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3 years ago

What is the next step for this construction?

Mathematics

1 answer:

Bezzdna [24]3 years ago

7 0

Answer:

Step-by-step explanati

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Need to write steps and solve thanks

nalin [4]

Answer:

The coordinates of endpoint V is (7,-27)

Solution:

Given that the midpoint of line segment UV is (5,-11) And U is (3,5).

To find the coordinates of V.

The formula for mid-point of a line segment is as follows,

Midpoint of UV is $\frac{x_{1}+x_{2}}{2}$ , $\frac{y_{1}+y_{2}}{2}$

As per the formula, $\frac{x_{1}+x_{2}}{2}$ =5, $\frac{y_{1}+y_{2}}{2}$ =-11

Here $x_{1}=3; y_{1}=5$

Substituting the value of $x_{1}$ we get,

$\frac{3+x_{2}}{2}$ =5

$3+x_{2}=5\times2$

$x_{2}=10-3$

$x_{2}=7$

Substituting the value of $x_{2}$ we get,

$\frac{5+y_{2}}{2}$ =-11

$5+y_{2}=-11\times2$

$y_{2}=-22-5$

$y_{2}=-27$

So, the coordinates of V is (7,-27)

7 0

3 years ago

A submarine travels 6.6 km due West from its base and then turns and travels due South for 3.9

geniusboy [140]

Answer:

$7.7$ km

Step-by-step explanation:

The submarine's path from its base forms a right triangle when its final position is "connected" to the base. We know that the right triangle has legs of $6.6$ km and $3.9$ km, and we need to find the length of its hypotenuse. To do so, we can use the Pythagorean Theorem, which states that in a right triangle, $a^{2}+b^{2} =c^{2}$ , where $a$ and $b$ are the lengths of the triangle's legs and $c$ is the length of the triangle's hypotenuse. In this case, we know what $a$ and $b$ are, and we need to solve for c, so after substituting the given values of $a$ and $b$ into $a^{2}+b^{2} =c^{2}$ to solve for c, we get:

$a^{2}+b^{2} =c^{2}$

$6.6^{2} +3.9^{2} =c^{2}$ (Substitute $a=6.6$ and $b=3.9$ into the equation)

$43.56+15.21=c^{2}$ (Evaluate the squares on the LHS)

$58.77=c^{2}$ (Simplify the LHS)

$c^{2}=58.77$ (Symmetric Property of Equality)

$\sqrt{c^{2} } =\sqrt{58.77}$ (Take the square root of both sides of the equation)

$c=7.7,c=-7.7$ (Simplify)

$c=-7.7$ is an extraneous solution because you can't have negative distance, if that makes sense, so therefore, the submarine is approximately $7.7$ km away from its base. Hope this helps!