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3 years ago

H=K+log(A/C) isolate a... make a the subject

1 answer:

6 0

$H=K+\log\left(\dfrac{A}{C}\right)\\\\
\log\left(\dfrac{A}{C}\right)=H-K\\\\
\dfrac{A}{C}=10^{H-K}\\\\
\boxed{A=C\cdot10^{H-K}}$

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8x-29 + 6x-11 + 3x-1 solve for x

elena-s [515]

Answer:

Step-by-step explanation:

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3 years ago

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Point D is in the interior of

Natalija [7]

Answer:

Angle ABC,

Step-by-step explanation:

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3 years ago

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A quadratic equation is shown below:

sweet [91]

Answer:

Part A:

( 1.8333, -0.08333)

Part B:

x = 2 or x = 5/3

Step-by-step explanation:

The quadratic equation

$3x^{2}-11x+10=0$ has been given.

Part A:

We are required to determine the vertex. The vertex is simply the turning point of the quadratic function. We shall differentiate the given quadratic function and set the result to 0 in order to obtain the co-ordinates of its vertex.

$\frac{d}{dx}(3x^{2}-11x+10)=6x-11$

Setting the derivative to 0;

6x - 11 = 0

6x = 11

x = 11/6

The corresponding y value is determined by substituting x = 11/6 into the original equation;

y = 3(11/6)^2 - 11(11/6) + 10

y = -0.08333

The vertex is thus located at the point;

( 1.8333, -0.08333)

Find the attached

Part B:

We can use the quadratic formula to solve for x as follows;

The quadratic formula is given as,

$x=\frac{-b+/-\sqrt{b^{2}-4ac } }{2a}$

From the quadratic equation given;

a = 3, b = -11, c = 10

We substitute these values into the above formula and simplify to determine the value of x;

$x=\frac{11+/-\sqrt{11^{2}-4(3)(10) } }{2(3)}=\frac{11+/-\sqrt{1} }{6}\\\\x=\frac{11+/-1}{6}\\\\x=\frac{11+1}{6}=2\\\\x=\frac{11-1}{6}=\frac{5}{3}$

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4 years ago

Discrete Math

andrezito [222]

Answer:

Part c: Contained within the explanation

Part b: gcd(1200,560)=80

Part a: q=-6 r=1

Step-by-step explanation:

I will start with c and work my way up:

Part c:

Proof:

We want to shoe that bL=a+c for some integer L given:

bM=a for some integer M and bK=c for some integer K.

If a=bM and c=bK,

then a+c=bM+bK.

a+c=bM+bK

a+c=b(M+K) by factoring using distributive property

Now we have what we wanted to prove since integers are closed under addition. M+K is an integer since M and K are integers.

So L=M+K in bL=a+c.

We have shown b|(a+c) given b|a and b|c.

Part b:

We are going to use Euclidean's Algorithm.

Start with bigger number and see how much smaller number goes into it:

1200=2(560)+80

560=80(7)

This implies the remainder before the remainder is 0 is the greatest common factor of 1200 and 560. So the greatest common factor of 1200 and 560 is 80.

Part a:

Find q and r such that:

-65=q(11)+r

We want to find q and r such that they satisfy the division algorithm.

r is suppose to be a positive integer less than 11.

So q=-6 gives:

-65=(-6)(11)+r

-65=-66+r

So r=1 since r=-65+66.

So q=-6 while r=1.

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3 years ago

Which of the following is not an inequality? A 435 ≥ 5p + 4 B g - 7 < 100 C 18x + 9y D a > 6b

sleet_krkn [62]

Answer:

C. 18x + 9y: Because there isn't an relation between the two, all the others have and inequality sign meaning you can solve them, however C does not.

Step-by-step explanation:

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3 years ago

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