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3 years ago

Which fraction is the bigger fractured 2/10 or 23/100

2 answers:

4 0

Answer: 23/100

Find the greatest common factor (gcf)

Which is 100 so 23/100 is done

2/10 multiply 10 times 10 to get 100

Multiply 2 times 10 which is 20/100

23/100 is greater

Anastaziya [24]3 years ago

4 0

23/100 is the bigger fraction

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Your friend has 100$ when he goes to the fair and 20$ on food. Rides at the fair cost $2 per ride. Which function can be used to

andreev551 [17]

Y= 100+20-2X is the equation

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4 years ago

What fractions in this list are less than 1/2? 3/6,7/15,4/9,5/8

Vedmedyk [2.9K]

All we need to do is to find a common denominator for both 1/2 and the other numbers individually:

1/2 = 3/6 (so 3/6 is not less than 1/2 - both numbers are equal)
1/2 = 15/30 and 7/15 = 14/30 (so 7/15 is less than 1/2)
1/2 = 9/18 and 4/9 = 8/18 (so 4/9 is less than 1/2)
1/2 = 4/8 (so 5/8 is not less - it's greater than 1/2)

Answer: From the list above only <u>7/15 and 4/9</u> are less than 1/2.

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4 years ago

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jek_recluse [69]

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4 years ago

Read 2 more answers

Which expression is equivalent to (4(5^3)1/2?

Feliz [49]

Answer:

$10\sqrt{5}$

Step-by-step explanation:

In this problem, we are given expression in exponent form and we have to simplify the expression.

The expression is $[4(5^{3} )]^{\frac{1}{2} }$

Now, we know that $5^{3} = 5 \times 5 \times 5 = 125$

So, $4(5^{3} ) = 4 \times 125 = 500$

Now, 500 can be written as 100 × 5 = 10² × 5

Therefore, $[4(5^{3} )]^{\frac{1}{2} } = (500)^{\frac{1}{2} } = 10\sqrt{5}$ {Since 500 = 100 × 5 = 10² × 5} (Answer)

6 0

4 years ago

Consider the function f(x)=x2e4x. f(x) has two inflection points at x = C and x = D with C≤D where C is and D is Finally for eac

fredd [130]

Answer:

the inflection points are

$x = -0.85, -0.14$

So,

$C = -0.85 \\D = -0.14$

It is concave down at the intervals

$(-\infty , -0.85] \cup [-0.14,\infty)$

And it would be concave up at

$(-0.85 , -0.14)$

Step-by-step explanation:

Remember that to find inflection points you need to find where

$f''(x) = 0$

Since

$f(x) = x^2 e^{4x}$

Then using the product and the chain rule you have that

$f'(x) = 2e^{4x} x(2x+1)$

And then, using again the chain rule and the product rule you have that

$f''(x) = 2e^{4x} (8x^2 +8x+1)$

Therefore you have to solve the equation

$8x^2 +8x+1 = 0$

Using the quadratic equation you get that there are two solutions, so the inflection points are

$x = -0.85, -0.14$

So,

$C = -0.85 \\D = -0.14$

Now remember that a function is concave up if the derivative is greater than zero and concave down if the derivative is less than zero. Therefor you have to solve these inequalties

$8x^2 +8x+1 < 0 \\8x^2 +8x+1 \geq 0$

And you would get that is concave down at the intervals

$(-\infty , -0.85] \cup [-0.14,\infty)$

And it would be concave up at

$(-0.85 , -0.14)$

8 0

3 years ago