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2 years ago

A4 Find length AC idk how to do this someone help me

Mathematics

2 answers:

Alecsey [184]2 years ago

8 0

Answer:

AC ≈ 16.8 cm ( to 1 dec. place )

Step-by-step explanation:

Using the sine ratio in the right triangle

sin71° = $\frac{opposite}{hypotenuse}$ = $\frac{AB}{AC}$ = $\frac{15.9}{AC}$

Multiply both sides by AC

AC × sin71° = 15.9 ( divide both sides by sin71° )

AC = $\frac{15.9}{sin71}$ ≈ 16.8 cm

Olegator [25]2 years ago

7 0

Answer:

16.8 cm

Step-by-step explanation:

Sin (theta)=opposite/HYPOTENUSE

Where theta=71 and opposite=15.9

Putting in the above formula

Sin 71=15.9/AC

AC=15.9/sin71

AC=15.9/0.945

AC=16.8 cm

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2 years ago

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2 years ago

The coordinate of centroid of a triangle whose vertices are (1,3,-2), (4,5,0), (6,3,9) is = ……………….

Dominik [7]

Answer:

$C = (\frac{11}{3},\frac{11}{3},\frac{7}{3})$

Step-by-step explanation:

Given

$(x_1,y_1,z_1) = (1,3,-2)$

$(x_2,y_2,z_2) = (4,5,0)$

$(x_3,y_3,z_3) = (6,3,9)$

Required

Determine the coordinates of the centroid

Represent the coordinates with C.

C is calculated as follows:

$C = (\frac{1}{3}(x_1+x_2+x_3),\frac{1}{3}(y_1+y_2+y_3),\frac{1}{3}(z_1+z_2+z_3}))$

Substitute values of x and y in the given equation

$C = (\frac{1}{3}(1+4+6),\frac{1}{3}(3+5+3),\frac{1}{3}(-2+0+9}))$

$C = (\frac{1}{3}(11),\frac{1}{3}(11),\frac{1}{3}(7}))$

$C = (\frac{11}{3},\frac{11}{3},\frac{7}{3})$

<em>The above is the coordinate of the centroid</em>

8 0

2 years ago

Sally made a box to hold her jewlery collection. She used 42 inches of wood to build the sides of the box. If the box is 9 inche

Juliette [100K]

The length is 12, because if it's 9 inches wide it has two sides of 9 inches which is 18. Then subtract that from the intial value 42 to get 24. After, you divide 24 by 2 due to there being 2 sides of the box and 24 is the sum of both sides. Hopethis helps!

8 0

2 years ago

Solve the inequality 2x>30+5/4x

insens350 [35]

Answer:

Step-by-step explanation:

$2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} } \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0$

$x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }$

case~2

$if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0$

8 0

1 year ago