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3 years ago

What adds to be 0 and multiplies to be 3

1 answer:

3 0

There are no 'real' numbers that can do that.

This will be a good question to practice on when you
start learning "imaginary" numbers.

The pair of numbers that answer this question are

+ i√3
and
- i√3 .

The symbol ' i ' is the symbol used for √(-1) .

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696 + 66 = ?<br>Use the number bond to help you make a hundred. 1)<br>696 + 66

Ghella [55]

762 because if you add it will give you the answer

7 0

3 years ago

Put them in order from least to greasiest 3/5 1/4 1/2 2/5

alexira [117]

1/4 < 2/5 < 3/5 < 1/2

3 0

3 years ago

Read 2 more answers

a function f(x) is described by the ordered pair (x,y). if given the function is translated 2 units to the right and then reflec

OleMash [197]

The ordered pairs would be (x-2, -y).

Translating a graph to the right subtracts the number of units from the x-coordinate.

Reflecting across the x-axis negates the y-coordinate.

3 0

3 years ago

Solve for d.<br> d+ 4 = 16.21<br> d=

lutik1710 [3]

Answer:

d = 12.21

Step-by-step explanation:

1) Subtract 4 from both sides.

d = 16.21 - 4

2) Simplify 16.21 - 4 to 12.21.

d = 12.21

Check the answer!

⇒ d + 4 = 16.21

1. Let d = 12.21.

⇒ 12.21 + 4 = 16.21

2. Simplify 12.21 + 4 to 16.21

16.21 = 16.21

Done!

Thanks,

Eddie

8 0

1 year ago

The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Appr

Kobotan [32]

Answer:

% change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

$T = 2.5 x + 0.5 x^{2}$

We will approximate this distance using the relation

$f (x + dx) = f (x)+ f' (x)dx$

dx = 26 - 25 = 1

T' = 2.5 + x

Therefore

$f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x$

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× $25^{2}$ + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × $25^{2}$

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

% change in stopping distance = $\frac{27.5}{375}$ × 100

% change in stopping distance = 7.34 %

6 0

3 years ago