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alexgriva [62]
3 years ago
6

55.757 round to tenths

Mathematics
2 answers:
Leno4ka [110]3 years ago
8 0
55.8 is the answer

Please make me brainliest.
Ksivusya [100]3 years ago
8 0
55.8 is 55.757 rounded to tenths
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I need help please answers please
motikmotik
We have two points:
(-3,5) and (2,-3)

I
An equation in oint-slope form has this shape:
y-y₀=m(x-x₀)
m=slope

1) we compute the slope (m) between two points:
Given two points (x₁,y₁) and (x₂,y₂) the slope between these pointw will be.
m=(y₂-y₁)/(x₂-x₁)

In this case, our slope would be:
m=(-3-5)/(2+3)=-8/5

2) we choose one of these points and replace it in the equation, the result will be the same:
(-3,5)
<span>y-y₀=m(x-x₀)
</span>y-5=-8/5(x+3)

Answer: y-5=-8/5(x+3)
8 0
4 years ago
The difference between each pair of consecutive terms in this arithmetic sequence is 1 1/4 . Find the next term in the sequence.
Ksenya-84 [330]

Answer:

7 3/4

Step-by-step explanation:

6 1/2 +1 1/4=7 3/4

To find any number in a sequence given the distance between each number, just add the distance to the last number in the sequence

3 0
3 years ago
Please answer!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Ilya [14]

Answer:

The length of the hypotenuse is 26 inches

Step-by-step explanation:

The leg(brace) attached to the wall is 10 inches

Perpendicular = 10 inches

Your surfboard sits on a leg that is 24 inches.

Base = 24 inches

Formula : Hypotenuse^2 = Perpendicular^2 +Base^2

Hypotenuse^2 = 10^2+24^2

Hypotenuse = \sqrt{10^2+24^2}

Hypotenuse = 26

Hence  the length of the hypotenuse is 26 inches

3 0
3 years ago
6 less than the number y
Nonamiya [84]

Answer:

6 less a number y can be represented by the expression 6 - y

3 0
3 years ago
Is it true that the integral LaTeX: \int x^2e^{2x}dx ∫ x 2 e 2 x d x can be evaluated using integration by parts? If so, state t
Marysya12 [62]

Answer:

Yes the integral can be evaluated by integration by parts as solved below.

Step-by-step explanation:

\int x^{2}e^{2x}dx

Taking algebraic function as first function and exponential function as second function we have

\int x^{2}e^{2x}dx=x^{2}\int e^{2x}dx-\int (x^{2})'\int e^{2x}dx\\\\=x^{2}\frac{e^{2x}}{2}-\int 2x\times \frac{e^{2x}}{2}dx\\\\\frac{x^{2}e^{2x}}{2}-\int xe^{2x}dx\\\\Now\\\\\int xe^{2x}dx=x\int e^{2x}dx-\int 1\cdot \int e^{2x}dx\\\\=\frac{xe^{2x}}{2}-\int \frac{e^{2x}}{2}dx\\\\\frac{xe^{2x}}{2}-\frac{e^{2x}}{4}\\\\\therefore \int x^{2}e^{2x}dx=\frac{x^{2}e^{2x}}{2}-\frac{xe^{2x}}{2}+\frac{e^{2x}}{4}

5 0
3 years ago
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