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vampirchik [111]

3 years ago

12

Find the quotient -14/7

1 answer:

AVprozaik [17]3 years ago

4 0

The quotient is -2.

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Find the value of x show your work

Ksivusya [100]

Answer:

x≈13.08

Step-by-step explanation:

We use the pythagora's theorem

$a^{2} +b^2=c^2\\a=5\\b=x\\c=14\\5^2+x^2=14^2\\x^2=196-25\\x^2=171\\x=3\sqrt{19} =13.08$

4 0

3 years ago

What is 270° converted to radians?<br><br> A.) pi/6<br> B.) 3/2<br> C.) 3pi/2<br> D.) 3

Aleksandr-060686 [28]

Answer:

the answer C) 3pi/2 semoga membantu

3 0

3 years ago

the temperature is 75 degrees fahrenheit. if the temperature decreases 15 degrees, what is the new temperature

lisabon 2012 [21]

Answer: 60 degrees!

Step-by-step explanation:

75 - 15 = 60

7 0

3 years ago

Find the perimeter and area of a rectangle with a width of 2 feet and a length 3 times as long.

rusak2 [61]

Answer:

Length = Width * 3

Width = 2 ft

Length = 2 ft * 3 = 6 ft

Perimeter = (L * 2) + (W * 2)

P = 16 ft

Area = L * W = 12 square feet

5 0

3 years ago

Read 2 more answers

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