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NeTakaya
2 years ago
7

Triangle DEF has vertices located at D (2, 1), E (3,5), and F (6,2).

Mathematics
1 answer:
IRINA_888 [86]2 years ago
3 0

Answer:

(A)

Length of DE = √17

EF = √18

DF = √17

(B) Slope of DE = 4

EF = 1

DF = 1/4

(C) Isosceles triangle.

Step-by-step explanation:

See attached image.

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Length = 2 feet, width = 15 inches, height = 8 inches<br><br><br><br> V= _____________
I am Lyosha [343]
Volume = length * width ( aka base ) * height


2 feet = 24 inches.

24*15*8=2880 inches.
6 0
2 years ago
Read 2 more answers
D<br> Evaluate<br> arcsin<br> (6)]<br> at x = 4.<br> dx
sineoko [7]

Answer:

\frac{1}{2\sqrt{5} }

Step-by-step explanation:

Let, \text{sin}^{-1}(\frac{x}{6}) = y

sin(y) = \frac{x}{6}

\frac{d}{dx}\text{sin(y)}=\frac{d}{dx}(\frac{x}{6})

\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

\frac{d}{dx}\text{sin(y)}=\text{cos}(y)\frac{dy}{dx} ---------(1)

\frac{1}{6}=\text{cos}(y)\frac{dy}{dx}

\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}

cos(y) = \sqrt{1-\text{sin}^{2}(y) }

          = \sqrt{1-(\frac{x}{6})^2}

          = \sqrt{1-(\frac{x^2}{36})}

Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

Or \frac{d}{dx}[\text{sin}^{-1}(\frac{x}{6})]=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

                   =\frac{1}{2\sqrt{5}}

4 0
2 years ago
Which relationship describes angles 1 and 2?
Alchen [17]
They are adjacent angles

7 0
2 years ago
Read 2 more answers
What is the standard form of the equation of a line for which the length of the normal segment to the origin is 8 and the normal
jeka57 [31]
Slope of line = tan(120) = -tan(60) = - &radic;3
Distance from origin = 8

Let equation be Ax+By+C=0
then -A/B=-&radic;3, or
B=A/&radic;3.
Equation becomes
Ax+(A/&radic;3)y+C=0

Knowing that line is 8 units from origin, apply distance formula
8=abs((Ax+(A/&radic;3)y+C)/sqrt(A^2+(A/&radic;3)^2))
Substitute coordinates of origin (x,y)=(0,0)  =>
8=abs(C/sqrt(A^2+A^2/3))
Let A=1 (or any other arbitrary finite value)
solve for positive solution of C
8=C/&radic;(4/3) => C=8*2/&radic;3 = (16/3)&radic;3

Therefore one solution is
x+(1/&radic;3)+(16/3)&radic;3=0
or equivalently
&radic;3 x + y + 16 = 0

Check:
slope = -1/&radic;3  .....ok
distance from origin
= (&radic;3 * 0 + 0 + 16)/(sqrt(&radic;3)^2+1^2)
=16/2
=8  ok.

Similarly C=-16 will satisfy the given conditions.

Answer  The required equations are
&radic;3 x + y = &pm; 16 
in standard form.

You can conveniently convert to point-slope form if you wish.




4 0
3 years ago
The data list shows the scores of ten students in Mr. Smith's math class. 61, 67, 81, 83, 87, 88, 89, 90, 98, 100 What is the st
cluponka [151]

Answer:

12.3

Step-by-step explanation:

Step 1

We find the mean

The data list shows the scores of ten students in Mr. Smith's math class. 61, 67, 81, 83, 87, 88, 89, 90, 98, 100

Mean = Sum of terms/Number of terms

Number of terms = 10

Mean = 61 + 67 + 81 + 83 + 87 + 88 + 89 + 90 + 98 + 100/10

Mean = 844/10

Mean = 84.4

Step 2

Standard deviation

The formula for sample standard deviation =

√(x - Mean)²/n - 1

= √[(61 - 84.4)² + (67 - 84.4)² + (81 - 84.4)² + (83 - 84.4)² + (87 - 84.4)² + (88 - 84.4)² + (89 - 84.4)² + (90 - 84.4)² + (98 - 84.4)² + (100 - 84.4)²]/10 - 1

=√ 547.56 + 302.76 + 11.56 + 1.96 + 6.76 + 12.96 + 21.16 + 31.36 + 184.96 + 243.36/10 - 1

= √1364.4/9

= √151.6

= 12.31259518

Approximately to the nearest tenth = 12.3

The standard deviation = 12.3

6 0
2 years ago
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