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3 years ago

Can someone give me the answer to this please?

Mathematics

2 answers:

andre [41]3 years ago

7 0

Answer:

ok so its C bc 8 x 0.5 = 4 and so there are 4 long rectangles so multiply by 4 which is 16 bc 4 x 4 = 16 next find the area of the 2 squares 0.5 x 0.5 =0.25

multiply that by 2 because there are 2 squares which is 0.5 and so

16+0.5= 16.5 in²

Step-by-step explanation:

GalinKa [24]3 years ago

5 0

Answer:

Step-by-step explanation:

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yawa3891 [41]

Answer:

W = 4

Step-by-step explanation:

4/6 = 6/9

2/3 = 2/3

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3 years ago

Given the three vertices A(−2, −3), B(2, 3), and C(4, −7), what are the coordinates of D that make quadrilateral ABDC a square?

lorasvet [3.4K]

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hope this helped

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3 years ago

Which trigonometric ratios are correct for triangle ABC? Check all that apply

sergij07 [2.7K]

Consider an angle M with measure m≠90°, in a right triangle.

Let

OPP denote the length of the side opposite to M,
ADJ denote the length of the side adjacent to M, and
HYP denote the hypotenuse.

then:

Sin(M) = OPP/HYP
Cos(M)= ADJ/HYPP
Tan(M)=OPP/ADJ

Back to our problem,

using the Pythagorean we can find the length of AB:

$|AB|^2+|AC|^2=|BC|^2\\\\|AB|^2+9^2=18^2\\\\|AB|^2=18^2-9^2\\\\|AB|^2=(18-9)(18+9)=9 \cdot 27=9 \cdot 9 \cdot3\\\\|AB|=9 \sqrt{3} 
$

$Sin(C)= \frac{OPP}{HYP}=\frac{9 \sqrt{3} }{18}= \frac{ \sqrt{3} }{2}$
$Cos(B)= \frac{ADJ}{HYP}= \frac{9 \sqrt{3}}{18}= \frac{ \sqrt{3}}{2}$
$Tan(C)= \frac{OPP}{ADJ}= \frac{9 \sqrt{3} }{9}= \sqrt{3}$
$Sin(B)= \frac{OPP}{HYP}= \frac{9}{18}= \frac{1}{2}$
$Tan(B)= \frac{OPP}{ADJ}= \frac{9}{9 \sqrt{3}}= \frac{1}{ \sqrt{3} }= \frac{ \sqrt{3}}{3}$

Answer: 1, 3, 4

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3 years ago