evaluate the following without using a calculator. (LOGARITMA)
1. evaluate the following without using a calculator. (LOGARITMA)
Jawaban:
ya n dak tau tanya kok sama saya
2. Evaluate Limit Involving Trigonometric Function
Jawaban:
0
Penjelasan dengan langkah-langkah:
[tex]lim_{x \to0} \frac{secx - 1}{x} \\ = lim_{x \to0} \frac{ \frac{1}{cosx} - 1}{x} \\ = lim_{x \to0} \frac{ \frac{1 - cosx}{cosx} }{x} \\ = lim_{x \to0} \frac{1 - cosx}{xcosx}. \frac{1 + cosx}{1 + cosx} \\ = lim_{x \to0} \frac{1 - {cos}^{2} x}{xcosx(1 + cosx)} \\ = lim_{x \to0} \frac{{sin}^{2} x}{xcosx(1 + cosx)} \\ = lim_{x \to0} \frac{sinx}{x}. \frac{sinx}{cosx} . \frac{1}{1 + cosx} \\ = 1. \frac{sin0}{cos0} . \frac{1}{1 + cos0} \\ = 1.0. \frac{1}{2} \\ = 0[/tex]
Jawab:
[tex]\displaystyle \lim_{x\to0}\frac{\sec x-1}{x}=0[/tex]
Penjelasan dengan langkah-langkah:
[tex]\displaystyle \lim_{x\to0}\frac{\sec x-1}{x}=\lim_{x\to0}\frac{\frac1{\cos x}-1}{x}\\\lim_{x\to0}\frac{\sec x-1}{x}=\lim_{x\to0}\frac{1-\cos x}{x\cos x}\\\lim_{x\to0}\frac{\sec x-1}{x}=\lim_{x\to0}\frac{1-\cos x}{x}\cdot\lim_{x\to0}\frac1{\cos x}\\\lim_{x\to0}\frac{\sec x-1}{x}=\lim_{x\to0}\frac{2(1-\cos x)}{2x}\cdot\lim_{x\to0}\frac1{\cos x}\\\lim_{x\to0}\frac{\sec x-1}{x}=\lim_{x\to0}\frac{2\sin^2\frac12x}{x}\cdot\lim_{x\to0}\frac1{\cos x}\\\lim_{x\to0}\frac{\sec x-1}{x}=\lim_{x\to0}\sin\frac12x\cdot\lim_{x\to0}\frac{2\sin\frac12x}{x}\cdot\lim_{x\to0}\frac1{\cos x}\\\lim_{x\to0}\frac{\sec x-1}{x}=\sin\frac12(0)\cdot2\cdot\frac12\cdot\frac1{\cos 0}\\\lim_{x\to0}\frac{\sec x-1}{x}=0\cdot1\cdot1\\\boxed{\boxed{\lim_{x\to0}\frac{\sec x-1}{x}=0}}[/tex]
3. evaluate the following
Jawaban:
yu Wi fi sporty dan rggjaija
Penjelasan dengan langkah-langkah:
uaiaisnsmks
Jawaban:
a. [tex] - \frac{11}{12} [/tex]
b. [tex] - \frac{9}{14} [/tex]
maaf kalo salah ya
4. Complete the following sentences using the correct answers!
Jawaban:
to cook, has, she, they, my, it
Penjelasan:
5. respond to the following statements using the expressions
Answer:
________
2.you had better call her now
3. I would like to congratulate you for your new novel that will be lounched next month.
4. It's amazing. it's beautiful, isn't it?
5. Thank you for your appreciation
6. I hope your test will run well and you'll get a good result.
*Thanks*
6. determine the solving of 3x+4y-20=0 & 4x=3y+10 by using substitution
4x=3y+10 x=3y+10/4 3x+4y-20=0 3(3y+10/4)-20=0 9y+30/4-20=0 9y+30-80=0 9y-50=0 9y=50 y=50/9
7. Evaluate the each following -6², (-9)³, -3³, (-5)² tolong jawaban nya kk
-36, -729, -27, 25
Maaf kalo salah
Jawab:
36, (-729), (-27), 25
Penjelasan dengan langkah-langkah:
-6²= (-6)x(-6)=36
(-9)³= (-9)x(-9)x(-9)=81x(-9)=(-729)
(-3)³=(-3)x(-3)x(-3)=9x(-3)=(-27)
(-5)²=(-5)x(-5)=25
8. the meaning of substitution
substituion is a method of changing one thing with another.
