Answer:
The quotient is 3x - 11 + 60/(x + 5) ⇒ 2nd answer
Step-by-step explanation:
* We will use the long division to solve the problem
- The dividend is 3x² + 4x + 5
- The divisor is x + 5
- The quotient is the answer of the division
- If the divisor not a factor of a dividend, the quotient has
a remainder
* Lets solve the problem
- At first divide the first term in the dividend by the first term in
the divisor
∵ 3x² ÷ x = 3x
- Multiply the divisor by 3x
∴ 3x (x + 5) = 3x² + 15x
-Subtract this expression from the dividend
∴ 3x² + 4x + 5 - (3x² + 15x) = 3x² + 4x + 5 - 3x² - 15x = -11x + 5
- Divide the first term -11x in the new dividend by the first
term x in the divisor
∴ -11x ÷ x = -11
- Multiply the divisor by -11
∴ -11(x + 5) = -11x - 55
-Subtract this expression from the new dividend
∴ -11x + 5 - (-11x - 55) = -11x + 5 + 11x + 55 = 60
∴ The quotient is 3x - 11 with remainder = 60
* The quotient is 3x - 11 + 60/(x + 5)
Answer:
I can’t just tell you the answer cause that’s all your looking for let me know what u need help with
Step-by-step explanation:
Answer:
A. 2,400π cubic units
Step-by-step explanation:
Answer:
The correct options are;
1) ΔBCD is similar to ΔBSR
2) BR/RD = BS/SC
3) (BR)(SC) = (RD)(BS)
Step-by-step explanation:
1) Given that RS is parallel to DC, we have;
∠BDC = ∠BRS (Angles on the same side of transversal)
Similarly;
∠BCD = ∠BSR (Angles on the same side of transversal)
∠CBD = ∠CBD = (Reflexive property)
Therefore;
ΔBCD ~ ΔBSR Angle, Angle Angle (AAA) rule of congruency
2) Whereby ΔBCD ~ ΔBSR, we therefore have;
BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1
1 + SC/BS = 1 + RD/BR = SC/BS = 1 + BR/RD - 1
SC/BS = RD/BR
Inverting both sides
BR/RD = BS/SC
3) From BR/RD = BS/SC the above we have by cross multiplication;
BR/RD = BS/SC gives;
BR × SC = RD × BR → (BR)(SC) = (RD)(BR).
