Cual es el factor de 10?

Mathematics

1 answer:

Katarina [22]3 years ago

8 0

Answer:

The factors of 10 are 1, 2, 5, and 10.

Translation: Los factores de 10 son 1, 2, 5 y 10.

Step-by-step explanation:

1x10= 10

2x5= 10

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What point on the parabola is closest to the point

Thepotemich [5.8K]

Answer:

Suppose the closest point is at p=(x0,y0), and set q=(−2,−3). Then the tangent to the parabola at p is perpendicular to ℓ, the line through p,q. and we end up numerically computing the roots from here.

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2 years ago

Find ST if S(-3, 10) and T(-2, 3)

FrozenT [24]

We have the following points and their coordinates:

$\begin{gathered} S(-3,10), \\ T(-2,3). \end{gathered}$

We must compute the distance ST between them.

The distance ST between the two points is given by:

$ST=\sqrt[]{(x_S-x_T)^2+(y_S-y_T)^2_{}},$

where (xS,yS) are the coordinates of the point S and (xT,yT) are the coordinates of the point T.

Replacing the coordinates of the points in the formula above, we find that:

$\begin{gathered} ST=\sqrt[]{(-3_{}-(-2)_{})^2+(10_{}-3_{})^2_{}}, \\ ST=\sqrt[]{1^2+7^2}, \\ ST=\sqrt[]{50}\text{.} \end{gathered}$

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10 months ago

Suppose you want to build a small walkway around your garden so that there is someplace to walk around and work on your vegetabl

wolverine [178]

Answer:

Approximately 3.5 feet - Option B

Step-by-step explanation:

Imagine that you have this walkway around the garden, with dimensions 30 by 20 feet. This walkway has a difference of x feet between it's length, and say the dimension 30 feet. In fact it has a difference of x along both dimensions - on either ends. Therefore, the increases length and width should be 30 + 2x, and 20 + 2x, which is with respect to an increases area of 1,000 square feet.

( 30 + 2x ) $*$ ( 20 + 2x ) = 1000 - Expand "( 30 + 2x ) $*$ ( 20 + 2x )"

600 + 100x + 4 $x^2$ = 1000 - Subtract 1000 on either side, making on side = 0

4 $x^2$ + 100x - 400 = 0 - Take the "quadratic equation formula"

( Quadratic Equation is as follows ) - $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ,

$x=\frac{-100+\sqrt{100^2-4\cdot \:4\left(-400\right)}}{2\cdot \:4}:\quad \frac{5\left(\sqrt{41}-5\right)}{2}$ ,

$x=\frac{-100-\sqrt{100^2-4\cdot \:4\left(-400\right)}}{2\cdot \:4}:\quad -\frac{5\left(5+\sqrt{41}\right)}{2}$

There can't be a negative width of the walkway, hence our solution should be ( in exact terms ) $\frac{5\left(\sqrt{41}-5\right)}{2}$ . The approximated value however is 3.5081...or approximately 3.5 feet.