substitusi adalah cara untuk mengganti/mengubah satu hal dengan yang lainnyajawabannya pengganti atau digantikan atau semisalnya
9. Complete the following table using the suitable expression!
b. could you turn on the AC? accepting : alright refusing: I'm sorry but I'm busy right now
c. could you close the door for me please? accepting : ok! refusing: I can't im doing something else!
d. could you help me with my homework? accepting: sure thing. refusing: I'm sorry but I haven't done it too
e. could you help me clean the whiteboard? accepting: will do! refusing: im afraid I couldn't help you right now
10. 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]
11. Quiz Math Part 3 Determine value of x and y form following linear equations using the substitution method. 2x + 3y = 4 4x + 3y = 13 Happy answer
X= 4 1/2 Y= - 5/3 smoga membantu ya
12. State the following trigonometric equations[tex] \cos(x + 35 ) = 0.5[/tex]35 itu degree ya.35 darjah
Jawab:
25⁰, 265⁰
Penjelasan dengan langkah-langkah:
13. respond to the following statements using the expressions
7. Wow, really? Where will we move?
8. For Real? Im so nervous
9.wow! congratulations for winning the third place this year's singing contest
14. Complete the sentences using the following verb
Jawaban:
Translate
Lengkapi Kalimat Menggunakan kata kerja
im sorry but where is the sentence?
15. By using a suitable algebraic identity, evaluate each of the following. (a) 904² (b) 791²(c) 603 x 597 (d) 99x 101
Penjelasan dengan langkah-langkah:
a. 904²=904+904
b. lakukan hal yg sama
c. 603 dikali 597 =359.991
d. 9.999
16. Evaluate the each following: -√225
Jawaban:
-√225= -15
smg membantu kak :>
17. . Evaluate the integral xe^4x²+1 dx
Integral dari [tex]\int x.e^{4x^2+1}~dx[/tex] adalah [tex]\frac{e^{4x^2+1}}{8} + C[/tex]Integral dari [tex]\int x.e^{4x^2}+1~dx[/tex] adalah [tex]\frac{e^{4x^2}}{8}+x + C[/tex]
.
PENDAHULUANIntegral adalah operasi invers atau operasi kebalikan dari turunan. Integral dapat memuat fungsi Aljabar dan fungsi trigonometri. Pada Integral Aljabar ,terbagi dalam 3 penyelesaian.
.
[tex]\textbf{Penyelesaian dengan cara Umum : } \\\int f(x) = \int ax^n~dx= \frac{a}{n+1}x^{n+1} \\\\\int f(x) = \int k(x^n)~dx = k \int x^n~dx= \frac{k}{n+1}x^{n+1} \\\\\int (f(x) \pm g(x))~dx = \int f(x)~dx \pm \int g(x)~dx \\\\[/tex]
.
[tex]\textbf{Penyelesaian dengan cara Subtitusi : } \\\int f(x).g(x)~dx = \int \frac{f(x)}{g'(x)}.u~du \\\\dimana : \\u : fungsi~g(x) \\f(x) : fungsi~berderajat~n \\g(x) : fungsi~berderajat~(n+1) \\\\[/tex]
.
[tex]\textbf{Penyelesaian dengan cara Parsial : } \\\int f(x).g(x)~dx \to \int u~dv = u.v - \int v~du \\\\dimana : \\u : fungsi~f(x)~atau~g(x)~yang~dapat~diturunkan~hingga~nol. \\dv : fungsi~f(x)~atau~g(x)~yang~tidak~memiliki~limitasi~turunan.[/tex]
Mari simak penyelesaian berikut!
.
