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3 years ago

What is 4.5w=5.1w-30 step by step please

Mathematics

2 answers:

vampirchik [111]3 years ago

8 0

4.5w = 5.1w -30

4.5w - 5.1w = 5.1w - 5.1w -30 (subtracted 5.1w on both sides)

4.5w - 5.1w = -30 (next combine like terms)

-0.6w = -30 (next do the inverse operation of w)

w = -30 ÷ -0.6

w = 50 (it became positive cause minus and minus = positive)

so your final answer is: w = 50

Hope that helps:D

Whitepunk [10]3 years ago

3 0

4.5w= 5.1w-30
⇒ 5.1w- 4.5w= 30
⇒ 0.6w= 30
⇒ w= 30/0.6
⇒ w= 50

The final answer is w=50~

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Select all ratios equivalent to 1:8.

DochEvi [55]

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Step-by-step explanation:

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3 years ago

Tell which set or sets of numbers below belongs to: natural numbers, whole numbers, integers, rational numbers, irrational numbe

Misha Larkins [42]

1/3 belongs to the rational set and to the real set.

<h3>To which sets does the number below belong?</h3>

Here we have the number 1/3.

First, remember that we define rational numbers as these numbers that can be written as a quotient between two integers.

Here 1 is an integer and 3 is an integer, then 1/3 is a rational number.

Also, the combination between the rational set and the irrational set is the set of the real numbers, then 1/3 is also a real number.

Then, concluding:

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2 years ago

Miguel buys 100 feet of fence to enclose a rectangular area of his backyard so his dog can run freely. What is the maximum area,

sdas [7]

Suppose the length of the triangle is x, if the perimeter of the rectangle is 100 ft, the width of the rectangle will be (50-x) ft.
Area of rectangle will be:
A=length*width
A=x(50-x)
A=50x-x^2
at maximum area, dA/dx=0
thus
dA/dx=50-2x=0
solving for x we get
2x=50
x=25
thus for maximum area length=25 ft
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5 0

3 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is:

$(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})$

$(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]$

$(x,y) = (10.452, 34.531)\,[km]$

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

$\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}$

$\theta \approx 16.840^{\circ}$

And the distance from the camp is calculated by the Pythagorean Theorem:

$r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}$

$r = 36.078\,km$

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

5 0

3 years ago

Solve the equation. Check for extraneous solutions<br><br> please help!!!

lozanna [386]

!z!=a→z=a or z=-a; z=4x+3, a=9+2x

!4x+3!=9+2x
1) 4x+3=9+2x
Solving for x:
4x+3-3-2x=9+2x-3-2x
2x=6
2x/2=6/2
x=3

Checking for extraneous solution:
!4x+3!=9+2x
x=3→!4(3)+3!=9+2(3)
!12+3!=9+6
!15!=15
15=15 Ok, then x=3 is not a extraneous solution

2) 4x+3=-(9+2x)
Solving for x:
4x+3=-9-2x
4x+3-3+2x=-9-2x-3+2x
6x=-12
6x/6=-12/6
x=-2

Checking for extraneous solution:
!4x+3!=9+2x
x=-2→!4(-2)+3!=9+2(-2)
!-8+3!=9-4
!-5!=5
5=5 Ok, then x=-2 is not a extraneous solution

Answer:
x = -2 or 3

4 0

3 years ago