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3 years ago

13

14ft = ?Yd and ?Ft. Show Work.

1 answer:

pav-90 [236]3 years ago

3 0

4.67 yards is 14 ft

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Write an expression equivalent to b^10/b^5 in the form b^m.

Mars2501 [29]

B^5 when you’re dividing exponents, just subtract!

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2 years ago

How do I do this problem?

Elenna [48]

Roots can be written as fraction exponents. An exponent of 1/2 means the square root.

$(12 x^{4})^{1/2} = \sqrt{12 x^{4} } \\ \\ \sqrt{12 x^{4} } = \sqrt{3*4*x^{2}*x^{2}} = 2x^{2} \sqrt{3}$

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2 years ago

LAST ONE !! ILL MARK BRAINLEST AGAIN‍♀️ win win lol <br> Q: what is MAB?

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The Correct Answer Is
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3 years ago

Please help me on this problem

Anna71 [15]

Answer:

The pairs of integer having two real solution for $ax^{2} -6x+c = 0$ are

$a = -4, c = 5$
$a = 1, c = 6$
$a = 2, c = 3$
$a = 3, c = 3$

Step-by-step explanation:

Given

$ax^{2} -6x+c = 0$

Now we will solve the equation by putting all the 6 pairs so we get the following

$-3x^{2} -6x-5 = 0$ for $a = -3 , c=-5$

$-4x^{2} -6x+5 = 0$ for $a = -4 , c=5$

$1x^{2} -6x+6 = 0$ for $a = 1 , c=6$

$2x^{2} -6x+3 = 0$ for $a = 2 , c=3$

$3x^{2} -6x+3 = 0$ for $a = 3 , c=3$

$5x^{2} -6x+4 = 0$ for $a = 5 , c=4$

The above all are Quadratic equations inn general form $ax^{2} +bx+c=0$

where we have a,b and c constant values

So for a real Solution we must have

$Disciminant , b^{2} -4\timesa\timesc \geq 0$

for $a = -3 , c=-5$ we have

$Discriminant =-24$ which is less than 0 ∴ not a real solution.

for $a = -4 , c=5$ we have

$Discriminant = 116$ which is greater than 0 ∴ a real solution.

for $a = 1 , c=6$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 2 , c=3$ we have

$Discriminant =12$ which is greater than 0 ∴ a real solution.

for $a = 3 , c=3$ we have

$Discriminant =0$ which is equal to 0 ∴ a real solution.

for $a = 5 , c=4$ we have

$Discriminant =-44$ which is less than 0 ∴ not a real solution.

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3 years ago

40% of a number is 68.What is the number?

Talja [164]

The number is 17 hope it helps

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2 years ago