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3 years ago

LESSON 23-1 PRACTICE

Mathematics

2 answers:

Lostsunrise [7]3 years ago

7 0

Answer:

pretty sure its 2

Step-by-step explanation:

monitta3 years ago

6 0

Answer:

Step-by-step explanation:

2 tuijnbb

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A 25-foot-long footbridge has two diagonal supports that meet in the center of the bridge. Each support makes a 65∘ angle with a

GuDViN [60]

Consider the diagram. From SOH CAH TOA, you know that
sin(65°) = (12.5 ft)/x
x = (12.5 ft)/sin(65°)
x ≈ 13.8 ft

5 0

3 years ago

LOOK HERE IF YOU ANSWER CORRECTLY YOU PASS 8th grade

Helen [10]

Answer:

$3,700

Step-by-step explanation:

16% of 2500=400

400×3(years)=1200

2500+1200=3700

5 0

3 years ago

Ezra's dad is building a cover for his sandbox. The sandbox is in the shape of a kite as shown. A kite has a width of 6 feet and

Tpy6a [65]

Answer:

Area of the sandbox cover = 12 ft²

Step-by-step explanation:

The area of a kite is given as the product of the two diagonals of the kite divided by 2.

Area of kite = (xy/2)

where x and y are the diagonals of the kite.

The sandbox cover, in shape of a kite has a width of 6 ft and a height of 4 ft.

These two dimensions are the diagonals of the kite. Hence, the area of a kite is

A = (6×4/2) = 12 ft²

Hope this Helps!!!

5 0

3 years ago

Find the equation of a line passing through (-4,-3) and perpendicular to<br> 3 x + 2 y = 14.

Reil [10]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$3x+2y=14\implies 2y=-3x+14\implies y=\cfrac{-3x+14}{2}\implies y = \cfrac{-3x}{2}+\cfrac{14}{2} \\\\\\ y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{3}{2}}x+7\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so therefore

$\stackrel{~\hspace{5em}\textit{perpendicular lines have \underline{negative reciprocal} slopes}~\hspace{5em}} {\stackrel{slope}{\cfrac{-3}{2}} ~\hfill \stackrel{reciprocal}{\cfrac{2}{-3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{2}{-3}\implies \cfrac{2}{3}}}$

so we're really looking for the equation of a line whose slope is 2/3 and passes through (-4 , -3)

$(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{2}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{2}{3}}(x-\stackrel{x_1}{(-4)}) \implies y+3=\cfrac{2}{3}(x+4) \\\\\\ y+3=\cfrac{2}{3}x+\cfrac{8}{3}\implies y=\cfrac{2}{3}x+\cfrac{8}{3}-3\implies y=\cfrac{2}{3}x-\cfrac{1}{3}$

3 0

2 years ago

Suppose given △ABD and △CBD.

patriot [66]

Answer:

The required result is proved with the help of angle bisector theorem.

Step-by-step explanation:

Given △ABD and △CBD, AE and CE are the angle bisectors. we have to prove that $\frac{AD}{AB}=\frac{DC}{CB}$

Angle bisector theorem states that an angle bisector of an angle of a Δ divides the opposite side in two segments that are proportional to the other two sides of triangle.

In ΔADB, AE is the angle bisector

∴ the ratio of the length of side DE to length BE is equal to the ratio of the line segment AD to the line segment AB.

$\frac{DE}{EB}=\frac{AD}{AB}$ → (1)

In ΔDCB, CE is the angle bisector

∴ the ratio of the length of side DE to length BE is equal to the ratio of the line segment CD to the line segment CB.

$\frac{DE}{EB}=\frac{CD}{CB}$ → (2)

From equation (1) and (2), we get

$\frac{AD}{AB}=\frac{CD}{CB}$

Hence Proved.

5 0

3 years ago