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3 years ago

7

Abbey`s cat weighs 18 pounds. Her dog weighs 2 pounds more than half her cat`s weight. Write and evaluate an expression to show

how much Abbey`s dog weighs.

2 answers:

KATRIN_1 [288]3 years ago

4 0

18 divide by 2 = 9 + 2 = 11 pounds

8_murik_8 [283]3 years ago

4 0

Answer:

Step-by-step explanation:

Let cat weigh c pounds= 18

Let dog weigh d pounds=2+c/2

d = 2+ 18/2 = 2+9 = 11

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<h3>To ProvE :- </h3>

1 + 3 + 5 + ..... + (2n - 1) = n²

<u>Method</u><u> </u><u>:</u><u>-</u>

If P(n) is a statement such that ,

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Step 1 : <u>Put </u><u>n </u><u>=</u><u> </u><u>1</u><u> </u><u>:</u><u>-</u><u> </u>

$\sf\longrightarrow LHS = \boxed{\sf 1 } \\$

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Step 2 : <u>Assume </u><u>that </u><u>P(</u><u>n)</u><u> </u><u>is </u><u>true </u><u>for </u><u>n </u><u>=</u><u> </u><u>k </u><u>:</u><u>-</u>

$\sf\longrightarrow 1 + 3 + 5 + \dots + (2k - 1 ) = k^2$

Add (2k +1) to both sides .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)=k^2+(2k+1)$

RHS is in the form of ( a + b)² = a²+b²+2ab .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1)= (k +1)^2$

Adding and subtracting 1 to LHS .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+1) + 1 -1 = (k +1)^2 \\$

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+(2k+2) - 1 = (k +1)^2$

Take out 2 as common .

$\sf\longrightarrow 1 + 3+5+\dots+(2k-1)+\{2(k+1)-1\}= (k +1)^2$

P(n) is true for n = k + 1 .

Hence by the principal of Mathematical Induction we can say that P(n) is true for all natural numbers 'n' .

<em>*</em><em>*</em><em>Edits</em><em> are</em><em> welcomed</em><em>*</em><em>*</em>

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