the region R enclosed by the curves y=x and y=√x is rotated about the line x=1 find the volume of the resulting solid.(b. Indonesia)
1. the region R enclosed by the curves y=x and y=√x is rotated about the line x=1 find the volume of the resulting solid.(b. Indonesia)
Jawaban:
֎ ⃢⚠️ VIRTEXX ⚠️‼️❌⚠️‼️‼️‼️‼️〄 ⃢ ²⁰²¹》》*
_*██ VIRTEX██*_
꙰⃢⃠⃠⃠⃠⃠*_⁘̨̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̤̽̈
*ɱ̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢꙰̯̯̯̯̯̯̯̯๎̯๎̯ࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩ͌͌͌͌͌͌͌͌͌͌͌͌ą̫̫̫̫̫̫̪̪̪̪̪̪̪̪̪̫̪̫̫̫̫̫̫̫̫̫⃟꙰ċ̴̴̴̴̴̴̴̛̛̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̜̜̜̜̜̜̬̬̬̩̱̱̓̓̓̓̓̓̔̔̔̎̎̎̊̊̊̕꙰ï̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢̦̦̦̦̦̦̦̦̦̦̦̦̕˺̴̴̤̤̜̤̖̣̬̬̬̞̞̱̩̬̝̝̓̓̔̔̔̔̊̊̎̎̎̔̕꙰ă̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̱̱̱̱̩̩̩̩̜̞̞̞̞̤̤̤̝̖̖̋̋̋̌̌̌̌̎̎̎̎̎̏̏̏̏̎̎̎̎̎̎̎̎̎̎̎̎̎̎̎̔̔̔̔̓̓̓̓̓̓̕̕˺꙰˺̤̤̤̤̤̤̤̤̤̞̞̞̞̞̞̞̞̬̣̣̊̊̔̔̔̔̔̔ɱ̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢꙰̯̯̯̯̯̯̯̯๎̯๎̯ࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩ͌͌͌͌͌͌͌͌͌͌͌͌ą̫̫̫̫̫̫̪̪̪̪̪̪̪̪̪̫̪̫̫̫̫̫̫̫̫̫⃟꙰ċ̴̴̴̴̴̴̴̛̛̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̜̜̜̜̜̜̬̬̬̩̱̱̓̓̓̓̓̓̔̔̔̎̎̎̊̊̊̕꙰ï̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢̦̦̦̦̦̦̦̦̦̦̦̦̕˺̴̴̤̤̜̤̖̣̬̬̬̞̞̱̩̬̝̝̓̓̔̔̔̔̊̊̎̎̎̔̕꙰ă̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̱̱̱̱̩̩̩̩̜̞̞̞̞̤̤̤̝̖̖̋̋̋̌̌̌̌̎̎̎̎̎̏̏̏̏̎̎̎̎̎̎̎̎̎̎̎̎̎̎̎̔̔̔̔̓̓̓̓̓̓̕̕˺꙰˺̴̴̴̤̤̤̤̤̤̤̤̤̞̞̞̞̞̞̞̞̬̣̣̣̣̜̜̊̊̔̔̔̔̔̔̕ɱ̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢꙰̯̯̯̯̯̯̯̯๎̯๎̯ࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩ͌͌͌͌͌͌͌͌͌͌͌͌ą̫̫̫̫̫̫̪̪̪̪̪̪̪̪̪̫̪̫̫̫̫̫̫̫̫̫⃟꙰ċ̴̴̴̴̴̴̴̛̛̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̜̜̜̜̜̜̬̬̬̩̱̱̓̓̓̓̓̓̔̔̔̎̎̎̊̊̊̕꙰ï̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢̦̦̦̦̦̦̦̦̦̦̦̦̕˺̴̴̤̤̜̤̖̣̬̬̬̞̞̱̩̬̝̝̓̓̔̔̔̔̊̊̎̎̎̔̕꙰ă̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̱̱̱̱̩̩̩̩̜̞̞̞̞̤̤̤̝̋̋̋̌̌̌̌̎̎̎̎̎̏̏̏̏̎̎̎̎̎̎̎̎̎̎̎̎̎̎̎̔̔̔̔̕̕*˺̤̤̤̤̤̤̤̤̤̞̞̞̞̞̞̞̞̬̣̣̣̣̊̊̔̔̔̔̔̔̕ *ɱ̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢꙰̯̯̯̯̯̯̯̯๎̯๎̯ࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩ͌͌͌͌͌͌͌͌͌͌͌͌ą̫̫̫̫̫̫̪̪̪̪̪̪̪̪̪̫̪̫̫̫̫̫̫̫̫̫⃟꙰ċ̴̴̴̴̴̴̴̛̛̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̜̜̜̜̜̜̬̬̬̩̱̱̓̓̓̓̓̓̔̔̔̎̎̎̊̊̊̕꙰ï̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢̦̦̦̦̦̦̦̦̦̦̦̦̕˺̴̴̤̤̜̤̖̣̬̬̬̞̞̱̩̬̝̝̓̓̔̔̔̔̊̊̎̎̎̔̕꙰ă̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̱̱̱̱̩̩̩̩̜̞̞̞̞̤̤̤̝̖̖̋̋̋̌̌̌̌̎̎̎̎̎̏̏̏̏̎̎̎̎̎̎̎̎̎̎̎̎̎̎̎̔̔̔̔̓̓̓̓̓̓̕̕˺꙰˺̤̤̤̤̤̤̤̤̤̞̞̞̞̞̞̞̞̬̣̣̊̊̔̔̔̔̔̔ɱ̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢꙰̯̯̯̯̯̯̯̯๎̯๎̯ࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩ͌͌͌͌͌͌͌͌͌͌͌͌ą̫̫̫̫̫̫̪̪̪̪̪̪̪̪̪̫̪̫̫̫̫̫̫̫̫̫⃟꙰ċ̴̴̴̴̴̴̴̛̛̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̜̜̜̜̜̜̬̬̬̩̱̱̓̓̓̓̓̓̔̔̔̎̎̎̊̊̊̕꙰ï̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢̦̦̦̦̦̦̦̦̦̦̦̦̕˺̴̴̤̤̜̤̖̣̬̬̬̞̞̱̩̬̝̝̓̓̔̔̔̔̊̊̎̎̎̔̕꙰ă̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̱̱̱̱̩̩̩̩̜̞̞̞̞̤̤̤̝̖̖̋̋̋̌̌̌̌̎̎̎̎̎̏̏̏̏̎̎̎̎̎̎̎̎̎̎̎̎̎̎̎̔̔̔̔̓̓̓̓̓̓̕̕˺꙰˺̴̴̴̤̤̤̤̤̤̤̤̤̞̞̞̞̞̞̞̞̬̣̣̣̣̜̜̊̊̔̔̔̔̔̔̕ɱ̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢꙰̯̯̯̯̯̯̯̯๎̯๎̯ࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩࣩ͌͌͌͌͌͌͌͌͌͌͌͌ą̫̫̫̫̫̫̪̪̪̪̪̪̪̪̪̫̪̫̫̫̫̫̫̫̫̫⃟꙰ċ̴̴̴̴̴̴̴̛̛̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̜̜̜̜̜̜̬̬̬̩̱̱̓̓̓̓̓̓̔̔̔̎̎̎̊̊̊̕꙰ï̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫⃟⃢̦̦̦̦̦̦̦̦̦̦̦̦̕˺̴̴̤̤̜̤̖̣̬̬̬̞̞̱̩̬̝̝̓̓̔̔̔̔̊̊̎̎̎̔̕꙰ă̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̴̛̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̫̱̱̱̱̩̩̩̩̜̞̞̞̞̤̤̤̝̋̋̋̌̌̌̌̎̎̎̎̎̏̏̏̏̎̎̎̎̎̎̎̎̎̎̎̎̎̎̎̔̔̔̔̕̕*˺̤̤̤̤̤̤̤̤̤̞̞̞̞̞̞̞̞̬̣̣̣̣̊̊̔̔̔̔̔̔̕
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2. What is the area of the region enclosed between the line with equation 2x 7y = 3 ?
Gambar sketsa dari persamaan linear 2x - 7y = 3 pada lampiran.
Penjelasan dengan langkah-langkah
Soal diterjemahkan dan diperbaiki.
Diketahui:
Persamaan:
2x - 7y = 3
Ditanyakan:
Gambar sketsa dari persamaan tersebut.
