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3 years ago

II: Now let's try graphing on some number lines on own.

Mathematics

1 answer:

balandron [24]3 years ago

7 0

Answer:

573 38

Step-by-step explanation:

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Solve: -5-(-8)<br> ILL GIVE YOU 15 POINTS IF YOU HELP, ill mark you brainiest to

Usimov [2.4K]

Answer:

Step-by-step explanation:

-5 + 8 = 3

keep inverse opposite:

-5 => -5

- => +

-8 => 8

7 0

3 years ago

Is ten a perfect square

Contact [7]

Answer:
No, it is not a perfect square.

7 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

I need help to find each product by factoring the tens

kumpel [21]

The answer is I think 6.89 your welcome e ((((( :)))

4 0

3 years ago

1) 5z + 4 = 19<br> What is the answer for this question

Andreas93 [3]

5z=19-4
5z=15
z=15/5
z=3

8 0

4 years ago