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3 years ago

7

X=3(180-x) What is thiz

2 answers:

Crank3 years ago

7 0

Ok it is so easy it is 489467000

AlexFokin [52]3 years ago

5 0

X=135 you have to distibute the three to the 180 and the -x and you get x=540-3x then you add the -3x to the other x which is called combining like terms now you have 4x=540 then you divide the four on both sides to get the x alone now you have x=135

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Use the elimination method to solve the system of equations. Choose the correct ordered pair<br>

UkoKoshka [18]

Answer:

Answer is B

Step-by-step explanation:

$12x - y = 25 \\ 9x + y = 17 \\ x = \frac{17 - y}{9} \\ 12( \frac{17 - y}{9} ) - y = 25 \\$

$\frac{204}{9} - \frac{12}{9} y - y = 25 \\ \frac{204}{9} - \frac{21y}{9} = 25 \\ - \frac{21y}{9} = 25 - \frac{204}{9} \\ - \frac{21y}{9} = 2.33 \\ - 21y = 20.997 \\ y = - 0.99$

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Bezzdna [24]

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Step-by-step explanation:

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Read 2 more answers

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

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