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3 years ago

Similiar triangles can have congruent angles and sides

2 answers:

5 0

This is true. Since they could be considered similar even when they are congruent, since in similar triangles, the measurements are proportional. In congruent triangles, this proportion is just one. or 1/1

Hope this helps!

denpristay [2]3 years ago

4 0

Answer:

The statement is true.

Step-by-step explanation:

The statement is true.

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Find the equation of the line that passes through (–3, 2) and the intersection of the lines x–2y=0 and 3x+y+5=0.

ipn [44]

Answer:

$\frac{19}{7}$ x+ $\frac{11}{7}$ y+5=0

Step-by-step explanation:

the intersection of x-2y=0 and 3x+y+5 is ( $\frac{-10}{7}$ ; $\frac{-5}{7}$ )

=> the line : $\frac{19}{7}$ x+ $\frac{11}{7}$ y+5=0

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3 years ago

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Find the perimeter of the shaded region. Round your answer to the nearest hundredth. Help me por favor

timurjin [86]

Answer:

19.15

Step-by-step explanation:

so find the circumference of the circle using c=2*pi*r it's like 18.849. then find the perimeter of the box so p=2l+2w and it's like 38. so subtract the circle circumference from 38 so 38-18.849=19.15

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3 years ago

Need help ASAP ? <br>3/4×20

In-s [12.5K]

3/4 x 20

3/4 x 20/1

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4 x 1 = 4

60 / 4 = 15

8 0

3 years ago

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Create a square root with at least 3 transformations

Fittoniya [83]

Answer:

Step-by-step explanation:

Start with f(x) = √x.

1) Multiplying this by "c" will either stretch or shrink the graph: stretch, if "c" is > 0; shrink, if 0 < c < 1.

2) Replacing 'x' with 'x - h' will translate the original graph h units to the right, if h is positive;

Replacing 'x' with 'x - h' will translate the original graph h units to the left, if h is negative.

3) Adding k to f(x) = √x will translate the graph upward by k units if k is positive and downward by k units if k is negative.

8 0

3 years ago

Here are the endpoints of the segments BC, FG, and JK.<br> B, −67

yulyashka [42]

$~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{-6}~,~\stackrel{y_1}{7})\qquad C(\stackrel{x_2}{-4}~,~\stackrel{y_2}{4})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ BC=\sqrt{[-4 - (-6)]^2 + [4 - 7]^2}\implies BC=\sqrt{(-4+6)^2+(-3)^2} \\\\\\ BC=\sqrt{2^2+(-3)^2}\implies \boxed{BC=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points}$

$F(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad G(\stackrel{x_2}{1}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ FG=\sqrt{[1 - (-2)]^2 + [-2 - (-4)]^2}\implies FG=\sqrt{(1+2)^2+(-2+4)^2} \\\\\\ FG=\sqrt{9+4}\implies \boxed{FG=\sqrt{13}} \\\\[-0.35em] ~\dotfill\\\\ ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ J(\stackrel{x_1}{4}~,~\stackrel{y_1}{2})\qquad K(\stackrel{x_2}{5}~,~\stackrel{y_2}{-2})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}$

$JK=\sqrt{[5 - 4]^2 + [-2 - 2]^2}\implies JK=\sqrt{1^2+(-4)^2}\implies \boxed{JK=\sqrt{17}} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \overline{BC}\cong \overline{FG}~\hfill$

4 0

2 years ago