Can any one help A B. C

On July 3rd, a family left for vacation. They drove 431.2 miles on the first day, then they drove an average of 826.4 miles each

Alexxx [7]

431.2+826.4d=2910.4
431.2 is only the first day therefore it doesn't need the variable because the variable represents the number of days. 826.4 even says "each remaining day" referring to what the variable represents.

6 0

2 years ago

Pleaseeeeeeeee helllllllpppppoo

kodGreya [7K]

I'm pretty sure the answer is c

4 0

2 years ago

Andrew and Sam start counting from 150 to 200 including. Andrew counts by fives, and Sam counts by threes. What is the ratio of

Agata [3.3K]

Answer:

11/17

Step-by-step explanation:

The question says to find the ratio by which Andrew counts to the ratio Sam counts.

Andrew counts by 5s:

150,155,160,165,170,175,180,185,190,195,200. His total count is 11

Sam counts by 3s:

150,153,156,159,162,165,168,171,174,177,180,183,186,189,192,195,198. His total number of count is 17

The ratio of their count is 11:17

= 11/17

3 0

3 years ago

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-y= 4x + 1

ivolga24 [154]

Answer:

I believe the answer would be C.

Step-by-step explanation:

This is because after plugging in the numbers (2 in replace of x) (-9 in replace of y) for the problem -2y- 6x =6... the statement was determined true.... so, therefore, I believe the correct answer is C

6 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' .

**Edits are welcomed**

8 0

2 years ago

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