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Step2247 [10]
3 years ago
12

The friendly old woman sold 38 long-

Mathematics
1 answer:
dem82 [27]3 years ago
3 0

Answer:

Step-by-step eWrite 2.9 as  

2.9

1

Multiply both numerator and denominator by 10 for every number after the decimal point

2.9 × 10

1 × 10

=  

29

10

Reducing the fraction gives

29

10xplanation:

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Art [367]
The slope is 2-2/6-6 which is zero . So the slope is no slope
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3 years ago
Solve the oblique triangle where side a has length 10 cm, side c has length 12 cm, and angle beta has measure thirty degrees. Ro
Ugo [173]

Answer:

Side\ B = 6.0

\alpha = 56.3

\theta = 93.7

Step-by-step explanation:

Given

Let the three sides be represented with A, B, C

Let the angles be represented with \alpha, \beta, \theta

[See Attachment for Triangle]

A = 10cm

C = 12cm

\beta = 30

What the question is to calculate the third length (Side B) and the other 2 angles (\alpha\ and\ \theta)

Solving for Side B;

When two angles of a triangle are known, the third side is calculated as thus;

B^2 = A^2 + C^2 - 2ABCos\beta

Substitute: A = 10,  C =12; \beta = 30

B^2 = 10^2 + 12^2 - 2 * 10 * 12 *Cos30

B^2 = 100 + 144 - 240*0.86602540378

B^2 = 100 + 144 - 207.846096907

B^2 = 36.153903093

Take Square root of both sides

\sqrt{B^2} = \sqrt{36.153903093}

B = \sqrt{36.153903093}

B = 6.0128115797

B = 6.0 <em>(Approximated)</em>

Calculating Angle \alpha

A^2 = B^2 + C^2 - 2BCCos\alpha

Substitute: A = 10,  C =12; B = 6

10^2 = 6^2 + 12^2 - 2 * 6 * 12 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 36 + 144 - 144 *Cos\alpha

100 = 180 - 144 *Cos\alpha

Subtract 180 from both sides

100 - 180 = 180 - 180 - 144 *Cos\alpha

-80 = - 144 *Cos\alpha

Divide both sides by -144

\frac{-80}{-144} = \frac{- 144 *Cos\alpha}{-144}

\frac{-80}{-144} = Cos\alpha

0.5555556 = Cos\alpha

Take arccos of both sides

Cos^{-1}(0.5555556) = Cos^{-1}(Cos\alpha)

Cos^{-1}(0.5555556) = \alpha

56.25098078 = \alpha

\alpha = 56.3 <em>(Approximated)</em>

Calculating \theta

Sum of angles in a triangle = 180

Hence;

\alpha + \beta + \theta = 180

30 + 56.3 + \theta = 180

86.3 + \theta = 180

Make \theta the subject of formula

\theta = 180 - 86.3

\theta = 93.7

5 0
3 years ago
Does anyone know how to solve this??
sergij07 [2.7K]
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Georgia [21]

Step-by-step explanation:

The system of equations for eq 1 which is 3x + y = 118 represents the Green High School which filled three buses(with a specific number of students identified as x) and a van(with a specific number of students identified as y) with a total of 118 students.

for eq 2; 4x + 2y = 164; represents Belle High School which filled four buses(with a specific number of students identified as x) and two vans(with a specific number of students identified as y) with a total of 164 students.

The solution represents the specific number of students in the buses and vans in eq1 and eq 2 with x being 36 students and y being 10 students.

substituting 36 for x and 10 for y in eq 1;

3(36) + 10 = 108 + 10 = 118 total students for Green High School

substituting 36 for x and 10 for y in eq2;

4(36) + 2(10) = 144 + 20 = 164 total students for Belle High school

6 0
2 years ago
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