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3 years ago

What estimate can you make for 2/3 and 1/6?

Mathematics

2 answers:

Leona [35]3 years ago

8 0

2/3= About 66% give or take a few decimals
1/6=About 16% also give or take a few decimals

OverLord2011 [107]3 years ago

3 0

The answer should be 5/6

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An equation in the form ax2+bx+c=0 is solved by the quadratic formula. The solution to the equation is shown below.x=−7±√ "572"

jolli1 [7]

Answer:

B: a = 1, b= 7, c = -2

Step-by-step explanation:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0$

Given:

$x=\dfrac{-7\pm\sqrt{57}}{2}$

Comparing the terms of the given x-value with those of the quadratic formula:

$\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}=\dfrac{-7\pm\sqrt{57}}{2}$

Therefore:

$2a=2 \implies a=1$
$b = 7$
$b^2-4ac=57$

Using the found values of a and b to solve for c:

$\implies b^2-4ac=57$

$\implies (7)^2-4(1)c=57$

$\implies 49-4c=57$

$\implies -4c=57-49$

$\implies -4c=8$

$\implies c=-2$

In summary: a = 1, b = 7, c = -2

$\implies x^2+7x-2=0$

Therefore, option B is the correct solution.

5 0

2 years ago

Read 2 more answers

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

Read 2 more answers

Ten people reported the number of hours they worked last week.

taurus [48]

First arrange them in ascending order:-

17 20 24 25 30 32 35 38 40 42
The median = mean of middle 2 numbers = (30+32)/2 = 31
Lower quartile = 24 and upper quartile = 38
Inter quartile range = 38 - 24 = 14

Its the first choice.

4 0

3 years ago

Read 2 more answers

If anyone knows the answer to this question please let me know! thanks

Bond [772]

Answer:

D. 10 units

Step-by-step explanation:

Use pythagorean theorem

$a^2 + b^2 = c^2\\6^2+8^2= c^2$