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3 years ago

Rp+s=t in terms of r, s,and t

Mathematics

1 answer:

expeople1 [14]3 years ago

8 0

Answer:

p=(t-s)/r

Step-by-step explanation:

This is an equation so all we have to do is solve backwards step by step:

rp+s=t

rp+s-s=t-s

rp/r=(t-s)/r

p=(t-s)/r

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Which of the following functions is graphed below?

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Answer:

Step-by-step explanation:

C is the solution

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3 years ago

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What are the potential solutions of log4x+log4(x+6)=2?

lions [1.4K]

The potential solutions of $log_4x+log_4(x+6)=2$ are 2 and -8.

<h3>Properties of Logarithms</h3>

From the properties of logarithms, you can rewrite logarithmic expressions.

The main properties are:

Product Rule for Logarithms - $log_{b}(a*c)=log_{b}a+log_{b}c$
Quotient Rule for Logarithms - $log_{b}(\frac{a}{c} )=log_{b}a-log_{b}c$
Power Rule for Logarithms - $log_{b}(a^c)=c*log_{b}a$

The exercise asks the potential solutions for $log_4x+log_4(x+6)=2$ . In this expression you can apply the Product Rule for Logarithms.

$log_4x+log_4(x+6)=2\\ \\ x*(x+6)=4^2\\ \\ x^2+6x=16\\ \\ x^2+6x-16=0$

Now you should solve the quadratic equation.

Δ= $b^2-4ac=36-4*1*(-16)=36+64=100$ . Thus, x will be $x_{1,\:2}=\frac{-6\pm \:\sqrt{100} }{2\cdot \:1}=\frac{-6\pm \:10}{2}$ . Then:

$x_1=\frac{-6+10}{2}=\frac{4}{2} =2\\ \\ \:x_2=\frac{-6-10}{2}=\frac{-16}{2} =-8$

The potential solutions are 2 and -8.

Read more about the properties of logarithms here:

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2 years ago

15 times a number increased

Minchanka [31]

Answer:

Step-by-step explanation:

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3 years ago

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What is the area of a cross section that is parallel to face CDHG ?

Dennis_Churaev [7]

Answer: Area of cross section that is parallel to face CDHG is 432 cm².

Step-by-step explanation:

Since we have given that

There is a cross section that is parallel to face CDHG.

So, Length of cross section would be 36 cm

Width of cross section would be 12 cm.

So, Area of cross section would be

$Length\times breadth\\\\=36\times 12\\\\=432\ cm^2$

Hence, Area of cross section that is parallel to face CDHG is 432 cm².

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3 years ago

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Evaluate the integral of the quotient of the cosine of x and the square root of the quantity 1 plus sine x, dx.

VMariaS [17]

Answer:

∫((cos(x)*dx)/(√(1+sin(x)))) = 2√(1 + sin(x)) + c.

Step-by-step explanation:

In order to solve this question, it is important to notice that the derivative of the expression (1 + sin(x)) is present in the numerator, which is cos(x). This means that the question can be solved using the u-substitution method.

Let u = 1 + sin(x).

This means du/dx = cos(x). This implies dx = du/cos(x).

Substitute u = 1 + sin(x) and dx = du/cos(x) in the integral.

∫((cos(x)*dx)/(√(1+sin(x)))) = ∫((cos(x)*du)/(cos(x)*√(u))) = ∫((du)/(√(u)))

= ∫(u^(-1/2) * du). Integrating:

(u^(-1/2+1))/(-1/2+1) + c = (u^(1/2))/(1/2) + c = 2u^(1/2) + c = 2√u + c.

Put u = 1 + sin(x). Therefore, 2√(1 + sin(x)) + c. Therefore:

∫((cos(x)*dx)/(√(1+sin(x)))) = 2√(1 + sin(x)) + c!!!

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3 years ago

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