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telo118 [61]
3 years ago
15

Ashley had a summer lemonade stand where she sold small cups of lemonade for $1.25 and large cups for $2.50. if ashley sold a to

tal of 155 cups of lemonade for $265, how many cups of each type did she sell?
Mathematics
1 answer:
Levart [38]3 years ago
5 0
The answer is 98 small and 57 large cups.

s - the number of small cups
l - the number of large cups

<span>Ashley sold a total of 155 cups: s + l = 155
</span><span>Ashley earned</span><span>for $265: 1.25 * s + 2.50 * l = 265

</span>s + l = 155
1.25 * s + 2.50 * l = 265
________
s = 155 - l
1.25 * s + 2.50 * l = 265
________
1.25 * (155 - l) + 2.50 * l = 265
193.75 - 1.25 * l + 2.50 * l = 265
193.75 + 1.25 * l = 265
1.25 * l = 265 - 193.75
1.25 * l = 71.25
l = 71.25 / 1.25
l = 57

______
s = 155 - l
l = 57

s = 155 - 57
s = 98

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What is the product of all constants $k$ such that the quadratic $x^2 + kx +15$ can be factored in the form $(x+a)(x+b)$, where
Setler79 [48]

Answer:

k=-16,k=-8,k=8,k=16

Step-by-step explanation:

We are given quadratic equations as

x^2+kx+15

and it can be factored as

=(x+a)(x+b)

now, we can multiply factor term

(x+a)(x+b)=x^2+(a+b)x+ab

now, we can compare

x^2+(a+b)x+ab=x^2+kx+15

so, we get

k=a+b

ab=15

we are given that

'a' and 'b' are integers

so, we can find all possible factors

15=(-1\times -15),(1\times 15)

15=(-3\times -5),(3\times 5)

so, we can find k

At (-1\times -15):

k=a+b

we can plug values

k=-1-15

k=-16

At (1\times 15):

k=a+b

we can plug values

k=1+15

k=16

At (-3\times -5):

k=a+b

we can plug values

k=-3-5

k=-8

At (3\times 5):

k=a+b

we can plug values

k=3+5

k=8

So, values of k are

k=-16,k=-8,k=8,k=16

6 0
3 years ago
Help me out please will mark you BRAINLIST
melisa1 [442]

Answer:

D. 18.68

Step-by-step explanation:

27\div \frac{5}{3}-6+53\times (\frac{2}{5})^2

Applying PEMDAS as order of operations.

Solving the exponents first [(\frac{2}{5})^2=\frac{4}{25}]

=27\div \frac{5}{3}-6+53\times \frac{4}{25}

Multiplying [53\times \frac{4}{25}=8.48]

=27\div \frac{5}{3}-6+8.48

Dividing [27\div \frac{5}{3}=16.2]

=16.2-6+8.48

Adding and subtracting.

=18.68

6 0
3 years ago
1. (5pts) Find the derivatives of the function using the definition of derivative.
andreyandreev [35.5K]

2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

8 0
2 years ago
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shutvik [7]

Answer: fgm

Step-by-step explanation:

6 0
3 years ago
What is a reasonable height for a 5-story building?<br> 16 mm<br> 16 km<br> 16 cm<br> 16 m
Bogdan [553]

Answer:

16 km !

Step-by-step explanation:

all the others are FAR too small.. they wouldnt even be one story up ! and 16m  isnt 5 stories either... km would be right !

i hope this helps !!

4 0
3 years ago
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