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Volgvan
2 years ago
8

If EF=3x-12, FG=4x-15, and EG=29, enter the values of x, EF, and FG.

Mathematics
1 answer:
anzhelika [568]2 years ago
4 0

Answer:

X =8

EF=12

FG=17

Step-by-step explanation:

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PLEASE PLEASE I NEED HELP ASAP
Sidana [21]
AE = AC = 4

m<CAB = 60 (equilateral triangle)
m<CAE = 90 (square)

m<BAE = 150 (= 60 + 90)

Triangle BAE is isosceles since AB = AE;
therefore, m<AEB = m<ABE.

m<AEB + m<ABE + m<BAE = 180

m<AEB + m< ABE + 150 = 180

m<AEB + m<AEB = 30

m<AEB = 15

In triangle ABE, we know AE = AB = 4;
we also know m<BAE = 150, and m<AEB = 15.

We can use the law of sines to find BE.

BE/(sin 150) = 4/(sin 15)

BE = (4 sin 150)/(sin 15)

BE = 7.727

3 0
3 years ago
Y=6*2x for x=-2 evaluate the function
atroni [7]

Answer:

Y = -24

Step-by-step explanation:

For this problem, evaluate simply means for each x, replace the value with -2. So, let's do that.

Y = 6*2x

Y = 6*2(-2)

Y = 6*-4

Y= -24

Cheers.

6 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}}

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Lelu [443]

Answer:

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Step-by-step explanation:

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3 years ago
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Fantom [35]

Answer: English please

Step-by-step explanation:

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