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ElenaW [278]
3 years ago
8

Why does multiplying numbers by 10 move the decimal point to the right , but multiplying by 0.10 moves the decimal point to the

left?
Helppp!
Mathematics
2 answers:
lesya692 [45]3 years ago
4 0

Answer:

Step-by-step explanation:

because if you multiply by ten, the number gets bigger and makes more zeros at the end and pushes the decimal point farther

because when you multiply .10 it is the same as multiplying by a tenth

Dafna11 [192]3 years ago
3 0

Answer:

Because multiplying by 0.10 is the same as dividing by 10.  It is a decimal smaller than 1 its the same thing as multiplying something by a fraction because when you multiply a number by a fraction the number gets smaller.

So multiplying by 10 and multiplying by 0.10 are inverses of each other.

Step-by-step explanation:

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165°<br> 145<br> x =<br> Please Help!!!!
Art [367]

Answer:

15°

Step-by-step explanation:

The 165° and <em>x</em> forms a straight line or a supplementary angle. Supplementary angles add up to 180°. Therefore:

165+x=180

Solve for <em>x:</em>

<em />165+x=180\\x=15\textdegree<em />

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3 years ago
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n an orchard, there are 13 apple trees, 21 orange trees, and a cherry tree. What is the ratio of orange trees to cherry trees?
Yuri [45]

Answer:

21 : 1

Step-by-step explanation:

4 0
3 years ago
Help me with my math pls :)
MrMuchimi

Answer:

4(x-8)=-32+4: x=1

2(x-2)=-3-2x: x1/4

3+5x=5(x+2)-7: x=0

8x+38=-3(-6-4x) x=5

Step-by-step explanation:

4(x-8)=-32+4

add subtract the numbers

-32+4=-28

4(x-8)=-28

Divide both sides by 4

\frac{4(x-8)}{4}=\frac{-28}{4}

Simplify

x-8=-7

Add 8 on both sides

x+8-8=-7+8

Simplify

x=1

2(x-2)=-3-2x

  • Expand 2(x-2)
  • apply distributive law: a(b-c)=ab-ac

a=2 b=x c=2

=2x-2·2

  • multiply the numbers

2·2=4

=2x-4

2x-4=-3x-2

  • Add 4 to both sides

2x-4+4=-3-2x+4

  • simplify

2x=-2x+1

  • Add 2x on both sides

2x+2x=2x+1+2x

  • Simplify

4x=1

  • Divide by 4 on both sides

4x÷4=1÷4

  • Simplify

x=1/4

3+5x=5(x+2)-7

<em> We move all terms to the left: </em>

<em> 3+5x-(5(x+2)-7)=0 </em>

<em> We calculate terms in parentheses: -(5(x+2)-7), so: </em>

<em> 5(x+2)-7 </em>

<em> We multiply parentheses </em>

<em>5x+10-7 </em>

<em> We add all the numbers together, and all the variables </em>

<em> 5x+3 </em>

<em> Back to the equation: </em>

<em> -(5x+3) </em>

<em>  We get rid of parentheses </em>

<em> 5x-5x-3+3=0 </em>

<em> We add all the numbers together, and all the variables </em>

<em> =0 </em>

<em> x=0/1 </em>

<em> x=0</em>

8x+38=-3(-6-4x)

Simplifying

8x + 38 = -3(-6 + -4x)

Reorder the terms:

38 + 8x = -3(-6 + -4x)

38 + 8x = (-6 * -3 + -4x * -3)

38 + 8x = (18 + 12x)

Solving

38 + 8x = 18 + 12x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-12x' to each side of the equation.

38 + 8x + -12x = 18 + 12x + -12x

Combine like terms: 8x + -12x = -4x

38 + -4x = 18 + 12x + -12x

Combine like terms: 12x + -12x = 0

38 + -4x = 18 + 0

38 + -4x = 18  

Add '-38' to each side of the equation.

38 + -38 + -4x = 18 + -38

Combine like terms: 38 + -38 = 0

0 + -4x = 18 + -38

-4x = 18 + -38

Combine like terms: 18 + -38 = -20

-4x = -20

Divide each side by '-4'.

x = 5

Simplifying

x = 5

Hope this helps

4 0
3 years ago
What are the coordinates of the vertices of the polygon in the graph that are Quadrant III?
Ratling [72]

Answer:

D. (-5,-1), (-2,-3)

Step-by-step explanation:

Just by looking at the answer choices, you can eliminate A, B, and C because they have some positive numbers. You should know that in Quadrant 3, there are only negative numbers.

<em>Eliminating answers is the key to getting the correct answer!</em>

8 0
3 years ago
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A particular airport has beacons on the ground 5 km out from the start of each runway. An airplane is directly over this beacon
AleksAgata [21]

Answer:

The direct distance from the plane to the start of the runway is approximately 5.0623 km

Step-by-step explanation:

The locations of the beacons on the runway, x = 5 km from the start of each runway

The angle of elevation of the airplane from the runway, θ = 9°

Let 'R' represent the direct distance from the plane to the start of the runway

The horizontal distance from the start of the runway to the beacon, 'x', the distance from the plane to the start of the runway, 'R', and the vertical height of the plane, 'h' form a right triangle with rides x, R, and h

By trigonometric ratios, we have;

cos\angle \theta = \dfrac{The \ length \ of \ the \ adjacent\ leg  \ to \ \angle \theta}{The \ length \ of \ the \ hypotenuse \ side} = \dfrac{x}{R}

Therefore, we have;

cos (9 ^{\circ}) = \dfrac{x}{R} = \dfrac{5}{R}

From which we have;

R=  \dfrac{5}{cos (9 ^{\circ}) } \approx 5.0623

The direct distance from the plane to the start of the runway, R ≈ 5.0623 km

7 0
3 years ago
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