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2 years ago

A theatre ticket costs £63 plus a booking fee of 3% what is the total cost of the ticket

Mathematics

2 answers:

Genrish500 [490]2 years ago

5 0

Answer:

64.89 is the total cost

Step-by-step explanation:

63 x .03 = 1.89

63 + 1.89 = 64.89

Sophie [7]2 years ago

5 0

Answer:64.89

Step-by-step explanation:

63*.03=1.89

63+1.89=64.89

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24. ABCDE and PQRST are regular pentagons.

SIZIF [17.4K]

Answer:

126°

Step-by-step explanation:

In quadrilateral DSRC, SR || DC (given)

AP = BQ = CR = DS = ET (given)

$\therefore$ CR bisects $\angle DCB$

$\therefore m\angle DCR=\frac{108\degree}{2}$

( $108\degree$ is the measure of an angle of pentagon)

$\therefore m\angle DCR=54\degree$

$\because m\angle DCR + m\angle SRC=180\degree$

(Measure of interior angles on the same side of transversal)

$\therefore 54\degree + m\angle SRC=180\degree$

$\therefore m\angle SRC=180\degree- 54\degree$

$\therefore m\angle SRC=126\degree$

3 0

2 years ago

Which expression produces a compatible number estimate for (StartFraction 4 over 7 EndFraction) (StartFraction 3 over 8 EndFract

Mariulka [41]

Answer:

(1/2)(1/2)(1/2)(1/2)

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Which of the following is the expansion of (3c + d2)6?

vovangra [49]

Answer: Option A) $729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$ is the correct expansion.

Explanation:

on applying binomial theorem, $(a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r$

Here a=3c, $b=d^2$ and n=6,

Thus, $(3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r$

⇒ $(3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6$

⇒ $(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}$

⇒ $(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}$