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3 years ago

Need help asapp!!!!!!!!!!!!

Mathematics

1 answer:

MissTica3 years ago

4 0

Answer:

Sorry I don't talk uwu voice I don't no wat to speak uwuuu I just yry one

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Help me out here please

VLD [36.1K]

Step-by-step explanation:

the answe to dis question is 2

3 0

3 years ago

Read 2 more answers

Kenyon ran for 8 miles each week while training. Here is his record of the number of miles he ran.

diamong [38]

Step-by-step explanation:

Let x be the miles Kenyon ran on tuesday

Let y be the miles Kenyon ran on thursday

Given,

Mon + Tues + Wed + Thu = 8mi

$1.2 + x + 1.6 + y = 8 \\ 2.8 + x + y = 8 \\ x + y = 8 - 2.8 \\ x + y = 5.2$

also given,

$y = x + 0.4 \\ - x + y = 0.4$

Now we have x + y = 5.2 as equation 1 and -x + y = 0.4 as equation 2

Using equation 1,

$x + y = 5.2 \\ x = 5.2 - y$

Substitute this equation into equation 2.

$- (5.2 - y) + y = 0.4 \\ - 5.2 + y + y = 0.4 \\ - 5.2 + 2y = 0.4 \\ 2y = 0.4 + 5.2 \\ 2y = 5.6 \\ y = 5.6 \div 2 \\ = 2.8mi$

Substitute y = 2.8 into equation 1.

$x + 2.8 = 5.2 \\ x = 5.2 - 2.8 \\ = 2.4mi$

Therefore Kenyon ran 2.4 miles on tuesday, and ran 2.8 miles on thursday.

8 0

2 years ago

Given that 'n' is a natural number. Prove that the equation below is true using mathematical induction.

LenaWriter [7]

<h3>To ProvE :- </h3>

1 + 3 + 5 + ..... + (2n - 1) = n²

Method :-

If P(n) is a statement such that ,

P(n) is true for n = 1
P(n) is true for n = k + 1 , when it's true for n = k ( k is a natural number ) , then the statement is true for all natural numbers .

$\sf\to \textsf{ Let P(n) : 1 + 3 + 5 + $\dots$ +(2n-1) = n$^{\sf 2}$ }$

Step 1 : Put n = 1 :-

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

$\sf\longrightarrow RHS = n^2 = 1^2 = \boxed{\sf 1 }$

Step 2 : Assume that P(n) is true for n = k :-

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

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' .