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3 years ago

Sally needs 1 3/4 yards of fabric to make a dress. She has 4 5/8 yards. How many yards of fabric will be left over?

1 answer:

4 0

Hey there,

Solution:
Fabric needed = 1 3 / 4
Fabric she has = 4 5 / 8
1 3 / 4 - make the denominator into 8
1 3 / 4 = 1 6 / 8
Fabric left = 4 5 / 8 - 1 6 / 8
= 37 / 8 - 14 / 8
= 23 / 8
= 2 7 / 8

Hope this helps :))

~Top♥

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One kitty weighs 2 pounds 4 ounces. Another kitten weighs 2 ounces less. What is the combined weight of the two kittens in ounce

My name is Ann [436]

70 ounces because there are 16 ounces in a pound so 16 x2 equals 32 + 2ounces equals 36 - 2 equals 34 , 34 + 36 = 70 ounces in total .

3 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 h and take the limit as h 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

Can somebody please help me with this question?

meriva

What, this is math right?

7 0

3 years ago

Can someone answer these questions please, i'm having trouble with them

fiasKO [112]

Answer:

1. $686.94

2. $735.03

3. $10707.55

4. $17631.94

5. $19635.72

Step-by-step explanation:

1st Question:

The interest rate is 7% for each year. This means that each year the person has to pay 7% more than the previous amount. So we need to multiply the initial amount by (0.07+1=1.07) in order to get the interest for the first year. if we want to find the second year's interests then we will have to multiply 2 (1.07)'s and so on.

in this case our function is: 600*(1.07)^t=P(t)

when t=2 P(2)=600*(1.07)^2=$686.94

2nd Question:

Function: 600*(1.07)^t=P(t)

when t=3 P(3)=600*(1.07)^3=$735.03

3rd Question:

initial value=$8500

1+0.08=1.08

Function: 8500*(1.08)^t=P(t)

t=3

P(3)=8500*(1.08)^3=$10707.55

4th Question:

initial value=$12000

1+1.08=1.08

t=5

Function: P(t)=12000*(1.08)^t

P(5)=12000*(1.08)^5=$17631.94

5th Question:

Function: 14000*(1.07)^t=P(t)

P(5)=14000*(1.07)^5

P(5)=$19635.72

4 0

2 years ago

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A small school has 42 boys. If the ratio of boys to girls is 7 : 1, how many, students are?

babunello [35]

There are a total of 48 students. Being 42 boys as mentioned earlier and 6 girls total attending too.

Please vote my answer brainliest! Thanks.

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3 years ago

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