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2 years ago

The Fighting Hens' record has improved every year. The 2016 team won 3

Mathematics

1 answer:

tankabanditka [31]2 years ago

3 0

Answer:

12321

Step-by-step explanation:

jeff

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Graph a line that is perpendicular to the given line. Determine the slope of the given line and the one you graphed in simplest

MArishka [77]

Answer:

What is the the given line?

Explanation:

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3 years ago

Square inch for 30 stones

Anna71 [15]

1 stone = 20 square inches

1 stone * 30 = 20 square inches * 30

30 stones = 600 square inches

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3 years ago

What is the length of the base if the area is 32 cmand the height is 4 cm

vovikov84 [41]

Answer:

128,8,28,36

Step-by-step explanation:

if u multiply 32 times 4 u get 128 but if u subtract them u get 28

but if u divide them then u get 8 but if u also add them u get 36

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3 years ago

If angle X is an acute angle with tan X = 8/7, what is the value of sec X?

iren [92.7K]

Answer: $\frac{\sqrt{113}}{7}$

Step-by-step explanation:

As X is an acute angle, all 6 trigonometric functions with an argument of X are positive.

Using the identity $1+\tan^{2} X=\sec^{2} X$ ,

$1+\left(\frac{8}{7} \right)^{2}=\sec^{2} X\\\\\sec^{2} X=\frac{113}{49}\\\\\therefore \sec X=\boxed{\frac{\sqrt{113}}{7}}$

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2 years ago

Please help?

Mazyrski [523]

Answer:

$23\sqrt{3}\ un^2$

Step-by-step explanation:

Connect points I and K, K and M, M and I.

1. Find the area of triangles IJK, KLM and MNI:

$A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\$

2. Note that

$A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2$

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

$A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2$

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

7 0

3 years ago