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kherson [118]
3 years ago
8

A baker is decorating the top of a round cake with cherries. The diameter of the cake is 9.5 inches. Each cherry is 0.75 inches

in diameter. About how many cherries will the baker need to decorate the circumference of the top of the cake? *
Mathematics
1 answer:
True [87]3 years ago
6 0

Answer:

A= 78.5 sq. in

Step-by-step explanation:

Step by step

1: A = pi * r^2

2: A = 3.14 * 25

In words.

1: Times the pi by r^2

2: The answer will come out as 3.14

3: Times the Answer that is 3.14 by 25

Answer: A= 78.5 sq. in

<em><u>Hope this helps.</u></em>

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What percent increase is 1/3 to 1/6
r-ruslan [8.4K]

Answer:

50% increase

Step-by-step explanation:

half of 6 is 3 thus 50%, and vise versa.

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2 years ago
Find the unknown measures. around lengths to the nearest hundredth and angle measures to the nearest degree.
qaws [65]

Answer:

choice 3)

JK = 3.51

WK = 9.15

∠J = 69°

Step-by-step explanation:

sin 21° = JK/9.8

0.3584 = JK/9.8

cos 21° = WK/9.8

0.9336 = WK/9.8

WK = 9.15

∠J = 90° - 21° = 69°

5 0
3 years ago
Quick algebra 1 question for 50 points!
Daniel [21]

Answer:

Scenario 1: Cameron works at a local grocery store after school. He is paid $8 for each hour(h) he works. And ammount of money can be (m)

considering theres no like first hour pay or something then it would be

Equation: m = 8h

Scenario 2: At a local bakery, one box of doughnuts(b) costs $5(c). Two boxes of doughnuts cost $10.

Equation: c = 5b where b is the ammount of doughnut boxes and c is the cost

<em>Scenario 1a: </em>Jim goes to the gym for 10 hours(h) It costed him $20(c) for those 10 hours.

Equation:  c = 2h

<em>Scenario 1b:</em><em> </em>Kayla goes to a for a ride on her bike and traveled 15 miles(m) and was out for 30 minutes. (t)

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8 0
1 year ago
How can i prove this property to be true for all values of n, using mathematical induction.
chubhunter [2.5K]

Proof -

So, in the first part we'll verify by taking n = 1.

\implies \: 1  =  {1}^{2}  =  \frac{1(1 + 1)(2 + 1)}{6}

\implies{ \frac{1(2)(3)}{6} }

\implies{ 1}

Therefore, it is true for the first part.

In the second part we will assume that,

\: {  {1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  =  \frac{k(k + 1)(2k + 1)}{6}  }

and we will prove that,

\sf{ \: { {1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} =  \frac{(k + 1)(k + 1 + 1) \{2(k + 1) + 1\}}{6}}}

\: {{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2}  =  \frac{(k + 1)(k + 2) (2k + 3)}{6}}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{k (k + 1) (2k + 1) }{6} +  \frac{(k + 1) ^{2} }{6}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{k(k+1)(2k+1)+6(k+1)^ 2 }{6}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{(k+1)\{k(2k+1)+6(k+1)\} }{6}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{(k+1)(2k^2 +k+6k+6) }{6}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{(k+1)(2k^2+7k+6) }{6}

{1}^{2} +  {2}^{2}  +  {3}^{2}  + ..... +  {k}^{2}  + (k + 1)^{2} = \frac{(k+1)(k+2)(2k+3) }{6}

<u>Henceforth, by </u><u>using </u><u>the </u><u>principle </u><u>of </u><u> mathematical induction 1²+2² +3²+....+n² = n(n+1)(2n+1)/ 6 for all positive integers n</u>.

_______________________________

<em>Please scroll left - right to view the full solution.</em>

8 0
2 years ago
in a sale a store gives a reduction of 25% off the ticket price. a summer coat is marked in the sale at £97.50 how much is the o
eduard
9+10 = 21 you stupid you stupid no I’m not
5 0
3 years ago
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