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3 years ago

Two sides and the non-included right angle of one right triangle are congruent to the corresponding parts of another

Mathematics

2 answers:

fgiga [73]3 years ago

7 0

Answer:

hypotenuse leg theorem or OHL

Step-by-step explanation:

Since the congruent angle is a right angle and it is not included, the two congruent sides of the triangles must include their hypotenuses and one of their legs.

The triangles would be congruent by the hypotenuse leg theorem, which states that if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of another right triangle, the triangles are congruent.

timurjin [86]3 years ago

7 0

Answer:

Step-by-step explanation:

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Endpoints of segment MN have coordinates (0, 0), (5, 1). The endpoints of segment AB have coordinates (1 1/22 , 2 1/4 ) and (−2

Nana76 [90]

Answer: $k=18\dfrac{8}{11}$ .

Step-by-step explanation:

If a line passing through two points, then

$Slope=\dfrac{y_2-y_1}{x_2-x_1}$

Endpoints of segment MN have coordinates (0, 0) and (5, 1).

Slope of MN $=\dfrac{1-0}{5-0}=\dfrac{1}{5}$

The endpoints of segment AB have coordinates $\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)$ and $\left(-2\dfrac{1}{4} , k\right)$ .

$A=\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)=\left(\dfrac{23}{22} ,\dfrac{9}{4}\right)$

$B=\left(-2\dfrac{1}{4} , k\right)=\left(-\dfrac{9}{4} , k\right)$ .

Slope of AB $=\dfrac{k-\frac{9}{4}}{-\frac{9}{4}-\dfrac{23}{22}}$

$=\dfrac{\frac{4k-9}{4}}{\frac{-99-46}{44}}$

$=\dfrac{4k-9}{4}\times \dfrac{44}{-145}$

$=4k-9\times \dfrac{11}{-145}$

$=\dfrac{44k-99}{-145}$

Product of slopes of two perpendicular segments is -1.

Slope of MN × Slope of AB = -1

$\dfrac{1}{5}\times \dfrac{44k-99}{-145}=-1$

$\dfrac{44k-99}{-725}=-1$

$44k-99=725$

$44k=725+99$

$k=\dfrac{824}{44}$

$k=\dfrac{206}{11}$

$k=18\dfrac{8}{11}$

Therefore, the value of k is $k=18\dfrac{8}{11}$ .

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4 years ago

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For the function f(x) = (x − 2)2 + 4, identify the vertex, domain, and range. a. The vertex is (–2, 4), the domain is all real n

12345 [234]

$f(x)=a(x-h)^2+k \Rightarrow \text{vertex}=(h,k)\\\\
f(x)=(x-2)^2+4 \Rightarrow \text{vertex}=(2,4)$

The range of $f(x)=a(x-h)^2+k$ is
$y\leq k$ for $a$
$y\geq k$ for $a>0$

The domain of any quadratic function is all real numbers.

In $f(x)=(x-2)^2+4$ , $a=1\ \textgreater \ 0$ , so the range is $y\geq4$

So it's D.

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Answer:

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Step-by-step explanation:

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IgorLugansk [536]

The correct answer for the stated problem above is D.

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Answer:

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