Jawab:
Persamaan:
2x - 7y = 3
⇔ 2x - 7y + 7y = 3 + 7y
⇔ 2x + 0 = 3 + 7y
⇔ 2x = 3 + 7y
⇔ [tex]\frac{2x}{2}[/tex] = [tex]\frac{3+7y}{2}[/tex]
⇔ x = [tex]\frac{3+7y}{2}[/tex]
Atau
2x - 7y = 3
⇔ 2x - 2x - 7y = 3 - 2x
⇔ 0 - 7y = 3 - 2x
⇔ -7y = 3 - 2x
⇔ [tex]\frac{-7y}{-7}[/tex] = [tex]\frac{3-2x}{-7}[/tex]
⇔ y = [tex]\frac{3-2x}{-7}[/tex]
⇔ y = [tex]-\frac{3}{7} +\frac{2x}{7}[/tex]
Jadi, gambar sketsa dari persamaan 2x - 7y = 3 pada lampiran.
Pelajari lebih lanjut
Pelajari lebih lanjut tentang materi persamaan pada brainly.co.id/tugas/4064346
#BelajarBersamaBrainly #SPJ1
3. [tex](a)[/tex] Find the integral [tex]\int\limits {(2x-2016)\sqrt{x+2015}} \, dx[/tex] [tex](b)[/tex] The region [tex]R[/tex] is bounded by [tex]y=\cos{2x},\;y=\sin{2x},\;x-axis,\;y-axis,\text{in quadrant 1}[/tex]. Sketch the curves, showing clearly the region [tex]R[/tex]. Find the exact area of [tex]R[/tex]
Jawab:
Penjelasan dengan langkah-langkah:
a.
[tex]let~u=\sqrt{x+2015}~->du=\frac{1}{2\sqrt{x+2015}}dx\\\\~~~~u^2=x+2015\\\\~~~~~x=u^2-2015\\\\\\\int\limits {(2x-2016)\sqrt{x+2015}} \, dx\\\\=\int {(2(u^2-2015)-2016)\sqrt{x+2015}} \, (2\sqrt{x+2015}du) \\\\=2\int\limits {(2(u^2-2015)-2016)(x+2015)} \, du\\\\=2\int\limits {(2(u^2-2015)-2016)u^2} \, du\\\\=2\int\limits {(2u^2-6046)u^2} \, du\\\\=2\int\limits {2u^4-6046u^2} \, du\\\\=2(\frac{2}{5}u^5-\frac{6046}{3}u^3)+C\\\\=4u^3(\frac{1}{5}u^2-\frac{3023}{3})+C\\[/tex]
[tex]\\=4(\sqrt{x+2015})^3(\frac{1}{5}(x+2015)-\frac{3023}{3})+C\\\\=\frac{4}{15}(\sqrt{x+2015})^3(3(x+2015)-5.3023)+C\\\\=\frac{4}{15}(\sqrt{x+2015})^3(3x-9070)+C[/tex]
b.
[tex]the~intersection~point~at~cos2x=sin2x~->x=\frac{\pi}{8}\\\\\\L=\int\limits^{\pi/8}_0 {cos2x-sin2x} \, dx+\int\limits^{\pi/2}_{\pi/8} {sin2x-cos2x} \, dx \\\\~~~=\frac{1}{2}sin2x+\frac{1}{2}cos2x|^{\pi/8}_0+(-\frac{1}{2}cos2x-\frac{1}{2}sin2x)|^{\pi/2}_{\pi/8}\\\\~~~=(\frac{1}{2}sin\frac{\pi}{4}+\frac{1}{2}cos\frac{\pi}{4}-\frac{1}{2}sin0-\frac{1}{2}cos0)+(-\frac{1}{2}cos\pi-\frac{1}{2}sin\pi+\frac{1}{2}cos\frac{\pi}{4}+\frac{1}{2}sin\frac{\pi}{4})\\[/tex]
[tex]\\~~~=\frac{1}{4}\sqrt{2}+\frac{1}{4}\sqrt{2}-0-\frac{1}{2}+\frac{1}{2}-0+\frac{1}{4}\sqrt{2}+\frac{1}{4}\sqrt{2}\\\\~~~=\sqrt{2}[/tex]
4. find the area of the region bounded by the curve y=sinx, the line y=x and the line x=Phi
Jawab:
Penjelasan dengan langkah-langkah:
Area
curve y = sin (x)
line y = x and x = π
[tex]\sf A = \int_{0}^{\pi} (x ~dx) - 2\int_{0}^{\frac{\pi}{2}} sin~ x ~ dx[/tex]
[tex]\sf A = [\frac{1}{2}x^2]_{0}^{\pi} + 2 [cos ~ x]_{0}^{\frac{\pi}{2}}[/tex]
[tex]\sf A = \frac{1}{2}\pi^2 + 2 [ cos \frac{\pi}{2} - cos ~ 0 ][/tex]
[tex]\sf A = \frac{1}{2}\pi^2 + 2 [ 0 - 1][/tex]
[tex]\sf A = \frac{1}{2}\pi^2 - 2[/tex]
5. The region bounded by the chord and the arc is ...RadiiArcApothemSegment
Jawaban:
The region bounded by the chord and the arc is segment
Penjelasan dengan langkah-langkah:
Segment (dalam bahasa Indonesia: Tembereng) adalah daerahh yang dibatasi oleh garis busur dan tali busur
