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gulaghasi [49]
4 years ago
10

Mike and Danielle each improve their yards by planting hospice and shrubs. They bought their supplies from the same store. Mike

spent $71 on for hospice and seven Travis. Danielle spent $53 on 13 hospice and three shrubs. Find the cost of one hospice and the cost of one shrub
Mathematics
2 answers:
gizmo_the_mogwai [7]4 years ago
6 0

Answer: each hospice costs $4 and each shrub costs $9

Step-by-step explanation:

Let x represent the cost of one hospice.

Let y represent the cost of one shrub.

Mike spent $71 on four hospice and seven Travis. This means that

4x + 7y = 71- - - - - - - - - - - - -1

Danielle spent $53 on 13 hospice and three shrubs. This means that

13x + 3y = 53- - - - - - - - - - - 2

Multiplying equation 1 by 13 and equation 2 by 4, it becomes

52x + 91y = 923

52x + 12y = 212

Subtracting, it becomes

79y = 711

y = 711/79

y = 9

Substituting y = 9 into equation 1, it becomes

4x + 7 × 9 = 71

4x + 63 = 71

4x = 71 - 63

4x = 8

x = 8/4 = 2

tangare [24]4 years ago
5 0

Answer: $71

Step-by-step explanation:

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chubhunter [2.5K]

Proof -

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

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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
A farm has chickens and cows. All the cows have 4 legs and all the chickens have 2 legs.
lora16 [44]

Answer:

21 chickens

Step-by-step explanation:

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3 0
3 years ago
Which fraction has a terminating decimal as its decimal expansion?
Tamiku [17]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Suppose that PR=47. Solve for x and find the lengths of PQ, and QR.
lidiya [134]

Answer:

x = 10

PQ = 27

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Step-by-step explanation:

You can add up PQ and QR and set it equal to PR.  Solve for x.

(2x + 7) + (2x) = 47

2x + 7 + 2x = 47

4x + 7 = 47

(4x + 7) - 7 = 47 - 7

4x = 40

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x = 10

Now, use the solved x-value to find PQ and QR.

PQ = 2x + 7

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6 0
4 years ago
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Greeley [361]

Answer:

Company B's offer is more cost effective

Step-by-step explanation:

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7 0
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