ - ~~ \stackrel{h}{5})^2~~ + ~~\stackrel{k}{116}~\hfill \underset{maximum~height}{\stackrel{vertex}{(5~~,~~\underset{\uparrow }{116})}}

6 0
2 years ago
Read 2 more answers
Can someone help please please
STatiana [176]

Answer: A: (6, 5)

Step-by-step explanation:

When graphing a system of equations, you can find the solution by finding where the graphs intersect. In this case, the graphs intersect at (6, 5).

To verify that this is indeed a solution to the system of equations, we can substitute (6, 5) into both of the given equations:

y = \frac{1}{2}x + 2

5 = \frac{1}{2} * 6 + 2

5 = 3 + 2

5 = 5

y = x - 1

5 = 6 - 1

5 = 5

Since both equations hold true when plugging in the given coordinate (6, 5), we know that this is a solution to the system of equations.

5 0
3 years ago
Let R be the triangular region in the first quadrant with vertices at.Points (0,0), (h,0), and (h,r), where r and h are positive
Romashka [77]

Answer:

The answer is "\pi  \int^{h}_{0}(\frac{r}{h} x)^2 \ dx"

Step-by-step explanation:

dv=(Area) \ thickness

    =\pi r^2 \ dx\\\\=\pi (\frac{r}{h} x)^2 \ dx  

V= \int^{h}_{0} \pi (\frac{r}{h} x)^2 \ dx\\\\

   =\pi  \int^{h}_{0}(\frac{r}{h} x)^2 \ dx

3 0
3 years ago
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