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laiz [17]
4 years ago
15

Answer please! I have been "getting" the correct answer, as far as I know. The answer which I got was 10x^2 - 4x. But I don't se

e how it's incorrect. PLEASE HELP!

Mathematics
1 answer:
GarryVolchara [31]4 years ago
4 0

Answer:

The answer is 10x² - 4x.

Step-by-step explanation:

You have to apply Distributive Law,

a(m + n) = am + an

So for this question :

2x(5x - 2)

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

= 10 {x}^{2}  - 4x

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4 3/10 divided by 3/5
inysia [295]

Answer:

7.16

Step-by-step explanation:

5 0
4 years ago
Name the set(s) of numbers to which -10 belongs.
NeX [460]
-10 belongs to integers, rational numbers, and real numbers.
6 0
3 years ago
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Which solution to the equation StartFraction 1 Over x minus 1 EndFraction = StartFraction x minus 2 Over 2 x squared minus 2 End
Darina [25.2K]

Answer:

x = - 4

Step-by-step explanation:

Given the expression

\frac{1}{x-1} = \frac{x-2}{2x^2-2}

This can also be expressed as;

\frac{1}{x-1} = \frac{x-2}{2(x-1)(x+1)}\\1 = \frac{x-2}{2(x+1)}

Cross multiply

2(x+1) = x - 2\\2x+2 = x-2

Add 2 to both sides

2x+2+2 = x-2+2\\2x+4 = x\\2x-x = -4\\x = -4

Hence the required extraneous solution is x = -4

4 0
3 years ago
Read 2 more answers
If a ball is thrown into the air with a velocity of 34 ft/s, its height (in feet) after t seconds is given by y = 34t − 16t2. Fi
Finger [1]

<u>ANSWER: </u>

If a ball is thrown into the air with a velocity of 34 feet per second, then velocity of the ball after 1 second is 2 feet per second

<u>SOLUTION: </u>

Given, a ball is thrown into the air with a velocity of 34 feet per second

Initial velocity (u) = 34 feet per second

And also given a relation between displacement and time = \mathrm{y}=34 \mathrm{t}-16 \mathrm{t}^{2} --- eqn 1

We need to find the velocity when t = 1 ; v = ?

We know that, v = u + at and \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}

where v is instantaneous velocity and u is initial velocity

a is acceleration

t is time interval  

s is displacement

using the displacement and time relation eqn (1) we get

Now, when t = 1, displacement s = 34(1) – 16(1)

\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}=34-16

34 \times 1+\frac{1}{2} \times a \times 1^{2}=18

34+\frac{a}{2}=18

\begin{array}{l}{\frac{a}{2}=18-34} \\\\ {\frac{a}{2}=-16} \\ {a=-16 \times 2} \\ {a=-32}\end{array}

here, -ve sign indicates that object is in deceleration . so acceleration is -32 ft/s

now put a value in v = u + at

v = 34 + (-32)(1)

v = 34 – 32

v = 2 ft/s

Hence, velocity of the ball after 1 second is 2 ft/s

6 0
3 years ago
Which form of a quadratic function is easiest to use when finding the maximum or minimum value of the function?
alexandr402 [8]

Answer:

ax² + bx + c

Step-by-step explanation:

The form of a quadratic equation that is easy to use when finding the maximum or minimum value of the function is ax² + bx + c.

Suppose a quadratic function:

f(x) = 2x² - 8x + 9

Use ( -b/2a ,  f(-b/2a) ).

-b/2a

a = 2

b = -8

-(-8)/2(2)

8/4

= 2

f(2) = 2(2)² - 8(2) + 9

f(2) = 2(4) - 8(2) + 9

f(2) = 8 - 16 + 9

f(2) = 1

The minimum value of this quadratic function is (2, 1).

It represents a minimum value because a > 0.

4 0
3 years ago
Read 2 more answers
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