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2 years ago

Please help me do number 5

Mathematics

1 answer:

Blizzard [7]2 years ago

7 0

Make more sense ou of this question and I can help you out more ok?

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Sharon has at most $25 to spend on her sister's birthday gift. She already bought a knitting machine for her, which cost $14.99.

Likurg_2 [28]

Answer:

$25 >= 14.99+2.75x

Step-by-step explanation:

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2 years ago

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Find the missing values on the diagram below. Assume that each line is evenly divided. I'm pretty confused on the right side of

belka [17]

Answer:

A = 72 B = 50% C = 40%

Step-by-step explanation:

Each space on the bottom is worth 10%.

Each space on the bottom is worth 36.

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2 years ago

This is a really critical thinking question.

SashulF [63]

Answer: hi

Step-by-step explanation:

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2 years ago

What is the equation of this graph?

Usimov [2.4K]

Answer:

The equation would be Y= -7/2x + 7

Step-by-step explanation:

With the points given (0,7) and (2,0) we find our slope which is: -7/2. We know that our change in the y values is -7 and the change in our x values is +2. Y IS ALWAYS OVER X SO WE GET -7/2.

To find our y intercept you just look at the graph of where there's a point on the y line. For example (0,7) the 7 would be our y intercept.

So we get Y= -7/2x +7

The slope is ALWAYS put next to the variable x.

Hope this helps! :)

(Side Note: You look at the graph from left to right, so our first point would be (0,7)!)

4 0

3 years ago

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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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