$56 c - 10 c^{2}$ is subtracted from the sum of $10 c^{2}$ and $(- 96 c^{2} - 23 c)$ . Find the desired result.

Open in App

Solution

Initially, let's find the sum of $10 c^{2}$ and $(- 96 c^{2} - 23 c)$
$= 10 c^{2} + (- 96 c^{2} - 23 c)$
$= 10 c^{2} - 96 c^{2} - 23 c$
$= - 86 c^{2} - 23 c$
Now $56 c - 10 c^{2}$ needs to be subtracted from $- 86 c^{2} - 23 c$
$= - 86 c^{2} - 23 c - (56 c - 10 c^{2})$
$= - 86 c^{2} - 23 c - 56 c + 10 c^{2}$
$= - 86 c^{2} + 10 c^{2} - 23 c - 56 c$
$= - 76 c^{2} - 79 c$

Suggest Corrections

Similar questions

12 x^{2} - 9 y + 129 - 16 x^{2}

9 m y + 10 m^{2} a n d - 9 x y - 20 m y^{2}

56 c - 10 c^{2}

10 c^{2}

- 96 c^{2} - 23 c

(i) solve 12x² - 9y + 129 - 16x²

(ii)find the sum of 9my + 10my² and -9xy - 20my²

(iii) 56c - 10c² is subtracted from the sum of 10c² and (-96c² -23c). Find the desired result.

\frac{1}{2}

^{10} C_{0} -

^{10} C_{1} +

2 \cdot^{10} C_{2} -

2^{2} \cdot^{10} C_{3}

2^{9} \cdot^{10} C_{10}

^{10} C_{1} +^{10} C_{2} +^{10} C_{3} + . . . +^{10} C_{9}

3^{1} \times^{10} C_{1} + 3^{2} \times^{10} C_{2} + 3^{3} \times^{10} C_{3} + . . . . + 3^{10} \times^{10} C_{10} =

Join BYJU'S Learning Program