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madreJ [45]
2 years ago
14

Select the correct answer.

Mathematics
1 answer:
Orlov [11]2 years ago
6 0

Answer: C 28.47° latitude and -7.35° longitude

Step-by-step explanation:

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What number is 300% of six
nadya68 [22]

Answer:

18

Step-by-step explanation:

100% of 6 is 6 so 300% (which does not actually exist) would be 3 times as much. 6x3=18

6 0
3 years ago
Read 2 more answers
A team has won 15 of 27 games that it has played what fraction of games have they won?A team has won 15 of the 27 games that he
svet-max [94.6K]

15/27 but if you want to simplify it it’s 5/9

5 0
3 years ago
Can you help me with both of them
Ksivusya [100]

Answer:

1. is  -6b+4            2. is  n= -6

Step-by-step explanation:

6 0
2 years ago
Find the vertex and length of the latus rectum for the parabola. y=1/6(x-8)^2+6
Ivan

Step-by-step explanation:

If the parabola has the form

y = a(x - h)^2 + k (vertex form)

then its vertex is located at the point (h, k). Therefore, the vertex of the parabola

y = \dfrac{1}{6}(x - 8)^2 + 6

is located at the point (8, 6).

To find the length of the parabola's latus rectum, we need to find its focal length <em>f</em>. Luckily, since our equation is in vertex form, we can easily find from the focus (or focal point) coordinate, which is

\text{focus} = (h, k +\frac{1}{4a})

where \frac{1}{4a} is called the focal length or distance of the focus from the vertex. So from our equation, we can see that the focal length <em>f</em> is

f = \dfrac{1}{4(\frac{1}{6})} = \dfrac{3}{2}

By definition, the length of the latus rectum is four times the focal length so therefore, its value is

\text{latus rectum} = 4\left(\dfrac{3}{2}\right) = 6

5 0
3 years ago
Find the exact length of the curve. x=et+e−t, y=5−2t, 0≤t≤2 For a curve given by parametric equations x=f(t) and y=g(t), arc len
Rama09 [41]

The length of a curve <em>C</em> parameterized by a vector function <em>r</em><em>(t)</em> = <em>x(t)</em> i + <em>y(t)</em> j over an interval <em>a</em> ≤ <em>t</em> ≤ <em>b</em> is

\displaystyle\int_C\mathrm ds = \int_a^b \sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2} \,\mathrm dt

In this case, we have

<em>x(t)</em> = exp(<em>t</em> ) + exp(-<em>t</em> )   ==>   d<em>x</em>/d<em>t</em> = exp(<em>t</em> ) - exp(-<em>t</em> )

<em>y(t)</em> = 5 - 2<em>t</em>   ==>   d<em>y</em>/d<em>t</em> = -2

and [<em>a</em>, <em>b</em>] = [0, 2]. The length of the curve is then

\displaystyle\int_0^2 \sqrt{\left(e^t-e^{-t}\right)^2+(-2)^2} \,\mathrm dt = \int_0^2 \sqrt{e^{2t}-2+e^{-2t}+4}\,\mathrm dt

=\displaystyle\int_0^2 \sqrt{e^{2t}+2+e^{-2t}} \,\mathrm dt

=\displaystyle\int_0^2\sqrt{\left(e^t+e^{-t}\right)^2} \,\mathrm dt

=\displaystyle\int_0^2\left(e^t+e^{-t}\right)\,\mathrm dt

=\left(e^t-e^{-t}\right)\bigg|_0^2 = \left(e^2-e^{-2}\right)-\left(e^0-e^{-0}\right) = \boxed{e^2-\frac1{e^2}}

5 0
2 years ago
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