✧❀⌨︎QUIZ⌨︎❀✧ [tex]1.) \bold{\boxed{solve\:for\:x,\:\:\sqrt{x+15}+\sqrt{x}=15}}\\\\2.)\bold{\boxed{gcf\:\:35y^4,\:\:14y^4,\:\:63y^4}}\\\\3.)\bold{\boxed{implicit\:derivative\:\frac{dy}{dx},\:\:\left(x-y\right)^2=x+y-1}}[/tex]

Berikut ini adalah pertanyaan dari Xxcrty4 pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Atas

✧❀⌨︎QUIZ⌨︎❀✧1.) \bold{\boxed{solve\:for\:x,\:\:\sqrt{x+15}+\sqrt{x}=15}}\\\\2.)\bold{\boxed{gcf\:\:35y^4,\:\:14y^4,\:\:63y^4}}\\\\3.)\bold{\boxed{implicit\:derivative\:\frac{dy}{dx},\:\:\left(x-y\right)^2=x+y-1}}

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Jawaban:

1. x = 49

2. jika dikali hasilnya = 30.870y¹²

jika ditambah hasilnya = 112y⁴

3.

 \frac{dy}{dx} \: = \frac{1 - 2x + 2y}{ - 2x + 2y - 1}

Penjelasan dengan langkah-langkah:

