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Murljashka [212]
3 years ago
14

What is the fewest number of expressions that a equation can contain?

Mathematics
2 answers:
nlexa [21]3 years ago
8 0
The fewest number of expressions is 3.
umka2103 [35]3 years ago
6 0
The fewest of expression is 4
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Find the equation of the line by given two points (-5, -2) and (1, 5). Slope
docker41 [41]

Answer:

The equation of the line is 6y = 7x + 23

Step-by-step explanation:

We begin by calculating the slope of the line;

mathematically, the slope is given by;

m = (y2-y1)/(x2-x1)

That would be;

m= (5-(-2))/(1-(-5))

m = 7/6

So the equation becomes;

y = 7/6x + b

We below need to get the value of the y-intercept b

We get this by making a substitution for any of the two points

Let’s say we make the substitution for (-5,-2) in which case y = -2 and x = -5

Thus, we have

-2 = 7/6(-5) + b

-2 = -35/6 + b

b = -2 + 35/6

b = (-12+ 35)/6 = 23/6

So we have;

y = 7/6x + 23/6

let’s multiply through by 6

6y = 7x + 23

7 0
3 years ago
Gracie has 5555 pet bunnies that love to eat whole carrots. She has 39393939 carrots to feed the bunnies. Gracie will use as man
Kipish [7]
39393939/5555 = 7091.61 so each gets 7091 carrots.
There will be 3434 carrots left over
8 0
3 years ago
Rewrite the quadratic function below in standard form y=3(x+2)(x-3)
babunello [35]
X=3(x^2-x_6)
3x^2-x-6
8 0
2 years ago
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Camille is attending a fundraiser. She pays for her admission and buys raffle tickets for $5 each. If she buys 10 raffle tickets
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To spend a total of $135, Camille would have bought 27 tickets at $5 each.
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3 years ago
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Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering
Svetllana [295]

Answer:

L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x

Step-by-step explanation:

We are asked to find the tangent line approximation for f(x)=\sqrt{10+x} near x=0.

We will use linear approximation formula for a tangent line L(x) of a function f(x) at x=a to solve our given problem.

L(x)=f(a)+f'(a)(x-a)

Let us find value of function at x=0 as:

f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}

Now, we will find derivative of given function as:

f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}

f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)

f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1

f'(x)=\frac{1}{2\sqrt{10+x}}

Let us find derivative at x=0

f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}

Upon substituting our given values in linear approximation formula, we will get:

L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)  

L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0

L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x

Therefore, our required tangent line for approximation would be L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x.

8 0
3 years ago
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