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Gennadij [26K]
3 years ago
11

The sum of a number and 7 is less than 12

Mathematics
2 answers:
Hitman42 [59]3 years ago
6 0
5
explanation:
12-7=5
MissTica3 years ago
6 0
Easy it’s 5 u have to subtract the two numbers
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A circle is the set of all points that are the same distance from one given point. find an example that contradicts this definit
lyudmila [28]

The accurate definition is that the circle that's closest to the 2 dimensional figure where all the set of point in the plane should be equal distance.

<h3>How to illustrate a circle?</h3>

A circle simply means the set of all the points that are the same distance from a given point.

In this case, the contradiction is that the definition should be applied to three dimensional space so that it will be a sphere.

The accurate definition is that the circle that's closest to the 2 dimensional figure where all the set of point in the plane should be equal distance. A example is a point or line.

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2 years ago
ITS A BIT BLURRY BUT IT WORKS SOMEONE PLEASE GET THIS DONE IN THE NEXT FEW MINUTES ILL MARK BRAINLIEST
ZanzabumX [31]

Answer:

a) m∠T

b) line W and line Z

c) ∠ZTU

d) ∠YTX

e) 90° or m∠5

f) m∠1 and m∠2

g) m∠2

h) 180° or m∠1 & m∠5 or m∠2 & m∠3 & m∠4

i) line WTY and line YTX

j) the angle bisector U

Step-by-step explanation:

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Emergency question! please help
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2 double stuffed oreos that have 150 calories has more calories if you add that up its 300

Step-by-step explanation:

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Which expression and diagram represent "three times a number"?
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Step-by-step explanation:

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Please help solve this system of equations
stepan [7]

Make a substitution:

\begin{cases}u=2x+y\\v=2x-y\end{cases}

Then the system becomes

\begin{cases}\dfrac{2\sqrt[3]{u}}{u-v}+\dfrac{2\sqrt[3]{u}}{u+v}=\dfrac{81}{182}\\\\\dfrac{2\sqrt[3]{v}}{u-v}-\dfrac{2\sqrt[3]{v}}{u+v}=\dfrac1{182}\end{cases}

Simplifying the equations gives

\begin{cases}\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81}{182}\\\\\dfrac{4\sqrt[3]{v^4}}{u^2-v^2}=\dfrac1{182}\end{cases}

which is to say,

\dfrac{4\sqrt[3]{u^4}}{u^2-v^2}=\dfrac{81\times4\sqrt[3]{v^4}}{u^2-v^2}

\implies\sqrt[3]{\left(\dfrac uv\right)^4}=81

\implies\dfrac uv=\pm27

\implies u=\pm27v

Substituting this into the new system gives

\dfrac{4\sqrt[3]{v^4}}{(\pm27v)^2-v^2}=\dfrac1{182}\implies\dfrac1{v^2}=1\implies v=\pm1

\implies u=\pm27

Then

\begin{cases}x=\dfrac{u+v}4\\\\y=\dfrac{u-v}2}\end{cases}\implies x=\pm7,y=\pm13

(meaning two solutions are (7, 13) and (-7, -13))

8 0
2 years ago
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