PEMBAHASANDIKETAHUI
[tex]\int x.e^{4x^2+1}~dx[/tex][tex]\int x.e^{4x^2}+1~dx[/tex].
DITANYA
[tex]Nilai~integral~...~?[/tex]
.
JAWAB
1. PENYELESAIAN PERTAMA.
Integral tersebut dapat diselesaikan dengan cara Subtitusi.
[tex]Misal : \\4x^2 +1 = u~~~\to~~~du = 8x~dx \\\\\boxed{dx = \frac{du}{8x}}[/tex]
.
[tex]\int x.e^{4x^2+1}~dx = \int x.e^{u}~\frac{du}{8x} \\\\~~~~~~~~~~~~~~~~~~~ = \frac{1}{8} \int e^u~du \\\\~~~~~~~~~~~~~~~~~~~ = \frac{1}{8} e^u + C \\\\~~~~~~~~~~~~~~~~~~~ = \frac{1}{8}e^{4x^2+1} + C \\\\\boxed{\boxed{\int x.e^{4x^2+1}~dx = \frac{e^{4x^2+1}}{8} + C}}[/tex]
.
2. PENYELESAIAN KEDUA.
Integral tersebut dapat diselesaikan dengan cara Subtitusi.
[tex]Misal : \\4x^2 = u~~~\to~~~du = 8x~dx \\\\\boxed{dx = \frac{du}{8x}}[/tex]
.
[tex]\int x.e^{4x^2}+1~dx = \int x.e^{u}~\frac{du}{8x} +\int 1~dx\\\\~~~~~~~~~~~~~~~~~~~~~ = \frac{1}{8} \int e^u~du+x +c_2 \\\\~~~~~~~~~~~~~~~~~~~~~ = \frac{1}{8} (e^u)+c_1+x+c_2 \\\\~~~~~~~~~~~~~~~~~~~~~ = \frac{1}{8}(e^{4x^2})+x + C \\\\\boxed{\boxed{\int x.e^{4x^2}+1~dx = \frac{e^{4x^2}}{8}+x + C}}[/tex]
.
KESIMPULAN
Jadi, nilai integral tersebut adalah
[tex]Jika~ \int x.e^{4x^2+1}~dx = \frac{e^{4x^2+1}}{8} + C[/tex][tex]Jika~ \int x.e^{4x^2}+1~dx = \frac{e^{4x^2}}{8}+x + C[/tex].
Catatan : Karena penulisan fungsi integral bernilai ambigu, silahkan sesuaikan dengan Integral dari persoalan yang dimaksud.
.
PELAJARI LEBIH LANJUTMateri Integral Parsial : brainly.co.id/tugas/28945863
Materi Integral Subtitusi : brainly.co.id/tugas/30095944
Materi Integral Trigonometri : brainly.co.id/tugas/28945391
.
_______________________________________________
DETAIL JAWABANKelas : 11
Mapel : Matematika
Materi : Integral Tak Tentu Fungsi Aljabar
Kode Kategorisasi : 11.2.10
18. complete the following sentensces using the correct pronoun.
Jawaban:
2. Them
3. Him
4. it
5. Her
6. Us
7. They
8. They
9. She
10. We
11. He
Saya minta maaf jika ada kesalahan saya sudah bantu sebisa saya semoga benar semua :)
19. Evaluate the following expression: 9 / 3 + 6 *5 = ......
Jawaban:
ga ngerti
Penjelasan:
cuman mau ambil balik poin :))
20. Using the substitution method ,solve the simultaneous equations 7x-2y:21 4x+y:57
7x - 2y = 21 4x + y = 57 That should be changed (just choose what do you want to change) : Y = -4x + 57 Enter to one equal: 7x - 2(-4x + 57) = 21 7x + 8x - 114 = 21 7x + 8x = 21 + 114 15x = 135 X = 9 And now, we gonna find the "y". 7(9) - 2y = 21 63 - 2y = 21 -2y = 21-63 -2y = -42 Y = -42/-2 = 21 Finished! And now you can prove it.
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