Please help lol !!!!

Mathematics

2 answers:

dangina [55]3 years ago

7 0

Answer:

63 degrees

Step-by-step explanation:

Since this is an isosceles triangle, the two bottom angles must be equal, while the top angle (in the way this specific triangle is oriented) must be different. Therefore, angle CDB is 63 degrees.

valentinak56 [21]3 years ago

5 0

Answer:

∠CDB = 63°

Step-by-step explanation:

∠CBD and ∠CDB opposite side lengths are congruent (indicated by the little markings)

Because their opposite sides are congruent they are too congruent

So ∠CBD = ∠CDB

because ∠CBD = 63° ∠CDB also equal 63°

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A figure is broken into a rectangle and a triangle. The triangle has a base of 2 and two-thirds feet and height of 3 feet. The r

ohaa [14]

Answer:

$b=2\frac23\:\sf ft$

Area of triangle = 4 ft²

Area of rectangle = $6\frac23\: \sf ft^2$

Area of irregular figure = $10\frac23\: \sf ft^2$

Step-by-step explanation:

$\begin{aligned}\sf Base\:of\:triangle\:(b) & = 5-2\frac13\\\\& = \dfrac{15}{3}-\dfrac73\\\\ & = \dfrac83\\\\ & = 2\frac23\:\sf ft \end{aligned}$

$\begin{aligned}\sf Area\:of\:a\:triangle & =\dfrac12 \sf \times base \times height\\\\& = \dfrac12 \times b \times 3\\\\ & = \dfrac12 \times \dfrac83 \times \dfrac31\\\\ & = \dfrac{24}{6}\\\\ & = 4\: \sf ft^2\end{aligned}$

$\begin{aligned}\sf Area\:of\:rectangle& =\sf length \times width\\\\& = 5 \times 1\frac13\\\\ & = \dfrac51\times \dfrac43\\\\& = \dfrac{20}{3}\\\\ & = 6\frac23\: \sf ft^2\end{aligned}$

$\begin{aligned} \sf Area\:of\:irregular\:figure & = \sf area\:of\:triangle+area\:of\:rectangle\\\\ & = 4 + 6\frac23\\\\ & = 10\frac23\: \sf ft^2\end{aligned}$