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3 years ago

6

Find the mean and median what is the length of the jumping frogs bodies shown below ( the lengths are centimeters ) - 20.3 , 12.

5 , 7.6 , 17.5 , 15.6 , 14.1 !

1 answer:

Reptile [31]3 years ago

4 0

Mean-20.3+12.5+7.9+17.5+15.6+14.1= 87.6
87.6/6=20.1
Median- 7.6+17.5/2= 12.55

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Write 2 fractions equivalent to 7/12

NeTakaya

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Step-by-step explanation:

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In JKL, k=4.1 cm,j=3.8 cm and angle J=Q3^ Find all possible values of angle K , to the nearest inch of a degree .

spayn [35]

Answer:

74.0°

Step-by-step explanation:

In triangle JKL, k = 4.1 cm, j = 3.8 cm and ∠J=63°. Find all possible values of angle K, to the nearest 10th of a degree

Solution:

A triangle is a polygon with three sides and three angles. Types of triangles are right angled triangle, scalene triangle, equilateral triangle and isosceles triangle.

Given a triangle with angles A, B, C and the corresponding sides opposite to the angles as a, b, c. Sine rule states that for the triangle, the following holds:

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In triangle JKL, k=4.1 cm, j=3.8 cm and angle J=63°.

Using sine rule, we can find ∠K:

$\frac{k}{sin(K)}=\frac{j}{sin(J)} \\\\\frac{4.1}{sin(K)}=\frac{3.8}{sin(63)} \\\\sin(K)=\frac{4.1*sin(63)}{3.8}\\\\sin(K)=0.9613\\\\K=sin^{-1} (0.9613)\\\\K=74.0^o \\$

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3 years ago

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Step-by-step explanation:

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2 years ago

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Snowcat [4.5K]

Answer:

The answer is the option B

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Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$x^{2}+18x+81=25$

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so

$a=1\\b=18\\c=56$

substitute in the formula

$x=\frac{-18(+/-)\sqrt{18^{2}-4(1)(56)}} {2(1)}$

$x=\frac{-18(+/-)\sqrt{100}} {2}$

$x=\frac{-18(+/-)10} {2}$

$x1=\frac{-18+10} {2}=-4$

$x2=\frac{-18-10} {2}=-14$

7 0

3 years ago

Read 2 more answers