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3 years ago

Ex plane how knowing 50 x 4 = 200 helps you find 500 x 400

Mathematics

2 answers:

Alex777 [14]3 years ago

8 0

Answer:

Step-by-step explanation:

if you know 50*4=200, you must also know that 50*10=500 and that 4*100=400 so 500*400= 50*10*4*100

multiplication is comutative so

50*4*10*100=200*1,000= 200, 000

leonid [27]3 years ago

3 0

Answer:

500×400=

5×4=20

=20+(4 of zero)0000

200000

=200,000

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What's 1.27 repeating as a fraction

svet-max [94.6K]

As a improper fraction, its 127/100

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3 years ago

Describe the transformations from the parent y=|x-2|+1 urgent pls answer.

Zanzabum

Answer:

find he domain by finding where the equation is defined. the range is the set of values that correspond with the domain.

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3 years ago

Which ratio is equivalent to 3/7? <br> A) 6 to 10<br> B) 9:12<br> C) 12/35<br> D) 7 to 3

Leona [35]

we are ratio as

$=\frac{3}{7}$

It will be equivalent to only those terms which would be multiple of this term

so, we will multiply top and bottom term by 5

and we get

$=\frac{3\times 5}{7\times 5}$

$=\frac{15}{7\times 5}$

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so, it is very similar to 12/35

so, it will be equivalent to 12/35

so, option-C.......Answer

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3 years ago

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in secon

Pavel [41]

Answer and explanation:

Given : The position of an object moving along an x axis is given by $x=3.24t-4.20t^2+1.07t^3$ where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(1)=3.24(1)-4.20(1)^2+1.07(1)^3$

$x(1)=3.24-4.20+1.07$

$x(1)=0.11$

b) At t= 2 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(2)=3.24(2)-4.20(2)^2+1.07(2)^3$

$x(2)=6.48-16.8+8.56$

$x(2)=-1.76$

c) At t= 3 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(3)=3.24(3)-4.20(3)^2+1.07(3)^3$

$x(3)=9.72-37.8+28.89$

$x(3)=0.81$

d) At t= 4 s

$x(t)=3.24t-4.20t^2+1.07t^3$

$x(4)=3.24(4)-4.20(4)^2+1.07(4)^3$

$x(4)=12.96-67.2+68.48$

$x(4)=14.24$

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

$\triangle x=x(4)-x(0)$

$\triangle x=14.24-0$

$\triangle x=14.24$

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

$\triangle x=x(4)-x(2)$

$\triangle x=14.24-(-1.76)$

$\triangle x=14.24+1.76$

$\triangle x=16$

4 0

3 years ago

The rectangle shown has a perimeter of 146 cm and the given area. Its length is 7 more than five times its width. Write and solv

dangina [55]

Answer:

System of equations:

L = 5W + 7

2W + 2L = P

L = 62 cm

W = 11 cm

Step-by-step explanation:

Given the measurements and key words/phrases in the problem, we can set up two different equations that can be used to find both variables, length and width, of the rectangle.

The formula for perimeter of a rectangle is: 2W + 2L = P, where W = width and L = length. We also know that the L is '7 more than five times its width'. This can be written as: L = 5W + 7. Using this expression for the value of 'L', we can use the formula for perimeter and solve for width:

2W + 2(5W + 7) = 146

Distribute: 2W + 10W + 14 = 146

Combine like terms: 12W + 14 = 146

Subtract 14 from both sides: 12W + 14 - 14 = 146 - 14 or 12W = 132

Divide 12 by both sides: 12W/12 = 132/12 or W = 11

Put '11' in for W in the equation for 'L': L = 5(11) + 7 or L = 55 + 7 = 62.

5 0

3 years ago