6. Find the area of the shaded region.
Sector area AOB :
θ/360° × πr²
= 58°/360° × π(5)²
= 29°/180° × 25π
= 145π/36
= 12.65 cm²
Missing side (AT) of triangle TAO:
tan(58°) = AT/5
AT = 5tan(58°)
AT= 8 cm
Area of triangle TAO:
1/2 × base × height
= 1/2 × AO × AT
= 1/2 × 5 × 8
= 20 cm²
Shaded region:
TAO - AOB
= 20 cm² - 12.65 cm²
= 7.35 cm²
7. What the applicant has enclosed along with the application letter
Penjelasan:
What the applicant has enclosed along with the application letter is Resume.
SEMOGAMEMBANTU
JADIKANJAWABANTERBAIK:)
BANTUSAYAUNTUKMENAIKANRANK:)
8. what is the function of sketch itself?
Can minimize errors in drawing or painting. Provides an overview of a drawing or painting theme. Can sharpen a painter's observations.
9. Pls help ThxThe diagram shows a square inscribed inside a circle, centre 0. Given that the area of the shaded region is 161 cm?, find the perimeter of the shaded region. Give your answer in the form (a + b) cm. [3] In the figure, ABCD is a square with side length equals 5 cm. Calculate the area of the shaded region. [3] B 'o D с
Jawab:
(4pi+16)cm
Penjelasan dengan langkah-langkah:
A= 16picm^2
16pi= 1/4 x pi x r^2
16= 1/4r^2
64=r^2
/64 = 8
r = 8
c= 1/4 x pi x (8x2)
c= 4pi
p= (4pi+16)cm
10. Hasan is designing a gate for a minimalist type of house. The actual length and height of the gate is 6 m and 2 m. If the length of the gate on the sketch is 15 cm, then the height of the gate on the sketch is...
Jawab:
Height gate on the sketch = 5 cm
Penjelasan dengan langkah-langkah:
example;
Height = H
Length = L
no. 1 = real gate
no. 2 = gate on the sketch
Length gate on cm = 6 m x 100 cm
= 600 cm
Height gate on cm = 2 m x 100 cm
= 200 cm
Formula;
[tex]\frac{a1}{a2}[/tex] = [tex]\frac{b1}{b2}[/tex]
[tex]\frac{height 1}{height 2} = \frac{length 1}{length 2}[/tex]
[tex]\frac{height1}{200 cm}[/tex]= [tex]\frac{15 cm}{600 cm}[/tex]
height 1 = 200 x [tex]\frac{15}{600}[/tex]
height 1 = [tex]\frac{3.000}{600}[/tex]
height 1 = 5 cm
So, Height gate on the sketch = 5 cm
11. Find the area of the blue region
Penjelasan dengan langkah-langkah:
Area of trapezium = (10 + 14) × 12 ÷ 2 cm²
Area of trapezium = 144 cm²
Area of triangle = 14 × 12 ÷ 2 cm²
Area of triangle = 84 cm²
Area of blue region = 144 cm² - 84 cm²
Area of blue region = 60 cm²
semoga membantu ya
12. Evaluating a Definite Integral Using a Geometric Formula In Exercises 23–32, sketch the region whose area is given by the definite integral. Then use a geometric formula to evaluate the integral (a>0, r>0).Bantuin Kalkulus nomor 32 donkkkk
Luas daerah yang dievaluasi adalah ½πr². Hal ini berarti:
[tex]\begin{aligned}\boxed{\vphantom{\Bigg|}\,\int_{-r}^{r}\sqrt{r^2-x^2}\,dx=\frac{1}{2}\pi r^2\,}\end{aligned}[/tex]
Diberikan integral:
[tex]\begin{aligned}\int_{-r}^{r}\sqrt{r^2-x^2}\,dx\end{aligned}[/tex]
dengan [tex]r > 0[/tex], yang akan dievaluasi dengan membuat sketsa daerahnya dan menghitung luasnya dengan rumus geometri.