cara nya ada digambar

makasih

semoga membantu

✏JAWABAN:1. Solve For x =[tex] \sqrt{x + 15} + \sqrt{x} = 15[/tex][tex] \sqrt{x + 15} = 15 - \sqrt{x} [/tex][tex]x + 15 = 225 - 30 \sqrt{x + x} [/tex][tex]30 \sqrt{x} = 225 - 15[/tex][tex]30 \sqrt{x} = 210[/tex][tex] \sqrt{ x } = 210 \div 30[/tex][tex] \sqrt{x} = 7[/tex][tex]x = 49[/tex]_______2. GCF =[tex]35y {}^{4} = 35 {}^{4} \times y[/tex][tex]14 {y}^{4} = 14 {}^{4} \times y[/tex][tex]63 {y}^{4} = 63 {}^{4} \times y[/tex][tex] = 14 {}^{4} \times y[/tex]_______3. Derivative =[tex](x - y) {}^{2} = x + y - 1[/tex][tex]\frac{d}{dx} ((x - y) {}^{2} ) = \frac{d}{dx} (x) + \frac{d}{dx} (y) - \frac{d}{dx} (1)[/tex][tex] \frac{d}{dg} ( {g}^{2} ) \times \frac{d}{dx} (x - y) = \frac{d}{dx} (x) + \frac{d}{dx} (y) - 0[/tex][tex]2g \times (1 - \frac{d}{dx} (y)) = 1 + \frac{d}{dy} (y) \times \frac{dy}{dx} [/tex][tex]2(x + y) \times (1 - \frac{d}{dx} (y)) = 1 + 1 \times \frac{dy}{dx} [/tex][tex]2x - 2x \times \frac{d}{dx} (y) - 2y + 2x \times \frac{d}{dx} (y) = 1 \times \frac{dy}{dx}[/tex][tex]2x - 2x \times \frac{d}{dy} (y) \times \frac{dy}{dx} - 2y + 2y \times \frac{d}{dy} (y) \times \frac{dy}{dx} = 1 + \frac{dy}{dx} [/tex][tex] - 2x \times \frac{dy}{dx} + 2y \times \frac{dy}{dx} - \frac{dy}{dx} = 1 - 2x + 2y[/tex][tex]( - 2 x + 2y - 1) \times \frac{dy}{dx} = 1 - 2x + 2y[/tex][tex] \frac{dy}{dx} = \frac{1 - 2x + 2y}{ - 2x + 2y - 1} [/tex]✏JAWABAN:1. Solve For x =[tex] \sqrt{x + 15} + \sqrt{x} = 15[/tex][tex] \sqrt{x + 15} = 15 - \sqrt{x} [/tex][tex]x + 15 = 225 - 30 \sqrt{x + x} [/tex][tex]30 \sqrt{x} = 225 - 15[/tex][tex]30 \sqrt{x} = 210[/tex][tex] \sqrt{ x } = 210 \div 30[/tex][tex] \sqrt{x} = 7[/tex][tex]x = 49[/tex]_______2. GCF =[tex]35y {}^{4} = 35 {}^{4} \times y[/tex][tex]14 {y}^{4} = 14 {}^{4} \times y[/tex][tex]63 {y}^{4} = 63 {}^{4} \times y[/tex][tex] = 14 {}^{4} \times y[/tex]_______3. Derivative =[tex](x - y) {}^{2} = x + y - 1[/tex][tex]\frac{d}{dx} ((x - y) {}^{2} ) = \frac{d}{dx} (x) + \frac{d}{dx} (y) - \frac{d}{dx} (1)[/tex][tex] \frac{d}{dg} ( {g}^{2} ) \times \frac{d}{dx} (x - y) = \frac{d}{dx} (x) + \frac{d}{dx} (y) - 0[/tex][tex]2g \times (1 - \frac{d}{dx} (y)) = 1 + \frac{d}{dy} (y) \times \frac{dy}{dx} [/tex][tex]2(x + y) \times (1 - \frac{d}{dx} (y)) = 1 + 1 \times \frac{dy}{dx} [/tex][tex]2x - 2x \times \frac{d}{dx} (y) - 2y + 2x \times \frac{d}{dx} (y) = 1 \times \frac{dy}{dx}[/tex][tex]2x - 2x \times \frac{d}{dy} (y) \times \frac{dy}{dx} - 2y + 2y \times \frac{d}{dy} (y) \times \frac{dy}{dx} = 1 + \frac{dy}{dx} [/tex][tex] - 2x \times \frac{dy}{dx} + 2y \times \frac{dy}{dx} - \frac{dy}{dx} = 1 - 2x + 2y[/tex][tex]( - 2 x + 2y - 1) \times \frac{dy}{dx} = 1 - 2x + 2y[/tex][tex] \frac{dy}{dx} = \frac{1 - 2x + 2y}{ - 2x + 2y - 1} [/tex]✏JAWABAN:1. Solve For x =[tex] \sqrt{x + 15} + \sqrt{x} = 15[/tex][tex] \sqrt{x + 15} = 15 - \sqrt{x} [/tex][tex]x + 15 = 225 - 30 \sqrt{x + x} [/tex][tex]30 \sqrt{x} = 225 - 15[/tex][tex]30 \sqrt{x} = 210[/tex][tex] \sqrt{ x } = 210 \div 30[/tex][tex] \sqrt{x} = 7[/tex][tex]x = 49[/tex]_______2. GCF =[tex]35y {}^{4} = 35 {}^{4} \times y[/tex][tex]14 {y}^{4} = 14 {}^{4} \times y[/tex][tex]63 {y}^{4} = 63 {}^{4} \times y[/tex][tex] = 14 {}^{4} \times y[/tex]_______3. Derivative =[tex](x - y) {}^{2} = x + y - 1[/tex][tex]\frac{d}{dx} ((x - y) {}^{2} ) = \frac{d}{dx} (x) + \frac{d}{dx} (y) - \frac{d}{dx} (1)[/tex][tex] \frac{d}{dg} ( {g}^{2} ) \times \frac{d}{dx} (x - y) = \frac{d}{dx} (x) + \frac{d}{dx} (y) - 0[/tex][tex]2g \times (1 - \frac{d}{dx} (y)) = 1 + \frac{d}{dy} (y) \times \frac{dy}{dx} [/tex][tex]2(x + y) \times (1 - \frac{d}{dx} (y)) = 1 + 1 \times \frac{dy}{dx} [/tex][tex]2x - 2x \times \frac{d}{dx} (y) - 2y + 2x \times \frac{d}{dx} (y) = 1 \times \frac{dy}{dx}[/tex][tex]2x - 2x \times \frac{d}{dy} (y) \times \frac{dy}{dx} - 2y + 2y \times \frac{d}{dy} (y) \times \frac{dy}{dx} = 1 + \frac{dy}{dx} [/tex][tex] - 2x \times \frac{dy}{dx} + 2y \times \frac{dy}{dx} - \frac{dy}{dx} = 1 - 2x + 2y[/tex][tex]( - 2 x + 2y - 1) \times \frac{dy}{dx} = 1 - 2x + 2y[/tex][tex] \frac{dy}{dx} = \frac{1 - 2x + 2y}{ - 2x + 2y - 1} [/tex]✏JAWABAN:1. Solve For x =[tex] \sqrt{x + 15} + \sqrt{x} = 15[/tex][tex] \sqrt{x + 15} = 15 - \sqrt{x} [/tex][tex]x + 15 = 225 - 30 \sqrt{x + x} [/tex][tex]30 \sqrt{x} = 225 - 15[/tex][tex]30 \sqrt{x} = 210[/tex][tex] \sqrt{ x } = 210 \div 30[/tex][tex] \sqrt{x} = 7[/tex][tex]x = 49[/tex]_______2. GCF =[tex]35y {}^{4} = 35 {}^{4} \times y[/tex][tex]14 {y}^{4} = 14 {}^{4} \times y[/tex][tex]63 {y}^{4} = 63 {}^{4} \times y[/tex][tex] = 14 {}^{4} \times y[/tex]_______3. Derivative =[tex](x - y) {}^{2} = x + y - 1[/tex][tex]\frac{d}{dx} ((x - y) {}^{2} ) = \frac{d}{dx} (x) + \frac{d}{dx} (y) - \frac{d}{dx} (1)[/tex][tex] \frac{d}{dg} ( {g}^{2} ) \times \frac{d}{dx} (x - y) = \frac{d}{dx} (x) + \frac{d}{dx} (y) - 0[/tex][tex]2g \times (1 - \frac{d}{dx} (y)) = 1 + \frac{d}{dy} (y) \times \frac{dy}{dx} [/tex][tex]2(x + y) \times (1 - \frac{d}{dx} (y)) = 1 + 1 \times \frac{dy}{dx} [/tex][tex]2x - 2x \times \frac{d}{dx} (y) - 2y + 2x \times \frac{d}{dx} (y) = 1 \times \frac{dy}{dx}[/tex][tex]2x - 2x \times \frac{d}{dy} (y) \times \frac{dy}{dx} - 2y + 2y \times \frac{d}{dy} (y) \times \frac{dy}{dx} = 1 + \frac{dy}{dx} [/tex][tex] - 2x \times \frac{dy}{dx} + 2y \times \frac{dy}{dx} - \frac{dy}{dx} = 1 - 2x + 2y[/tex][tex]( - 2 x + 2y - 1) \times \frac{dy}{dx} = 1 - 2x + 2y[/tex][tex] \frac{dy}{dx} = \frac{1 - 2x + 2y}{ - 2x + 2y - 1} [/tex]

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Last Update: Thu, 15 Jul 21