Gravity might be a stimulus that induces a tropic response in a plant. a. True b. False

Afina-wow [57]

Just subsitute values for x and get values for y
(x,y)
(0,-2)
(1,1)
(-1,-5)
(-2,-8)
here is the graph

8 0

3 years ago

Suppose that you have declared a numeric array named numbers, and two of its elements are numbers[1] and numbers[5]. you know th

mart [117]

Answer:

In the scenario in which you have declared a numeric array named numbers, and two of its elements are numbers[1] and numbers[5] you know that numbers[1] is smaller than numbers[5] and that there are exactly three elements between those two elements. Correct answer: a and c.

numbers[i] is the element of the array called numbers

Step-by-step explanation:

5 0

3 years ago

Suppose a ski jumper is 2,650 feet from the base of a ski jump and looks up to the top at an angle of elevation of 26 degrees. A

yuradex [85]

We can draw a right triangle to easily visualize the given problem. Since the ski jumper is 2650 ft from the base of the ski jump, we can consider this as one of the right triangle's legs.

What we're looking for is the height of the ski jump or the other leg of the right triangle. Since the ski jumper looks up at an angle of elevation of 26°, we can use the tangent function to find the missing height. You may refer to the image below to better understand the problem.

Recall that tanθ = (opposite side facing the angle)/(side adjacent to the angle). For this case, we have

$\tan26= \frac{x}{2650}\\2650(\tan26) = x\\x \approx1292.49$

Hence, the height of the ski jump (rounded off the nearest whole foot) is 1292 feet.

Answer: 1292 feet

4 0

4 years ago

How do i solve this ?

MariettaO [177]

Answer:

20 (im pretty sure)

Step-by-step explanation:

all angles should equal 360 so add them together and you get 340 so find what's missing—20

8 0

3 years ago

Eliminate the parameter and obtain the standard form of the rectangular equation. Ellipse: x = h + a cos(θ), y = k + b sin(θ) Us

gayaneshka [121]

Answer:

a) $(\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1$

b) $x=1+5 \,cos(\theta)\,,\,y=4\,sin(\theta)$

c) $\frac{(x-1)^2}{25}+ \frac{y^2}{16}=1$

Step-by-step explanation:

Start by isolating the trigonometric expression in both equations, ad then use the Pythagorean identity:

$cos^2(\theta)+sin^2(\theta)=1$

to obtain the standard equation of a conic.

$x=h+a\,cos(\theta)\\x-h=a\,cos(\theta)\\cos(\theta)=\frac{x-h}{a}$ $y=k+b\,sin(\theta)\\y-k=b\,sin(\theta)\\sin(\theta)=\frac{y-k}{b}$

then:

$cos^2(\theta)+sin^2(\theta)=1\\(\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1$

this is the equation of an ellipse centered at (h,k), and with horizontal axis length = 2a , and vertical axis length = 2b.

The parametric equations are those we started with:

$x=h+a \,cos(\theta)\,,\,y=k+b\,sin(\theta)$ but we need to find the appropriate parameters for the requested ellipse, as shown below.

For an ellipse of vertices (-4,0) a,d (6,0) and foci at (-2,0) and (4.0), we are dealing with an ellipse with major horizontal axis on the line y=0, and major diameter length of 10 units, so the parameter $a=5$ . The center of the ellipse is therefore at (1,0).

We recall that the vertices of a translated horizontal ellipse are located at $(a+h,k)$ and $(-a+h,k)$ , then k=0 to satisfy the information given (-4,0) & (6,0), and since $a=5$ , we deduce that $a+h = 6$ and therefore h=1.

To find "b" (the only parameter missing for the standard equation of the conic), we need the information on the foci (-2,0) and (4,0) which must equal (h-c,k) and (h+c,k) with k=0 and h=1 which then gives that c=3

Now using the formula for the parameter "c" of the foci: $c=\sqrt{a^2-b^2}$

$c=\sqrt{a^2-b^2}\\3=\sqrt{5^2-b^2}\\9=25-b^2\\b^2=25-9\\b^2=16\\b=4$

Then we can write the equation of this ellipse in standard form as:

$(\frac{x-h}{a})^2+ (\frac{y-k}{b})^2=1\\(\frac{x-1}{5})^2+ (\frac{y-0}{4})^2=1\\\\(\frac{x-1}{5})^2+ (\frac{y}{4})^2=1\\\frac{(x-1)^2}{25}+ \frac{y^2}{16}=1$

8 0

3 years ago