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Effectus [21]
3 years ago
15

Anyone know this?I need helppp

Mathematics
2 answers:
Fittoniya [83]3 years ago
8 0

Answer:

nope sorry

Step-by-step explanation:

EastWind [94]3 years ago
3 0

Answer:

just think hard about it. if you can't get any answers. just guess! try your best. good luck! its what i always do in these type of situations.

Step-by-step explanation:

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Allushta [10]

Answer:

I'm not really sure but it is either a or c

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Derivative of these questions<br>​
shtirl [24]

Answer:  \bold{y'=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)}

<u>Step-by-step explanation:</u>

y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\\\\\\\text{Rationalize the denominator:}\\y=\dfrac{1}{\sqrt{x+a}-\sqrt{x+b}}\bigg(\dfrac{\sqrt{x+a}+\sqrt{x+b}}{\sqrt{x+a}+\sqrt{x+b}}\bigg)\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{x+a-x-b}\\\\\\y=\dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}\\\\\\\text{Apply the derivative (derivative of the numerator (top) divided by the}\\\text{denominator (bottom) squared).}

\text{Derivative of }\sqrt{x+a}\quad \rightarrow \quad (x+a)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + a)^{-\frac{1}{2}}\\\\\text{Derivative of }\sqrt{x+b}\quad \rightarrow \quad (x+b)^{\frac{1}{2}}\quad = \quad \dfrac{1}{2}(x + b)^{-\frac{1}{2}}\\\\\\\text{Derivative of}\ \dfrac{\sqrt{x+a}+\sqrt{x+b}}{a-b}:\\\\= \dfrac{\dfrac{1}{2}(x + a)^{-\frac{1}{2}}+\dfrac{1}{2}(x + b)^{-\frac{1}{2}}}{(a-b)^2}

\text{Factor out }\dfrac{1}{2}\ \text{from the numerator:}\\\\\dfrac{(x+a)^{-\frac{1}{2}}+(x+b)^{-\frac{1}{2}}}{2(a-b)^2}\\\\\\\text{Move the terms with negative exponents to the denominator:}\\\\=\dfrac{1}{2(a-b)^2}\bigg(\dfrac{1}{\sqrt{x+a}}+\dfrac{1}{\sqrt{x+b}}\bigg)

3 0
3 years ago
Geometry Help! 30 POINTS! + BRAINLIEST
Triss [41]
SRT = 20   => STR = 20.
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5 0
4 years ago
Read 2 more answers
Course Home
NNADVOKAT [17]

Answer:

(a) The amount of pure acid in 60 oz of 20% acid solution is 12 oz.

(b) The amount of pure acid in 60 oz of 25% acid solution is 15 oz.

(c) The amount of pure acid in 60 oz of 30% acid solution is 18 oz.

(d) The amount of pure acid in 60 oz of 50% acid solution is 30 oz.

Step-by-step explanation:

* <em>Lets explain how to solve the problem</em>

- The solution is a mixture of two or more substances

- A container of liquid contains 60 oz of solution

- The solution contains pure acid by different concentrations

- (a) 20% (b) 25% (c) 30% (d) 50%

- We want to find the number of ounces of the pure acid in each

 concentration

- Assume that the solution is 100%

- That means 100% represents 60 oz of solution

a)

- The percentage of pure acid is 20%

∵    solution     ⇒     pure acid

        100%               20%

        60                    The number of ounces

- By using cross multiplication

∴ The number of ounces = \frac{60*20}{100}= 12

(a) The amount of pure acid in 60 oz of 20% acid solution is 12 oz.

b)

- The percentage of pure acid is 25%

∵    solution    ⇒      pure acid

        100%               25%

        60                    The number of ounces

- By using cross multiplication

∴ The number of ounces = \frac{60*25}{100}= 15

(b) The amount of pure acid in 60 oz of 25% acid solution is 15oz.

c)

- The percentage of pure acid is 30%

∵    solution     ⇒     pure acid

        100%               30%

        60                    The number of ounces

- By using cross multiplication

∴ The number of ounces = \frac{60*30}{100}= 18

(c) The amount of pure acid in 60 oz of 30% acid solution is 18 oz.

d)

- The percentage of pure acid is 30%

∵    solution     ⇒     pure acid

        100%               50%

        60                    The number of ounces

- By using cross multiplication

∴ The number of ounces = \frac{60*50}{100}= 30

(d) The amount of pure acid in 60 oz of 50% acid solution is 30 oz.

3 0
3 years ago
The units’ digit of a two-digit number is 5 more than the tens’ digit, and the number is three times as great as the sum of the
solniwko [45]

Answer:

The number is 27

Step-by-step explanation:

Let the 10s digit be x

Let the units digit be y

y = x + 5

10x + y = 3(x + y)                  Remove the brackets          

10x + y = 3x + 3y                 Substitute the x + 5 into the second equation for y

10x + x + 5 = 3x + 3(x + 5)   Remove the brackets on the right.

10x + x + 5 = 3x + 3x +15     Collect like terms on each side.

11x + 5 = 6x + 15                   Subtract 5 from both sides

11x + 5 - 5 = 6x + 15 - 5        Collect like terms

11x = 6x + 10                         Subtract 6x from both sides

11x - 6x = 6x - 6x + 10

5x =  10                                 Divide by 5

5x/5 = 10/5

x = 2

y = x + 5

y = 2 + 5

y = 7

3 0
3 years ago
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