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3 years ago

Not sure how to do the write in long form.

Mathematics

2 answers:

ohaa [14]3 years ago

8 0

"long" form is actually Standard Form. This is done by converting the expression to a number by eliminating (x 10ⁿ)

1. 1.00 x 10⁵ = 100,000 (a positive exponent makes the number bigger by moving the decimal 5 places to the right)

2. 1 x 10⁻¹² = 0.000000000001 ( a negative exponent makes the number smaller by moving the decimal 12 places to the left)

3. 4 x 10² = 400 (a positive exponent makes the number bigger by moving the decimal 2 places to the right)

tekilochka [14]3 years ago

5 0

1. 100000
2. 0.000000000001
3. 400

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Solve the inequality: (x+3)(x−4)<−12

melamori03 [73]

Answer:

0<x<1

Step-by-step explanation:

(x+3)(x−4)<−12

FOIL

x^2 -4x+3x -12 <-12

x^2 -x -12 <-12

Add 12 to each side

x^2 -x -12+12 <-12+12

x^2 -x <0

Factor out an x

x(x-1) <0

Using the zero product property

x(x-1) =0

x=0 x-1=0

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x<0

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0<x<1

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x>1

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3 years ago

PLEASE HELP !!!! Solve the system using either substitution or elimination: 3x+6y=6 x-5y=23

natta225 [31]

Answer:

Solve for the first variable in one of the equations, then substitute the result into the other equation. Point Form: ( 8 , − 3 ) Equation Form: x = 8 , y = − 3

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3 0

3 years ago

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The graph of which function has a minimum located at (4, –3)?

valentinak56 [21]

Answer:

None of the options is the answer to the question

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$f(x)=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

case A) we have

$f(x)=x^{2}+4x-11$

In this case the x-coordinate of the vertex will be negative

therefore

case A is not the solution

case B) we have

$f(x)=-2x^{2}+16x-35$

This case is a vertical parabola open downward (the vertex is a maximum)

The vertex is the point $(4,-3)$ <u>but is not a minimum</u>

see the attached figure

therefore

case B is not the solution

case C) we have

$f(x)=x^{2}-4x+5$

Convert into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-5=x^{2}-4x$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-5+4=(x^{2}-4x+4)$

$f(x)-1=(x^{2}-4x+4)$

Rewrite as perfect squares

$f(x)-1=(x-2)^{2}$

$f(x)=(x-2)^{2}+1$ --------> vertex form

The vertex is the point $(2,1)$

therefore

case C is not the solution

case D) we have

$f(x)=2x^{2}-16x+35$

Convert into vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)-35=2x^{2}-16x$

Factor the leading coefficient

$f(x)-35=2(x^{2}-8x)$

Complete the square. Remember to balance the equation by adding the same constants to each side

$f(x)-35+32=2(x^{2}-8x+16)$

$f(x)-3=2(x^{2}-8x+16)$

Rewrite as perfect squares

$f(x)-3=2(x-4)^{2}$

$f(x)=2(x-4)^{2}+3$ --------> vertex form

The vertex is the point $(4,3)$

therefore

case D is not the solution

The answer to the question will be the function

$f(x)=2x^{2}-16x+29$

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4 years ago

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√12 + √10 − √2) is

r-ruslan [8.4K]

Hello!

$\bf \sqrt{12} + \sqrt{10} - \sqrt{2} =$

$\bf \sqrt{ {2}^{2} \times 3} + \sqrt{10} - \sqrt{2} =$

$\bf \sqrt{ {2}^{2} } \sqrt{ 3} + \sqrt{10} - \sqrt{2} =$

$\bf \boxed{ 2 \sqrt{3} + \sqrt{10} - \sqrt{2}}$

Answer: (c) An irrational number

Good luck! :)

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3 years ago

2 Points

sergeinik [125]

Answer:

to get coordinate of mid point use (20+0)/2 for x coordinate,

A is correct answer!

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