What is the range of the function O

A 56 kg miniature horse weighs 268 fewer kilograms than shetland pony

Studentka2010 [4]

Let
x--------> Shetland pony's weigh
y--------> Miniature horse's weigh

we know that
 $y=56 kg$
$x=268+y \\ x=268+56 \\ x=324 kg$

the answer is
Shetland pony's weigh is $324 kg$

6 0

3 years ago

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Graph the function y = 4x4 – 8x2 + 4. Which lists all of the turning points of the graph?

Serhud [2]

Answer:

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

Step-by-step explanation:

STEP 1:

The equation at the end of step 1

$((4 (x^4)) - 2^3x^2) + 4$

STEP 2:

The equation at the end of step 2:

$(2^2x^4 - 2^3x^2) + 4$

STEP 3:

STEP 4: Pulling out like terms

4.1 Pull out like factors:

$4x^4 - 8x^2 + 4 = 4(x^4 - 2x^2 + 1)$

Trying to factor by splitting the middle term

4.2 Factoring $x^4 - 2x^2 + 1$

The first term is, $x^4$ its coefficient is 1.

The middle term is, $-2x^2$ its coefficient is -2.

The last term, "the constant", is +1.

Step-1: Multiply the coefficient of the first term by the constant 1 • 1 = 1

Step-2: Find two factors of 1 whose sum equals the coefficient of the middle term, which is -2.

-1 + -1 = -2 That's it

Step-3: Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -1 and -1

x4 - 1x2 - 1x2 - 1

Step-4 : Add up the first 2 terms, pulling out like factors :

x2 • (x2-1)

Add up the last 2 terms, pulling out common factors :

1 • (x2-1)

Step-5 : Add up the four terms of step 4 :

(x2-1) • (x2-1)

Which is the desired factorization

Trying to factor as a Difference of Squares:

4.3 Factoring: x2-1

Theory : A difference of two perfect squares, A2 - B2 can be factored into (A+B) • (A-B)

Proof : (A+B) • (A-B) =

A2 - AB + BA - B2 =

A2 - AB + AB - B2 =

A2 - B2

Note : AB = BA is the commutative property of multiplication.

Note : - AB + AB equals zero and is therefore eliminated from the expression.

Check : 1 is the square of 1

Check : x2 is the square of x1

Factorization is : (x + 1) • (x - 1)

Trying to factor as a Difference of Squares:

4.4 Factoring: x2 - 1

Check : 1 is the square of 1

Check : x2 is the square of x1

Factorization is : (x + 1) • (x - 1)

Multiplying Exponential Expressions:

4.5 Multiply (x + 1) by (x + 1)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (x+1) and the exponents are :

1 , as (x+1) is the same number as (x+1)1

and 1 , as (x+1) is the same number as (x+1)1

The product is therefore, (x+1)(1+1) = (x+1)2

Multiplying Exponential Expressions:

4.6 Multiply (x-1) by (x-1)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is (x-1) and the exponents are :

1 , as (x-1) is the same number as (x-1)1

and 1 , as (x-1) is the same number as (x-1)1

The product is therefore, (x-1)(1+1) = (x-1)2

Final result :

4 • (x + 1)2 • (x - 1)2

4 0

3 years ago

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How do you do this problem? Simple and concise explanation, please!

natita [175]

Step-by-step Answer:

This is a problem of partial fractions.

Step 1:

factor all denominators on the left-hand side (LHS).

LHS = a/[(x+1)(x-1)] + x/[(x-3)(x+1)]

Note the common factor (x+1) in both denominators, which makes the combined/common denominator [(x+1)(x-1)(x-3)]

Step 2:

multiply each term, top and bottom, by the factor "missing" from the common denominator.

The first term is missing (x-3), the second term is missing (x-1)

a(x-3)/[(x+1)(x-1)(x-3)] + x(x-1)/[(x+1)(x-1)(x-3)]

which can be simplified to:

[a(x-3)+x(x-1)]/[(x+1)(x-1)(x-3)]

Expand numerator:

[x^2 + ax-x -3a]/[(x+1)(x-1)(x-3)]

Step 3:

For the two expressions on each side of the equal sign (LHS and RHS) to be equivalent (for ALL values of x), the numerators and denominators must be identical when expanded, so

For denominator, we have factors [(x+1)(x-1)(x-3)], or b,c,d = -3, -1, 1 (in ascending order).

For the numerator, we need to have LHS = RHS

x^2 + (a-1)x -3a = x^2 + 2x -9

putting a=3 gives the

LHS = x^2 + (3-1)x -3(3) = x^2 + 2x - 9 which makes equality with the RHS.

So we have solved for all values required.

6 0

3 years ago

Rewrite the expression in the form 5^n (5^3)^2=?

Ira Lisetskai [31]

This problem is all about bases and exponents. Because we have a quotient and the bases are both 5's, that means that we can use the rule of exponents for quotients to rewrite and simplify:

That's the simplification as long as you are "allowed" to leave the exponent as a negative number.

7 0

3 years ago

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luvenia can row 4mph in still water. She takes as long to row 7 mi upstream as 21 mi downstream. how

xeze [42]

Answer:

The speed of the river is 2mph.

Step-by-step explanation:

I guess that we want to find the speed of the river.

First, remember the relation: speed*time = distance

If the speed of the river is Sr, when Luvenia moves downstream (in the same direction that the flow of the water) the total speed will be equal to the speed of Luvenia in still water plus the speed of the water:

Sd = 4mph + Sr

and at this speed, in a time T, she can move 21 miles, so we have:

Sd*T = (4mph + Sr)*T = 21 mi

When moving upstream, the speed will be:

Su = (4mph - Sr)

and in the same time T as before, she moves 7 miles, so we have the equation:

Su*T = (4mph - Sr)*T = 7 mi

Then we have two equations:

(4mph + Sr)*T = 21 mi

(4mph - Sr)*T = 7 mi

Now we can take the quotient of those two equations and get:

((4mph + Sr)*T)/((4mph - Sr)*T) = 21/7

The time T vanishes, and we can solve it for Sr.

(4mph + Sr)/(4mph - Sr) = 3

4mph + Sr = 3*(4mph - Sr) = 12mph - 3*Sr

4*Sr = 12mph - 4mph = 8mph

Sr = 8mph/4 = 2mph.

7 0

4 years ago

What is the range of the function O - O -8 O Osy O2 sy