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defon
3 years ago
14

Find the area of the figure. A) 114 cm2 B) 350 cm2 C) 418 cm2 D) 447 cm2

Mathematics
2 answers:
NISA [10]3 years ago
5 0
Square has 4 equal sides so it would be the same for all the corresponding sides.

area of square = s*s

                        = 19*19

                       = 361cm^2

area of triangle = \frac{hb}{2}

                         = 19*6

                         = 114/2

                         = 57cm^2

total area = 57+361=418cm^2
Strike441 [17]3 years ago
4 0

Answer:

Step-by-step explanation:

418 cm2

Separate the figure into a square and a triangle. Find the area of each and add.

s2 = area of square

192 = 361 cm

bh = area of triangle

(6)(19) = 57 cm2

361 + 57 = 418 cm2

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Answer:

Step-by-step explanation:

width= x

length= x+6

perimter  = 2x+ 2(x+6)

36= 2x+ 2x+ 12

36= 4x + 12

36-12= 4x

4x= 24

x= 6

therefore width= 6cm

length= 6+6 = 12

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I GIVES BRAINLIEST TO PEOPLES WHO ANSWERS
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A) $28.14       B) $8.14

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7 0
3 years ago
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Find the angle between the given vectors to the nearest tenth of a degree.
pochemuha
\bf \textit{angle between two vectors }\\ \quad \\
cos(\theta)=\cfrac{u \cdot v}{||u||\ ||v||} \implies \cfrac{\text{dot product}}{\text{product of magnitudes}}\\ \quad \\\\
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\theta=cos^{-1}\left[ \cfrac{\ \textless \ 8,4\ \textgreater \ \quad \cdot \quad \ \textless \ 9,-9\ \textgreater \ }{(\sqrt{8^2+4^2})(\sqrt{9^2+(-9)^2})} \right]


\bf \theta=cos^{-1}\left[ \cfrac{(8\cdot 9)+(4\cdot -9)}{(\sqrt{64+16})(\sqrt{81+81})} \right]\implies \theta=cos^{-1}\left[ \cfrac{36}{(\sqrt{80})(\sqrt{162})} \right]
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6 0
3 years ago
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Question 17
RoseWind [281]

Answer:

Perpendicular

Step-by-step explanation:

Remember that:

  • Two lines are parallel if their slopes are the same.
  • Two lines are perpendicular is their slopes are negative reciprocals.
  • And two lines are neither (a.k.a intersecting) if they are neither parallel nor perpendicular.

So, let's find the slope of QR and ST.

QR)

We can use the slope formula:

m=\frac{y_2-y_1}{x_2-x_1}

Let Q(-6, 11) be (x₁, y₁) and R(2, -1) be (x₂, y₂). Substitute:

m=\frac{-1-11}{2-(-6)}

Subtract:

m=-12/8=-3/2

So, the slope of QR is -3/2.

ST)

Let S(-4, 8) be (x₁, y₁) and T(-1, 10) be (x₂, y₂). Substitute:

m=\frac{10-8}{-1-(-4)}

Subtract:

m=2/3

So, the slope of ST is 2/3.

The negative reciprocal of -3/2 is 2/3.

And the negative reciprocal of 2/3 is -3/2.

So, the slopes are indeed negative reciprocals of each other.

So, QR and ST are perpendicular.

7 0
3 years ago
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