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Olenka [21]
3 years ago
13

Select proportional or not proportional to correctly classify the pair of ratios 1.8/7.2 and 0.4/1.6

Mathematics
1 answer:
andreev551 [17]3 years ago
8 0

Answer: proportional

Step-by-step explanation: just took the quiz

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Answer:my xbox name is lunabgdrg

Step-by-step explanation:

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What number can be multiplied by 10 to make 42,300?
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10*4230 that should do the trick for ya just multiply it and ya got answer
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4 years ago
A set of equations is given below:
ipn [44]
The correct answer is 'Equation F can be written as 2d + 1 = 3d + 7'. In order to find the answer to a system of equations, the two equations must be set equal to each other. For instance, if we had the equations x = y + 1 and x = 3y - 1, we would set the two equations equal to one another to find the answer.

x = y + 1
x = 3y - 1
y + 1 = 3y - 1
1 = 2y - 1
2 = 2y
1 = y

x = y + 1
x = 1 + 1
<span>x = 2

We can use this to solve the set of equations above.

</span><span>2d + 1 = 3d + 7
</span>1 = d + 7
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c = 2d + 1
c = 2(-6) + 1
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Hope this helps!
6 0
3 years ago
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Which Data set has the smallest IQR ?
mel-nik [20]

Answer:

<h2 /><h2>The interquartile range (IQR) is the difference between the upper (Q3) and lower (Q1) quartiles, and describes the middle 50% of values when ordered from lowest to highest. The IQR is often seen as a better measure of spread than the range as it is not affected by outliers. Interquartile Range. 25% of values.</h2>

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8 0
4 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
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