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pshichka [43]
3 years ago
12

I need help please and thank you

Mathematics
2 answers:
Ratling [72]3 years ago
8 0

What Do you need help with if you added a picture we cant see it

barxatty [35]3 years ago
8 0

Answer:

Can  you explain to me what you are trying to do

Step-by-step explanation:

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Mai will spend more than $23 on gifts. So far, she has spent $15. What are the possible additional amounts she will spend? Use c
mote1985 [20]

Answer:

c = ≥ (23 - 15) therefore c = ≥ 8

3 0
3 years ago
Uche pumps gasoline at a rate of 18\,\dfrac{\text{L}}{\text{min}}18 min L ​ 18, start fraction, start text, L, end text, divided
Fiesta28 [93]

Answer:

Uche's pumping rate is <u>300 mL/s</u>.

Step-by-step explanation:

Given:

Uche pumps gasoline at a rate of 18 L/min.

Now, to find Uche's pumping rate in mL/s.

The rate at which Uche pumps gasoline = 18 L/min.

So, to get the pumping rate in mL/s we use convert L/min to mL/s by using conversion factor:

<u>1 L/min = 16.6667 mL/s.</u>

18 L/min = 16.6667 × 18 mL/s.

18L/min = 300 mL/s.

Therefore, Uche's pumping rate is 300 mL/s.

8 0
3 years ago
Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
adoni [48]

Answer: The required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

Step-by-step explanation:

Since we have given that

y=\ln[x(2x+3)^2]

Differentiating log function w.r.t. x, we get that

\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}

Hence, the required derivative is \dfrac{8x^2+18x+9}{x(2x+3)^2}

3 0
3 years ago
Solve the inequality. 10k &lt;75 and 4-k is less than or equal to 0
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Is it 10k < 75 , 4-k?
3 0
3 years ago
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Which one is true about the diagram?
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It could be all it could be one but the true answer is I’m not sure which one
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3 years ago
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