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2 years ago

Rounding each number to the nearest tenth 2.65=

Mathematics

2 answers:

baherus [9]2 years ago

7 0

Answer:

the nearest tenth would be 2.70 because the tens place is above 5 so let it score!

Step-by-step explanation:

plz mark brainlest

mixas84 [53]2 years ago

3 0

Answer:

2.7

Step-by-step explanation:

that is correct....

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3x + 2 =11 (solve for x)

torisob [31]

Answer:

x=3

Step-by-step explanation:

first let's try to get just 3x on one side. to do this we can subtract 2 from both side of the equation.

3x+(2-2)=11-2

3x=9

now we have to isolate the variable we can divide both sides by 3:

3x/3=9/3

x=3

and you get the answer that x is 3

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3 years ago

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gizmo_the_mogwai [7]

Answer:can someone answer my question

Step-by-step explanation:

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3 years ago

Solve the proportion 20/3=?/6

Zepler [3.9K]

It equals 40/6. I hope this helps. Good luck!

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3 years ago

Read 2 more answers

Solve 25^2x+1 = 144

Sergeu [11.5K]

Answer:

X = 0.272

Step-by-step explanation:

Firstly, use a common log on the 25 to undo it. Secondly, use the log25(144) on the other side to get ~ 0.54. Move the 1 over with subtraction and then divide out the 2. This will leave you with X= 0.272

4 0

3 years ago

Given that the quadratic equation has equal roots (k^2-1)x^2-2kx-3k-1=0

Alik [6]

Answer:

(See explanation for further details)

Step-by-step explanation:

The real expression is:

$(k^{2}-1)\cdot x{{2} - 2\cdot k \cdot x - 3\cdot k + 1 = 0$

The general equation for the second-order polynomial is:

$x = \frac{2\cdot k \pm \sqrt{4\cdot k^{2}-4\cdot (k^{2}-1)\cdot (-3\cdot k + 1)}}{k^{2}-1}$

This condition must be observed for the case of a quadratic equation with equal roots:

$4\cdot k^{2} - 4\cdot (k^{2}-1)\cdot (-3\cdot k + 1) = 0$

$k^{2} + (k^{2}-1)\cdot (3\cdot k + 1) = 0$

$k^{2} + 3\cdot k^{3} - 3\cdot k - k^{2}-1 = 0$

$3\cdot k^{3} - 3\cdot k - 1 = 0$

7 0

3 years ago