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2 years ago

find the perimeter and area of the polygon with the given vertices. P(-3,4), Q(1,4), R(-3,-2), S(3,-2)

Mathematics

1 answer:

Elza [17]2 years ago

5 0

Answer:

OMG!!!!!!!!!OMG!!!OMG

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Zarrin [17]

Answer:

one-half

Step-by-step explanation:

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1 year ago

15 POINTS

ra1l [238]

D because using the y=mx+b the slope(mx) is 1/2 (rise over run) and it starts at point 0,-2(b) since y is always 0 and -2 is the x value

6 0

2 years ago

Find the length of the third side. If necessary, round to the nearest tenth

Lilit [14]

$\huge\bold{Given:}$

Length of the base = 8

Length of the hypotenuse = 17

$\huge\bold{To\:find:}$

The length of the third side '' $x$ ".

$\large\mathfrak{{\pmb{\underline{\orange{Solution}}{\orange{:}}}}}$

$\longrightarrow{\purple{x\:=\: 15}}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}$

Using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

$\longrightarrow{\blue{}}$ ${x}^{2}$ + (8)² = (17)²

$\longrightarrow{\blue{}}$ ${x}^{2}$ + 64 = 289

$\longrightarrow{\blue{}}$ ${x}^{2}$ = 289 - 64

$\longrightarrow{\blue{}}$ ${x}^{2}$ = 225

$\longrightarrow{\blue{}}$ $x$ = $\sqrt{225}$

$\longrightarrow{\blue{}}$ $x$ = $15$

Therefore, the length of the missing side $x$ is $15$ .

$\huge\bold{To\:verify :}$

$\longrightarrow{\green{}}$ (15)² + (8)² = (17)²

$\longrightarrow{\green{}}$ 225 + 64 = 289

$\longrightarrow{\green{}}$ 289 = 289

$\longrightarrow{\green{}}$ L.H.S. = R. H. S.

Hence verified.

$\huge{\textbf{\textsf{{\orange{My}}{\blue{st}}{\pink{iq}}{\purple{ue}}{\red{35}}{\green{♨}}}}}$

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1 year ago

Name two pairs of congruent angles.

larisa [96]

ZOSP= PSQ and

ZPSQ E ZOSR

4 0

2 years ago

Help on this question plss!

GuDViN [60]

Adding 4 to both side is wrong he has to be subtract 4 to both side

7 0

2 years ago

Read 2 more answers