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3 years ago

13

Is 3x-6y a direct variation? Identify the constant of variation

1 answer:

Alex777 [14]3 years ago

8 0

Answer:

1

Step-by-step explanation:

1

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Emily wants to know the volume of a beach ball. She deduces that the beach ball has diameter of 12 inches, and then uses the fol

Over [174]

Answer:

D. She used the beach's ball's diameter when she should have used the radius

Step-by-step explanation:

To find the volume of the beach ball, using the volume of a sphere is the right formula to use, which is ⁴/3πr³.

The formula she used is correct.

Since the diameter of the ball is assumed to be 12 inches, what is needed to find the volume is the radius.

Radius (r) = ½(diameter) = ½(12) = 6 in.

This is where Emily made a mistake.

She used the diameter of the beach ball instead of its radius (r) which is needed in the equation.

She should have gotten,

V = ⁴/3(3.14)(6)² = 904.32 cubic inches

7 0

3 years ago

The value of x-y is negative, and x and y are both positive. What can you conclude about the values of x

Stolb23 [73]

Well for two positive numbers subtracted to equal a negative number, the first number must have a value that is less than the second number. (Ex. 3-4= -1) In this case, you can conclude that x is less than y! (x<y)

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3 years ago

21) 0=-2y + 10 - 6x<br> 14 - 22y= 18x

zhenek [66]

Answer:

PlPlease be more spacific

Step-by-step explanation:

4 0

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

alexandr402 [8]

Answer:

let the grade points be a,b,c

a+b+c=258

b=c+10( since she noticed that on tests two and three,she scored 10 points fewer).

3 0

3 years ago

Read 2 more answers