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Alecsey [184]
2 years ago
5

Solve by factoring x^2-8x-20-0

Mathematics
2 answers:
agasfer [191]2 years ago
6 0

Answer:

(x-10)(x+2)

Step-by-step explanation:

Oksana_A [137]2 years ago
5 0
X^2-8x-20 and if you rewrite that it’s (x-10) (x+2)
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The minimum and maximum distances from a focus to a point on an ellipse occur when that point on the ellipse is an endpoint of t
Sauron [17]
The answer is True.

Explanation:
Let a =  major axis
Let b = minor axis
Let c =  focal length.

Consider the right focus, located a distance c from the center of the ellipse (at the origin).
From the right focus to the right point on the major axis is equal to a-c. This is the minimum distance.
From the right focus to the left point on the major axis is equal to a+c. This is the maximum distance. 
7 0
3 years ago
Read 2 more answers
If Florist B increases the cost per rose to $5.20,for what number of roses is it less expensive to order from Florist A? From Fl
Alenkinab [10]

Answer:

For<em> </em>38 roses at <em>$5.15</em> per rose, Florist A is Less Expensive then Florist B

For<em> </em>34 roses at <em>$5.20</em> per rose, Florist A is Less Expensive then Florist B

Step-by-step explanation:

The Full Question Reads:

<u><em>Derek wants to order some roses online.  Florist A charges $4.75 per blue rose plus $40 delivery charge.  Florist B charges $5.15 per red rose plus $25 delivery charge.  If Florist B increases the cost per rose to $5.20, for what number of roses is it less expensive to order from Florist A?  From Florist B?</em></u>

To begin we need to understand our given information and what we are actually looking for.

<u>Given Information</u>

Florist A:

charges $4.75 per blue rose

charges $40 per delivery

Florist B:

charges $5.15 per red rose

charges $25 per delivery

Note: in this problem we ignore the colour of the rose (red or blue) as it does not affect our solution or contributes to it. Therefore we shall call our first variable representing the number of roses as x.

The next step is to construct a system of equations representing our problem and our given information. Here we are looking at linear relationships so a linear function of the form:

y=ax+b   Eqn(1).

<em>will suffice where</em>

y: represents the total cost from the Florists (i.e. number of roses and delivery charges). Dependent Variable

x: represents the number of roses purchased from the Florists. Independent Variable

a: is our relationship factor between y and x

b: is our constant which in this problem is the <em>delivery charge</em> value for each florist.  

Since we have two Florists, we will construct two equations, one for each florist, by employing Eqn(1).  and our given information as follow:

Florist A:  y=4.75x+40   Eqn(2).

Florist B:  y=5.15x+25   Eqn(3).

By employing both equations above and writing them as an inequality <em>since we are looking for which value of </em>x<em> (i.e. number of roses) will the less expensive Florist be and then solving for </em>x<em> we have: </em>

<em>4.75x+40</em>

<em>4.75x-5.15x       Gathering all similar terms together</em>

<em>-0.4x                       Simplifying</em>

<em>x>\frac{-15}{-0.4}\\                             Solving for x </em>

<em>x>37.5</em>

<em>(note how < changes to > since any multiplication/devision process of negative sign in an Inequality will change the order of < to > and vice versa). </em>

Since a rose has to be sold as a whole and not half we will say that <em>x>38</em> So we can then plug in the value of x in Eqn(2) and Eqn(3) and find the cost of buying more than 38 roses from each:

Eqn(2):  Florist A:  y(x>38)=4.75*38+40=220.5  

Eqn(3):  Florist B:  y(x>38)=5.15*38+25=220.7

Which tells us that Florist A is less expensive than Florist B by $0.20 for a purchase of more than 38 roses.

Next the question tells us that Florist B increases the cost from $5.15 to $5.20 per rose, which in Eqn(3) denotes our a value and thus Eqn(3). now becomes:

Florist B:  y=5.20x+25   Eqn(3).

Applying the same method like before and solving for the value of x<em> we have: </em>

<em>4.75x+40</em>

<em>4.75x-5.20x       Gathering all similar terms together</em>

<em>-0.45x                       Simplifying</em>

<em>x>\frac{-15}{-0.45}\\                           Solving for x </em>

<em>x>33.3</em>

Similarly as a rose has to be sold as a whole and not half we will say that <em>x>34</em> So we can then plug in the value of x in Eqn(2) and Eqn(3) and find the cost of buying more than 34 roses from each:

Eqn(2):  Florist A:  y(x>34)=4.75*34+40=201.5  

Eqn(3):  Florist B:  y(x>34)=5.15*34+25=201.8

Which tells us that Florist A is again less expensive than Florist B by $0.30 for a purchase of more than 34 roses. It was also shown that as the cost of the rose increased by $0.05 the number of roses for purchase decreased.

4 0
3 years ago
Solve for x: Log4(x+8)=3
viktelen [127]
If log_{4}(x+8)=3
Then 4^3=x+8
64=x+8
64-8=x
56=x
3 0
3 years ago
The common factor to 18xy and 24 y
vladimir1956 [14]
18xy + 24y = 6y(3x + 8)

The common factor is 6y
6 0
3 years ago
F(x) = <img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%2B7%7D%20-%5Csqrt%7Bx%5E2%2B2x-15%7D" id="TexFormula1" title="\sqrt{x+7} -\
elena-s [515]

Answer:

x >= -7  ................(1a)

x >= 3   ...............(2a1)

Step-by-step explanation:

f(x) =  \sqrt{x+7}-\sqrt{x^2+2x-15}  .............(0)

find the domain.

To find the (real) domain, we need to ensure that each term remains a real number.

which means the following conditions must be met

x+7 >= 0  .....................(1)

AND

x^2+2x-15 >= 0 ..........(2)

To satisfy (1),  x >= -7  .....................(1a) by transposition of (1)

To satisfy (2), we need first to find the roots of (2)

factor (2)

(x+5)(x-3) >= 0

This implis

(x+5) >= 0 AND (x-3) >= 0.....................(2a)

OR

(x+5) <= 0 AND (x-3) <= 0 ...................(2b)

(2a) is satisfied with x >= 3   ...............(2a1)

(2b) is satisfied with x <= -5 ................(2b1)

Combine the conditions (1a), (2a1), and (2b1),

x >= -7  ................(1a)

AND

(

x >= 3   ...............(2a1)

OR

x <= -5 ................(2b1)

)

which expands to

(1a) and (2a1)   OR  (1a) and (2b1)

( x >= -7 and x >= 3 )  OR ( x >= -7 and x <= -5 )

Simplifying, we have

x >= 3  OR ( -7 <= x <= -5 )

Referring to attached figure, the domain is indicated in dark (purple), the red-brown and white regions satisfiy only one of the two conditions.

3 0
3 years ago
Read 2 more answers
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