Membuat Sketsa Daerah Integral
Kita tahu bahwa [tex]x^2 + y^2 = r^2[/tex] adalah persamaan lingkaran yang berpusat pada titik pusat koordinat [tex](0, 0)[/tex] dengan jari-jari [tex]r[/tex].
Maka, berdasarkan persamaan lingkaran tersebut, dapat diperoleh:
[tex]\begin{aligned}&y^2=r^2-x^2\\&\Rightarrow y=\pm\sqrt{r^2-x^2}\end{aligned}[/tex]
Artinya:
grafik [tex]y=\sqrt{r^2-x^2}[/tex] berbentuk setengah lingkaran pada arah sumbu-y positif, dangrafik [tex]y=-\sqrt{r^2-x^2}[/tex] berbentuk setengah lingkaran pada arah sumbu-y negatif.Oleh karena itu, daerah yang dievaluasi integralnya, yaitu [tex]y=\sqrt{r^2-x^2}[/tex] pada interval [tex][-r,r][/tex] berbentuk setengah lingkaran, dengan pusat (0,0) dan panjang jari-jari sebesar [tex]r[/tex]. Titik-titik potong pada sumbu koordinat adalah [tex](-r, 0)[/tex], [tex](0, r)[/tex], dan [tex](r, 0)[/tex].
Sketsa daerah pada sistem koordinat Cartesius terdapat pada gambar.
Evaluasi Integral Tentu dengan Rumus Geometri
Berdasarkan sketsa daerah yang telah digambarkan, luas daerah yang dievaluasi adalah setengah kali luas lingkaran dengan jari-jari [tex]r[/tex], yaitu ½πr².
Oleh karena itu, nilai integral tentu yang dievaluasi dapat dinyatakan dengan:
[tex]\displaystyle\int_{-r}^{r}\sqrt{r^2-x^2}\,dx=\frac{1}{2}\pi r^2[/tex]
[tex]\blacksquare[/tex]
________________
Tambahan:
Kita juga dapat menghitung integral tersebut dengan integral substitusi sebagai berikut.
[tex]\begin{aligned}&\int_{-r}^{r}\sqrt{r^2-x^2}\,dx\\&\quad\textsf{Substitusi trigonometri:}\\&\qquad x=r\sin u\\&\qquad\Rightarrow dx=r\cos u\,du\\&\quad\textsf{Interval $u$:}\\&\qquad u=\arcsin\left(\frac{x}{r}\right)\\&\qquad \Rightarrow \left[\arcsin\left(\frac{-r}{r}\right),\ \arcsin\left(\frac{r}{r}\right)\right]\\&\qquad \Rightarrow \left[\arcsin(-1),\ \arcsin(1)\right]\\&\qquad \Rightarrow \left[-\frac{\pi}{2},\ \frac{\pi}{2}\right]\\\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\int_{-\pi/2}^{\pi/2}r\cos u\sqrt{r^2-r^2\sin^2u}\,du\\&{=\ }\int_{-\pi/2}^{\pi/2}r\cos u\sqrt{r^2\left(1-\sin^2u\right)}\,du\\&{=\ }\int_{-\pi/2}^{\pi/2}r\cos u\sqrt{r^2\cos^2u}\,du\\&{=\ }\int_{-\pi/2}^{\pi/2}r\cos u\cdot r\cos u\,du\\&{=\ }\int_{-\pi/2}^{\pi/2}r^2\cos^2u\,du\\&{=\ }r^2\cdot\int_{-\pi/2}^{\pi/2}\cos^2u\,du\\&{=\ }r^2\cdot\int_{-\pi/2}^{\pi/2}\frac{1+\cos2u}{2}\,du\\&{=\ }\frac{1}{2}r^2\cdot\int_{-\pi/2}^{\pi/2}(1+\cos2u)\,du\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1}{2}r^2\left[u+\frac{1}{2}\sin2u\right]_{-\pi/2}^{\pi/2}\\&{=\ }\frac{1}{2}r^2\left[\left(\frac{\pi}{2}+\frac{1}{2}\sin(\pi)\right)-\left(-\frac{\pi}{2}+\frac{1}{2}\sin(-\pi)\right)\right]\\&{=\ }\frac{1}{2}r^2\left[\left(\frac{\pi}{2}+0\right)-\left(-\frac{\pi}{2}+0\right)\right]\\&{=\ }\frac{1}{2}r^2\left[\frac{\pi}{2}+\frac{\pi}{2}\right]\\&{=\ }\frac{1}{2}r^2\cdot\pi\\&{=\ }\boxed{\,\frac{1}{2}\pi r^2\,}\\\end{aligned}[/tex]
[tex]\blacksquare[/tex]
13. Find the volume of the solid obtained by rotating the region bounded by two parabolas
Penjelasan:
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14. what is the meaning of Sketch
Jawaban:
Sketch is a rough or unfinished drawing or painting, often made to assist in making a more finished picture.
