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3 years ago

One angle of a triangle has a measure of 66 degrees the measure of the third angle is 57 more than 1/2 of the measure

Mathematics

1 answer:

marissa [1.9K]3 years ago

7 0

The sum of all 3 angles is 180 degrees, you should be able to solve from here.

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Given that the following two geometric series(meetkundige reekse) are covergent 1+x+x²+x³+... And 1-x+x²-x³+... Determine the va

Oksanka [162]

The sets are (1,-1) or (-1,1)

Lets see

Take until 4terms

$\\ \rm\longmapsto S_4=1+1+1^2+1^3=1+1+1+1=4$ (1,-1)

$\\ \rm\longmapsto S_4=1+(-1)+(-1)^2+(-1)^3=1-1+1-1=0$ (-1,1)

$\\ \rm\longmapsto S_4=1-(-1)+(-1)^2-(-1)^3=1+1+1+1=4$ (1,-1)

$\\ \rm\longmapsto S_4=1-1+(1)^2-1^3=1-1+1-1=0$ (-1,1)

Hence verified

Another set can be (0,0)

Proof:-

$\\ \ast\rm\longmapsto S_3=1+0+0^2=1$

$\\ \ast\rm\longmapsto S_3=1-0+0^2=1$

Hence verified.

3 0

3 years ago

How do you answer this MATH

raketka [301]

X = 5 or x = 13.

If you need details, please ask!

7 0

3 years ago

Read 2 more answers

The work that Yi did to find the greatest common factor of 42 and 63

Lilit [14]

Answer:

Step-by-step explanation:

factors of 42: 1,42, 2,21, 3,14, 6,7

factors of 63: 1,63, 3,21, 7,9

7 0

3 years ago

Graph step by step: f(x)=-2x+1

Paul [167]

Step-by-step explanation:

Root:(1/2, 0)

Domain: x∈R

Range: y∈R

Vertical intercept: (0, 1)

4 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 C parameterized by a vector function r(t) = x(t) i + y(t) j over an interval a ≤ t ≤ b 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

x(t) = exp(t ) + exp(-t ) ==> dx/dt = exp(t ) - exp(-t )

y(t) = 5 - 2t ==> dy/dt = -2

and [a, b] = [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

3 years ago