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3 years ago

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А (b) Two ovens are used to bake banana cakes and chocolate cakes. Each oven will ding according to its setting time. A banana c

ake takes 50 minutes to bake wbile a chocolate cake takes 45 minutes to bake.
(i) If each cake is put into the ovens at the same time, what is the time taken until both ovens ding simultaneously?

(ii) When the both ovens ding, how many banana cakes and chocolate cakes are baked?

1 answer:

ruslelena [56]3 years ago

8 0

Answer:

(b)(i) LCM of 45 and 50 = 450

Hence, it take 450 minutes until both ovens ding simultaneously.

(b)(ii) 450÷50=9 banana cakes baked

450÷45=10 chocolate cakes baked.

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vladimir2022 [97]

Answer:

the first choice

Step-by-step explanation:

$- 5 {x}^{2} + 7x + 3$

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3 years ago

The diameter of a planet is about 10,652 mi. The diameter of the planet's moon is about 25% of the planet. What percent of the v

alekssr [168]

Answer:

$2\%$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

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x----> the volume of the planet's moon

y---> the volume of the planet

so

$z^{3}=\frac{x}{y}$

In this problem we have

$z=0.25$

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3 years ago

It cost 5 dollars for a child ticket and 8 dollars for a adult ticket. Total tickets sold were 110 bringing in 820 dollars? How

omeli [17]

Number of child tickets bought is 20

<h3><u>Solution:</u></h3>

Given that It cost 5 dollars for a child ticket and 8 dollars for a adult ticket

cost of each child ticket = 5 dollars

cost of each adult ticket = 8 dollars

Let "c" be the number of child tickets bought

Let "a" be the number of adult tickets bought

Total tickets sold were 110 bringing in 820 dollars

<em>Number of child tickets bought + number of adult tickets bought = 110</em>

c + a = 110 ----- eqn 1

<em><u>Also we can frame a equation as:</u></em>

Number of child tickets bought x cost of each child ticket + number of adult tickets bought x cost of each adult ticket = 820

$c \times 5 + a \times 8 = 820$

5c + 8a = 820 -------- eqn 2

Let us solve eqn 1 and eqn 2 to find values of "c" and "a"

From eqn 1,

a = 110 - c ------ eqn 3

Substitute eqn 3 in eqn 2

5c + 8(110 - c) = 820

5c + 880 - 8c = 820

-3c = - 60

c = 20

Therefore from eqn 3,

a = 110 - 20 = 90

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Therefore number of child tickets bought is 20

8 0

3 years ago

Dont really understand how to do this

bija089 [108]

Answers:

$t_{10} = -22 \ \text{ and } S_{10} = -85$

========================================================

Explanation:

$t_1 = \text{first term} = 5\\t_2 = \text{first term}-3 = t_1 - 3 = 5-3 = 2$

Note we subtract 3 off the previous term (t1) to get the next term (t2). Each new successive term is found this way

$t_3 = t_2 - 3 = 2-3 = -1\\t_4 = t_3 - 3 = -1-3 = -4$

and so on. This process may take a while to reach $t_{10}$

There's a shortcut. The nth term of any arithmetic sequence is

$t_n = t_1+d(n-1)$

We plug in $t_1 = 5 \text{ and } d = -3$ and simplify

$t_n = t_1+d(n-1)\\t_n = 5+(-3)(n-1)\\t_n = 5-3n+3\\t_n = -3n+8$

Then we can plug in various positive whole numbers for n to find the corresponding $t_n$ value. For example, plug in n = 2

$t_n = -3n+8\\t_2 = -3*2+8\\t_2 = -6+8\\t_2 = 2$

which matches with the second term we found earlier. And,

$tn = -3n+8\\t_{10} = -3*10+8\\t_{10} = -30+8\\t_{10} = \boldsymbol{-22} \ \textbf{ is the tenth term}$

---------------------

The notation $S_{10}$ refers to the sum of the first ten terms $t_1, t_2, \ldots, t_9, t_{10}$

We could use either the long way or the shortcut above to find all $t_1$ through $t_{10}$ . Then add those values up. Or we can take this shortcut below.

$Sn = \text{sum of the first n terms of an arithmetic sequence}\\S_n = (n/2)*(t_1+t_n)\\S_{10} = (10/2)*(t_1+t_{10})\\S_{10} = (10/2)*(5-22)\\S_{10} = 5*(-17)\\\boldsymbol{S_{10} = -85}$

The sum of the first ten terms is -85

-----------------------

As a check for $S_{10}$ , here are the first ten terms:

t1 = 5
t2 = 2
t3 = -1
t4 = -4
t5 = -7
t6 = -10
t7 = -13
t8 = -16
t9 = -19
t10 = -22

Then adding said terms gets us...

5 + 2 + (-1) + (-4) + (-7) + (-10) + (-13) + (-16) + (-19) + (-22) = -85

This confirms that $S_{10} = -85$ is correct.

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2 years ago

How can i do 3×3×11 6 grade

allsm [11]

You can start by doing 3×3 which is 9 than get the 9 and multiply it by 11 which would be 99

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3 years ago

Read 2 more answers