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zavuch27 [327]
3 years ago
15

Lin's father is paying for a $30 meal. He has a 15%-off coupon for the

Mathematics
1 answer:
Natali [406]3 years ago
4 0

Answer:

$27.03

Step-by-step explanation:

Multiply: .15 times 30 =4.5 (to get from a percent to a decimal, move the decimal twice to the right.

Subtract: 30 minus 4.5= 25.5

Multiply: 25.5 times 1.06= 27.03

$27.03

Hope this helped!! :)

Brainliest?!?!

Stay safe and have a wonderful day/afternoon/night!!!

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The ball bearing have volumes of 1.6cm cube and 5.4cm cube . Find the ratio of their surface area.
Korvikt [17]

Answer:

  4/9

Step-by-step explanation:

The scale factor for the linear dimensions of the ball bearings will be the cube root of the volume scale factor:

  k = ∛(1.6/5.4) = 2/3

Then the scale factor for the areas will be the square of this scale factor:

  ratio of surface area = (2/3)² = 4/9

_____

The area is the product of two linear dimensions, so its scale factor is the product of the linear dimension scale factors. That is, the scale factor for area is the square of the linear dimension scale factor.

Similarly, volume is the product of three linear dimensions, so its scale factor is the cube of the linear dimension scale factor.

8 0
3 years ago
Write the sentence as an inequality: Sixteen is no more than the product of 4 and x
n200080 [17]

Your sentence:

16 < 4x

4 0
3 years ago
Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 25 − t2 , [−1, 5]
Zina [86]

Answer: Absolute minimum: f(-1) = -2\sqrt{6}

              Absolute maximum: f(\sqrt{12.5}) = 12.5

Step-by-step explanation: To determine minimum and maximum values in a function, take the first derivative of it and then calculate the points this new function equals 0:

f(t) = t\sqrt{25-t^{2}}

f'(t) = 1.\sqrt{25-t^{2}}+\frac{t}{2}.(25-t^{2})^{-1/2}(-2t)

f'(t) = \sqrt{25-t^{2}} -\frac{t^{2}}{\sqrt{25-t^{2}} }

f'(t) = \frac{25-2t^{2}}{\sqrt{25-t^{2}} } = 0

For this function to be zero, only denominator must be zero:

25-2t^{2} = 0

t = ±\sqrt{2.5}

\sqrt{25-t^{2}} ≠ 0

t = ± 5

Now, evaluate critical points in the given interval.

t = -\sqrt{2.5} and t = - 5 don't exist in the given interval, so their f(x) don't count.

f(t) = t\sqrt{25-t^{2}}

f(-1) = -1\sqrt{25-(-1)^{2}}

f(-1) = -\sqrt{24}

f(-1) = -2\sqrt{6}

f(\sqrt{12.5}) = \sqrt{12.5} \sqrt{25-(\sqrt{12.5} )^{2}}

f(\sqrt{12.5}) = 12.5

f(5) = 5\sqrt{25-5^{2}}

f(5) = 0

Therefore, absolute maximum is f(\sqrt{12.5}) = 12.5 and absolute minimum is

f(-1) = -2\sqrt{6}.

8 0
3 years ago
If you buy 3 pairs of socks with an original cost of 9.99/pair at a discount of 15% with a sales tax of 5% how much do the socks
Delicious77 [7]
$26.75........................                                 
8 0
3 years ago
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Given the scatter plot, choose the function that best fits the data.
geniusboy [140]

Considering the scatter plot, the function that best fits the data is given by:

f(x) = 3x².

<h3>What is the function that best fits the data?</h3>

To find the function that best fits the data, we have to look at the scatter plot, which gives a series of points (x,y).

We have that:

  • When x increases, y also increases, hence the function is increasing, which removes the option f(x) = -3x.
  • The increase looks "faster" than a linear increase, hence a quadratic model should be used, thus removing option f(x) = 3x.

Hence the function that best fits the data is given by:

f(x) = 3x².

More can be learned about scatter plots at brainly.com/question/22968877

#SPJ1

5 0
2 years ago
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