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Snezhnost [94]
3 years ago
5

Please help me with this thanks

Mathematics
1 answer:
geniusboy [140]3 years ago
5 0
<span> When area is equal 1120. We can write an equation
(x+4)(-x+64)=1120
-x²-4x+64x+256=1120
-x²+60x+256-1120=0
-x²+60x-864=0
D=b² - 4ac= 3600-4*864=144, √D=12
x= (-b+/-√D)/2a
x=(-60+/-12)/(-2)
x=24, x=36
For x=24
(x+4)=24+4=28
(-x+64)=(-24+64)=40
For x=36
(x+4)=36+4=40
(x-64) =(-36+64)=28
So sides should be 28 and 40 in.
We did not get any extraneous solutions. They could be if we get negative length side, for example. They can come because a quadratic equation can
give positive and negative numbers because a^2  and (-a)^2 give the same positive number.
We chose to solve this equation using formula for quadratic equations, because this equation has too big numbers to solve it using other methods.



</span>
You might be interested in
{[8-3)x2]+[(5x6)-5]}÷5=
KIM [24]


the answer is 7

8-3=5

5x2=10

10+(5x6)-5=35

35/5=7

3 0
3 years ago
Blood is a buffer solution. When carbon dioxide is absorbed into the bloodstream, it produces carbonic acid and lowers the pH. T
Irina18 [472]

Answer:

x ≅ 20.10 torr

Step-by-step explanation:

Given that the equation which models blood pH in the question is;

pH(x)=6.1+log(800/x)

where;

pH = 7.7

x = partial pressure of carbon dioxide in arterial blood, measured in torr.

we are asked to find (x)

In order to do that, we use the given equation:

pH(x)=6.1+log(800/x)

since pH = 7.7

7.7 = 6.1 + log (800/x)

7.7 - 6.1 =  log (800/x)

1.6 =  log (800/x)

10^{1.6}= \frac{800}{x}

x= \frac{800}{10^{1.6}}

x = 20.09509145

x ≅ 20.10 torr

4 0
3 years ago
NEED HELP THIS IS SOOOOOO HARD NEED PROOF <br> WILL GIVE BRAINLIEST AND 5-STAR
Shkiper50 [21]

Answer:

J    1

   --------

    x^2 -x

Step-by-step explanation:

x+1

----------

x^3-x

Factor out an x in the denominator

x+1

----------

x(x^2-1)

We can factor the terms in the parentheses because it is a difference of squares

x+1

----------

x(x-1) (x+1)

Canceling the x+1 terms

1

----------

x(x-1)

Distribute in the denominator

1

--------

x^2 -x

7 0
3 years ago
Drag the expressions into the boxes to correctly complete the table.
lora16 [44]

Answer:

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

Step-by-step explanation:

The algebraic expressions are said to be the polynomials in one variable which consist of terms in the form ax^n.

Here:

n = non-negative integer

a = is a real number (also the the coefficient of the term).

Lets check whether the Algebraic Expression are polynomials or not.

Given the expression

x^4+\frac{5}{x^3}-\sqrt{x}+8

If an algebraic expression contains a radical in it then it isn’t a polynomial. In the given algebraic expression contains \sqrt{x}, so it is not a polynomial.

Also it contains the term \frac{5}{x^3} which can be written as 5x^{-3}, meaning this algebraic expression really has a negative exponent in it which is not allowed. Therefore, the expression x^4+\frac{5}{x^3}-\sqrt{x}+8 is not a polynomial.

Given the expression

-x^5+7x-\frac{1}{2}x^2+9

This algebraic expression is a polynomial. The degree of a polynomial in one variable is considered to be the largest power in the polynomial. Therefore, the algebraic expression is a polynomial is a polynomial with degree 5.

Given the expression

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi

in a polynomial with a degree 4. Notice, the coefficient of the term can be in radical. No issue!

Given the expression

\left|x\right|^2+4\sqrt{x}-2

is not a polynomial because algebraic expression contains a radical in it.

Given the expression

x^3-4x-3

a polynomial with a degree 3. As it does not violate any condition as mentioned above.

Given the expression

\frac{4}{x^2-4x+3}

\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}

Therefore, is not a polynomial because algebraic expression really has a negative exponent in it which is not allowed.

SUMMARY:

x^4+\frac{5}{x^3}-\sqrt{x}+8                               →    Not a Polynomial

-x^5+7x-\frac{1}{2}x^2+9                           →    A Polynomial

x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi              →    A Polynomial

\left|x\right|^2+4\sqrt{x}-2                                   →    Not a Polynomial

x^3-4x-3                                        →    A Polynomial

\frac{4}{x^2-4x+3}                                              →    Not a Polynomial

3 0
3 years ago
NEED HELP FAST!
mihalych1998 [28]
The answer is currently A so yep thats the answer ok
8 0
3 years ago
Read 2 more answers
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