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2 years ago

Which equation a line that goes through the points (1, 5) and (-2,2)

Mathematics

1 answer:

Korolek [52]2 years ago

6 0

Answer:

y= x+ 4

is the answer

<h2>hope it helped </h2>

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What is the solution to the equation?<br><br>√2x+10−6=2<br><br><br>*Picture of the Problem Below*

Klio2033 [76]

Answer:

x = 27

Step-by-step explanation:

Given

$\sqrt{2x+10}$ - 6 = 2 ( add 6 to both sides )

$\sqrt{2x+10}$ = 8 ( square both sides )

2x + 10 = 8² = 64 ( subtract 10 from both sides )

2x = 54 ( divide both sides by 2 )

x = 27

5 0

2 years ago

In the fraction 35/10 what is the product that lies between

RUDIKE [14]

The result of the fraction is 3.5

we can calculate this like this:
$\frac{35}{10} = \frac{5*7}{2*5} = \frac{7}{5}=3.5$

I think that the question is asking between which two natural numbers the result lies, and the result is 3 and 4: it's bigger than 3 and smaller than 4

8 0

2 years ago

Is one over eight irrational or rational

lukranit [14]

The answer to this is irrational

4 0

2 years ago

If p is a polynomial show that lim x→ap(x)=p(a

Lostsunrise [7]

Let p(x) be a polynomial, and suppose that a is any real number. Prove that

lim x→a p(x) = p(a) .

Solution. Notice that

2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .

So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x – 2.

Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| < 1, so −2 < x < 0. In particular |x| < 2. So

|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|

= 2|x|^3 + 5|x|^2 + |x| + 2

< 2(2)^3 + 5(2)^2 + (2) + 2

= 40

Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2 + x − 2| < ε/40 · 40 = ε.

3 0

2 years ago

Find the missing side length.

LekaFEV [45]

I don’t know how those answers exist but the correct answer here is 10, 16-6 = 10 yds.

6 0

2 years ago