Hope it helps!
15. Need this urgentlyThe diagram shows a square inscribed inside a circle, centre 0. Given that the area of the shaded region is 161 cm2, find the perimeter of the shaded region. Give your answer in the form (a + b) cm. [3]
Jawab:
O is also center of square, so angle O = 90° (intersection of diagonals)
shaded area = [tex]\frac{90}{360}[/tex] × area of circle
16π = [tex]\frac{90}{360}[/tex] × πr²
r² = 16 ÷ [tex]\frac{90}{360}[/tex]
r = [tex]\sqrt{64}[/tex]
r = 8 cm
perimeter = 2×radius + 1/4 circumference of circle
= 2r + 1/4 × π × 2r
= 2 × 8 + 1/4 × π × 2 × 8
= (16 + 4π) cm
16. The figure shows circle of radius 14 cm with one quadrant removed, touching the sides of square. Find @ the perimeter of the unshaded region, (i) the area of the unshaded region, (iii) the area of the shaded region.
di antara nilai p berikut yang memenuhi proporsi p:7 21:49 adalah
17. the weight of the body is more at polar region of the Earth then at the equatorial region why
Jawaban:
As the distance of the pole is less than the distance of the equator from the center of the earth, the force of attraction is higher on the body at poles than at the equator. Hence the weight of a body is greater at pole than at the equator
#ihopethishelps
18. Find the maximum and minimum point for y=3+4x³-x⁴. Sketch the graph of the curve.
Jawab:
Penjelasan dengan langkah-langkah:
[tex]y = 3+4x^3 - x^4\\\\y' = 12x^2 - 4x^3 = 0\\\\x^3 - 3x^2 = 0\\\\x^2(x-3) = 0\\\\x = 0 \;\cup\; 3\\\\y'' = 24x - 12x^2 = 0\\\\x^2 - 2x = 0\\\\x(x-2) = 0\\\\x = 0 \;\cup\; 2\\\\y''' = 24-24x = 0 => x = 1\\\\ y'''' = -24\\\\\begin{minipage}{15em}berdasarkan turunan keempat, ketiga, kedua, dan pertama :\\\\- f(x) terbuka keatas dan memiliki minimum pada interval $0 < x < 2$ \\\\- f(x) terbuka kebawah dan \\memiliki maksimum pada \\interval $x < 0$ atau $x > 2$ \\\\\end{minipage}[/tex]
[tex]\begin{minipage}{15 em}- f(x) memiliki perubahan kecekungan parabola singgung terbesar pada x = 1\\\\- berdasarkan turunan kedua maka f(x) memiliki maksimum lokal di x = 3\\\\- x = 0 pada interval $ 0< x < 2 $ bukan merupakan maksimum maupun minimum, tetapi secara keseluruhan tanpa interval x = 0 merupakan minimum f(x) (karena maksimum f(x) sudah ditemukan)\end{minipage}[/tex]
[tex]\begin{minipage}{15 em}- Nilai turunan keempat negatif dan mempengaruhi sifat turunan diatasnya \\- f(x) monoton turun ketika $x > 3$ dan monoton naik untuk \\$x < 0 \;\cap\; 0<x<3$ \\(netral pada x = 0)\end{minipage}[/tex]
maximum = f(3), "minimum" = f(0) = 3 = titik potong sumbu y
maximum = 3+4(3)³ - 3⁴ = 30
19. the region in indonesia that had touched off by tsunami was
aceh province, sumateraaceh, pangandaran, krakatau, pesisir sumatra, ciamis, kebumen
20. Polar bears live in the arctic region(-) …? Do Polar bears Ilive in the arctic region ?
Jawaban:
Polar bears live in the arctic region
(-) Polar bears don't not live in the arctic region
(?) Do polar bears live in the arctic region?
Jawaban:
(-) Polar bears don'tlive in the arctic region.
Penjelasan:
(+) Polar bears live in the arctic region.
(-) Polar bears don'tlive in the arctic region.
(?) Do polar bears live in the arctic region?
your welcome~
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