Give collinear points in space, how many lines can be drawn

Factor both quadratic expressions. (x 4 + 5x 2 - 36)(2x 2 + 9x - 5) = 0

Sonbull [250]

So focusing on x^4 + 5x^2 - 36, we will be completing the square. Firstly, what two terms have a product of -36x^4 and a sum of 5x^2? That would be 9x^2 and -4x^2. Replace 5x^2 with 9x^2 - 4x^2: $x^4+9x^2-4x^2-36$

Next, factor x^4 + 9x^2 and -4x^2 - 36 separately. Make sure that they have the same quantity inside of the parentheses: $x^2(x^2+9)-4(x^2+9)$

Now you can rewrite this as $(x^2-4)(x^2+9)$ , however this is not completely factored. With (x^2 - 4), we are using the difference of squares, which is $a^2-b^2=(a+b)(a-b)$ . Applying that here, we have $(x+2)(x-2)(x^2+9)$ . x^4 + 5x^2 - 36 is completely factored.

Next, focusing now on 2x^2 + 9x - 5, we will also be completing the square. What two terms have a product of -10x^2 and a sum of 9x? That would be 10x and -x. Replace 9x with 10x - x: $2x^2+10x-x-5$

Next, factor 2x^2 + 10x and -x - 5 separately. Make sure that they have the same quantity on the inside: $2x(x+5)-1(x+5)$

Now you can rewrite the equation as $(2x-1)(x+5)$ . 2x^2 + 9x - 5 is completely factored.

<h3><u>Putting it all together, your factored expression is $(x+2)(x-2)(x^2+9)(2x-1)(x+5)=0$ </u></h3>

5 0

3 years ago

A merchant sold a pen for $6.90 ,thereby making a profit of 15% on her cost calculate the cost of the pen to the merchant to the

NeTakaya

So you know the merchant made a 15% profit on the pen, so she bought it for a cheaper price. To find the cost of the pen before you have to take the price now, $6.90 and times it by 85%. You do 85% because you subtract the 15% she saved from 100% and you get 85%. So 6.90x.85= 5.865 which rounds to $5.87

7 0

3 years ago

Suppose that $3000 is placed in an account that pays 16% interest compounded each year. Assume that no withdrawals are made from

Papessa [141]

Answer:

a) $3480

b) $4036.8

Step-by-step explanation:

The compound interest formula is given by:

$A(t) = P(1 + \frac{r}{n})^{nt}$

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.

Suppose that $3000 is placed in an account that pays 16% interest compounded each year.

This means, respectively, that $P = 3000, r = 0.16, n = 1$

So

$A(t) = P(1 + \frac{r}{n})^{nt}$

$A(t) = 3000(1 + \frac{0.16}{1})^{t}$

$A(t) = 3000(1.16)^{t}$

(a) Find the amount in the account at the end of 1 year.

This is A(1).

$A(t) = 3000(1.16)^{t}$

$A(1) = 3000(1.16)^{1} = 3480$

(b) Find the amount in the account at the end of 2 years.

This is A(2).

$A(2) = 3000(1.16)^{2} = 4036.8$

4 0

3 years ago

An elementary school is offering 3 language classes: one in Spanish, one in French,and one in German. The classes are open to an

Alona [7]

Answer:

A. 0.5

B. 0.32

C. 0.75

Step-by-step explanation:

There are

28 students in the Spanish class,
26 in the French class,
16 in the German class,
12 students that are in both Spanish and French,
4 that are in both Spanish and German,
6 that are in both French and German,
2 students taking all 3 classes.

So,

2 students taking all 3 classes,
6 - 2 = 4 students are in French and German, bu are not in Spanish,
4 - 2 = 2 students are in Spanish and German, but are not in French,
12 - 2 = 10 students are in Spanish and French but are not in German,
16 - 2 - 4 - 2 = 8 students are only in German,
26 - 2 - 4 - 10 = 10 students are only in French,
28 - 2 - 2 - 10 = 14 students are only in Spanish.

In total, there are

2 + 4 + 2 + 10 + 8 + 10 +14 = 50 students.

The classes are open to any of the 100 students in the school, so

100 - 50 = 50 students are not in any of the languages classes.

A. If a student is chosen randomly, the probability that he or she is not in any of the language classes is

$\dfrac{50}{100} =0.5$

B. If a student is chosen randomly, the probability that he or she is taking exactly one language class is

$\dfrac{8+10+14}{100}=0.32$

C. If 2 students are chosen randomly, the probability that both are not taking any language classes is

$0.5\cdot 0.5=0.25$

So, the probability that at least 1 is taking a language class is

$1-0.25=0.75$

3 0

3 years ago

On a piece of paper, graph y < 3/4 x+2 Then determine which answer matches the graph drew.

OleMash [197]

Step-by-step explanation:

Given inequality is $y$ .

To graph that inequality, first we need to graph the line $y=\frac{3}{4}x+2$ . Then shade it for the inequality symbol <.

Graphing line can be done using two points.

plug x=0 into $y=\frac{3}{4}x+2$ , we get:

$y=\frac{3}{4}*0+2=2$

Hence first point is (0,2)

Similarly plug any random number for x say x=4, we get:

$y=\frac{3}{4}*4+2=5$

Hence first point is (4,5).

Now we just need to graph both points and join them by a dotted line. because we have < not <=.

Next part is to test for shading.

Which can be done using any test point which is not on the given line. Say (0,0)

plug x=0,y=0 into $y$ .

$0$

0<2

which is true, hence shading will be in direction of test point (0,0).

Hence final answer will be the choice which looks like the graph attached below;

3 0

3 years ago