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3 years ago

The sum of two and three times the difference of a number and five is equal to four times the difference a number and one.

Mathematics

1 answer:

Andru [333]3 years ago

4 0

Answer:

$x=-9$

Step-by-step explanation:

We translate the phrase into an equation:

$2+3(x-5)=4(x-1)\\$

Then we simplify:

$2+3(x-5)=4(x-1)\\2+3x-15=4x-4\\-13+3x=4x-4\\-9=x$

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Solve for x: 5x+3y=15

Gnoma [55]

Answer:

y = (-5/3)x - 5

Step-by-step explanation:

5x + 3y = =15

slope intercept for is: y = mx + b

3y = -5x=15

divide both sides by 3:

y = (-5/3)x - 5

8 0

3 years ago

How do I turn 279x21 into a distributive property

Vilka [71]

(200plus79)x(20plus 1)
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6 0

3 years ago

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago

URGENT PLEASE HELP ME, ITS DUE IN A HOUR

Olegator [25]

Answer:

Area of gray region = $8x^2+8x-7$

Step-by-step explanation:

The area covered by the furniture can be subtracted from the total area of the room to calculate the area of the gray region.

Total area of the room: $13x^2+7x-2$

Area taken up by furniture: $5x^2-x+5$

Area of gray region = total area of the room - area taken up by furniture

Area of gray region = $13x^2+7x-2-(5x^2-x+5)$

Remember to distribute the negative sign:

Area of gray region = $13x^2+7x-2-5x^2+x-5$

Combine like terms:

Area of gray region = $8x^2+8x-7$

7 0

3 years ago

Read 2 more answers

Find all possible values of α+

const2013 [10]

Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

recall angle sum indentity of tan:

$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore when k is 1 we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

when is 0:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

likewise by Opposite Angle Identity we obtain:

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

8 0

3 years ago

Read 2 more answers