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Alja [10]
3 years ago
5

What does X equal in X plus 5 plus 11x equals 12x plus y

Mathematics
2 answers:
aleksandr82 [10.1K]3 years ago
7 0

Answer:

x = all real numbers

Step-by-step explanation:

X plus 5 plus 11x equals 12x plus y

x+ 5 +11x=12x+y

We are solving for x

Combine like terms

12x +5 =12x+y

Subtract 12 x from each side

12x+5-12x =12x+y -12x

5 =y

x = all real numbers


zhuklara [117]3 years ago
4 0

Answer:

x = all real numbers

Step-by-step explanation:

X plus 5 plus 11x equals 12x plus y

x+ 5 +11x=12x+y

We are solving for x

Combine like terms

12x +5 =12x+y

Subtract 12 x from each side

12x+5-12x =12x+y -12x

5 =y

x = all real numbers

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Find the missing lengths in the triangle, if y = 34.10.
joja [24]

Answer:

11.36

Step-by-step explanation:

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4 years ago
What is the slope of у>- 2х + 3
charle [14.2K]

Answer:

-2x

Step-by-step explanation:

Graph

y > −2x + 3

Use the slope-intercept form to find the slope and y-intercept.

The slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.

y = mx + b

Find the values of m and b using the form y = mx + b. m = −2

b = 3

The slope of the line is the value of m, and the y-intercept is the value of b.

Slope: −2

intercept: (0, 3)

Graph a dashed line, then shade the area above the boundary line since y is greater than

−2x + 3.

y > −2x + 3

8 0
3 years ago
Solve the inequality
Irina18 [472]

Answer:

The answer for ur question is x> 4

4 0
2 years ago
Which of the following triangles are right triangles? select all apply
Alex Ar [27]

Answer:

The following Triangles are Right Triangles:

1.  Δ ABC Where AB = 76 in, BC = 357 in, and AC = 365 in.

2. Δ MIT Where MI = 123 cm, IT = 836 cm, and MT = 845 cm.

3. Δ MEL Where ME = 20 ft, EL = 99 ft, and ML = 101 ft.

Step-by-step explanation:

For a Triangle to be a Right Triangle it must Satisfy Pythagoras theorem.

i.e.

(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}

For Δ ABC Where AB = 76 in, BC = 357 in, and AC = 365 in.

AC² = 365² = 133225

AB² + BC² = 76² + 357² = 1333225

∴  AC² = AB² + BC²

Hence Δ ABC a Right Triangle.

For Δ MIT Where MI = 123 cm, IT = 836 cm, and MT = 845 cm.

MT² = 845² = 71405

MI² + IT² = 123² + 836² = 714025

∴ MT² = MI² + IT²

Hence Δ MIT a Right Triangle.

For  Δ MEL Where ME = 20 ft, EL = 99 ft, and ML = 101 ft.

ML² = 101² = 10201

ME² + EL² = 20² + 99² = 10201

∴ ML² = ME² + EL²

Hence Δ MEL a Right Triangle.

7 0
3 years ago
A)A cuboid with a square x cm and height 2xcm². Given total surface area of the cuboid is 129.6cm² and x increased at 0.01cms-¹.
Nutka1998 [239]

Answer: (given assumed typo corrections)


(V ∘ X)'(t) = 0.06(0.01t+3.6)^2 cm^3/sec.


The rate of change of the volume of the cuboid in change of volume per change in seconds, after t seconds. Not a constant, for good reason.



Part B) y'(x+Δx/2)×Δx gives exactly the same as y(x+Δx)-y(x), 0.3808, since y is quadratic in x so y' is linear in x.


Step-by-step explanation:

This problem has typos. Assuming:

Cuboid has square [base with side] X cm and height 2X cm [not cm^2]. Total surface area of cuboid is 129.6 cm^2, and X [is] increas[ing] at rate 0.01 cm/sec.


129.6 cm^2 = 2(base cm^2) + 4(side cm^2)

= 2(X cm)^2 + 4(X cm)(2X cm)

= (2X^2 + 8X^2)cm^2

= 10X^2 cm^2

X^2 cm^2 = 129.6/10 = 12.96 cm^2

X cm = √12.96 cm = 3.6 cm


so X(t) = (0.01cm/sec)(t sec) + 3.6 cm, or, omitting units,

X(t) = 0.01t + 3.6

= the length parameter after t seconds, in cm.


V(X) = 2X^3 cm^3

= the volume when the length parameter is X.


dV(X(t))/dt = (dV(X)/dX)(X(t)) × dX(t)/dt

that is, (V ∘ X)'(t) = V'(X(t)) × X'(t) chain rule


V'(X) = 6X^2 cm^3/cm

= the rate of change of volume per change in length parameter when the length parameter is X, units cm^3/cm. Not a constant (why?).


X'(t) = 0.01 cm/sec

= the rate of change of length parameter per change in time parameter, after t seconds, units cm/sec.

V(X(t)) = (V ∘ X)(t) = 2(0.01t+3.6)^3 cm^3

= the volume after t seconds, in cm^3

V'(X(t)) = 6(0.01t+3.6)^2 cm^2

= the rate of change of volume per change in length parameter, after t seconds, in units cm^3/cm.

(V ∘ X)'(t) = ( 6(0.01t+3.6)^2 cm^3/cm )(0.01 cm/sec) = 0.06(0.01t+3.6)^2 cm^3/sec

= the rate of change of the volume per change in time, in cm^3/sec, after t seconds.


Problem to ponder: why is (V ∘ X)'(t) not a constant? Does the change in volume of a cube per change in side length depend on the side length?


Question part b)


Given y=2x²+3x, use differentiation to find small change in y when x increased from 4 to 4.02.


This is a little ambiguous, but "use differentiation" suggests that we want y'(4.02) yunit per xunit, rather than Δy/Δx = (y(4.02)-y(4))/(0.02).


Neither of those make much sense, so I think we are to estimate Δy given x and Δx, without evaluating y(x) at all.

Then we want y'(x+Δx/2)×Δx


y(x) = 2x^2 + 3x

y'(x) = 4x + 3


y(4) = 44

y(4.02) = 44.3808

Δy = 0.3808

Δy/Δx = (0.3808)/(0.02) = 19.04


y'(4) = 19

y'(4.01) = 19.04

y'(4.02) = 19.08


Estimate Δy = (y(x+Δx)-y(x)/Δx without evaluating y() at all, using only y'(x), given x = 4, Δx = 0.02.


y'(x+Δx/2)×Δx = y'(4.01)×0.02 = 19.04×0.02 = 0.3808.


In this case, where y is quadratic in x, this method gives Δy exactly.

6 0
4 years ago
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