Simplify. could i get an explanation too? PLEASE HELP. DUE SOON

Divide:(10x²-3x+4)÷(2x-5)

Helen [10]

Answer:

$\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}$

Quotient: 5x+11

Remainder: 59

Step-by-step explanation:

I'm going to do long division.

The bottom goes on the outside and the top goes in the inside.

Setup:

---------------------------------

2x-5 | 10x^2 -3x +4

Starting the problem from the setup:

5x +11 (I put 5x on top because 5x(2x)=10x^2)

--------------------------------- (We are going to distribute 5x to the divisor)

2x-5 | 10x^2 -3x +4

-(10x^2 -25x) (We are now going to subtract to see what's left.)

-----------------------------------

22x +4 (I know 2x goes into 22x, 11 times.)

( I have put +11 on top as a result.)

-(22x -55) (I distribute 11 to the divisor.)

-----------------------

59 (We are done since the divisor is higher degree.)

The quotient is 5x+11.

The remainder is 59.

The result of the division is equal to:

$5x+11+\frac{59}{2x-5}$ .

We can actually use synthetic division as well since the denominator is linear.

Let's solve 2x-5=0 to find what to put on the outside of the synthetic division setup:

2x-5=0

Add 5 on both:

2x=5

Divide both sides by 2:

x=5/2

Or realize that 2x-5 is the same as 2(x-(5/2)) which you will have to do anyways if you choose this route:

So 5/2 will go on the outside:

5/2 | 10 -3 4

| 25 55

------------------------------

10 22 59

So we have:

$\frac{10x^2-3x+4}{2x-5}$

$=\frac{10x^2-3x+4}{2(x-\frac{5}{2})}=\frac{1}{2} \cdot \frac{10x^2-3x+4}{x-\frac{5}{2}}=\frac{1}{2}(10x+22+\frac{59}{x-\frac{5}{2}})$

Distribute the 1/2 back:

$\frac{10x^2-3x+4}{2x-5}=\frac{10x+22}{2}+\frac{59}{2(x-\frac{5}{2})}$

$\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}$

3 0

4 years ago

Please help answer the two questions in the photo

Alex

The given pentagon can be divided into two figures : a triangle and rectangle as shown in figure.

Let us find area of each figure separately.

Triangle:

Area of triangle is given by:

$A=\frac{1}{2}*b*h$

where b=base and h=height

Height of triangle :

h=(3x+5)-(2x+1)

h=x+4

Base = 2x-2

$A=\frac{1}{2}*(x+4)(2x-2)$

Area of triangle= (x-1)(x+4) = x²+3x-4

Area of rectangle:

Area of rectangle is given by:

A=l*b

$A=(2x-2)(2x+1)$

Area of rectangle = 4x²-2x-2

Area of pentagon = Area of triangle + Area of rectangle

Area of pentagon = x²+3x-4 + 4x²-2x-2

Area of pentagon = 5x²+x-6

If area of pentagon is 42 cm², then solving for x,

$42=5x^{2}+x-6$

$5x^{2}+x-48=0$

Factorising to get x,

x=-3.2 and x=3

If we take x as negative the side will be negative, so we neglect x=-3.2

So x=3 is the answer.

3 0

3 years ago

A spinner is spun 20 times, and the number of times the arrow lands

VARVARA [1.3K]

Answer:

Use language that's easy-to-understand. If there's a piece of vocabulary that you feel needs clarifying, define it!

Step-by-step explanation:

Color

Red

Purple

3 0

2 years ago

An automobile manufacturer has discovered that 20% of all the transmissions it installed in a particular style of truck are defe

Hatshy [7]

Answer:

0.148 = 14.8% probability that they will need to order at least one more new transmission

Step-by-step explanation:

For each transmission, there are only two possible outcomes. Either it is defective after a year of use, or it is not. The probability of a transmission being defective is independent of any other transmission. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

20% of all the transmissions it installed in a particular style of truck are defective after a year of use.

This means that $p = 0.2$

Sold seven trucks:

This means that $n = 7$

It has two of the new transmissions in stock. What is the probability that they will need to order at least one more new transmission?

This is the probability that at least 3 are defective, that is:

$P(X \geq 3) = 1 - P(X < 3)$

In which

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{7,0}.(0.2)^{0}.(0.8)^{7} = 0.2097$

$P(X = 1) = C_{7,1}.(0.2)^{1}.(0.8)^{6} = 0.3670$

$P(X = 2) = C_{7,2}.(0.2)^{2}.(0.8)^{5} = 0.2753$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.2097 + 0.3670 + 0.2753 = 0.852$

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.852 = 0.148$

0.148 = 14.8% probability that they will need to order at least one more new transmission

6 0

3 years ago

Could I get the answer, all work, and explanation for this problem? Thanks

shepuryov [24]

$\bf \sqrt{x+8}+\sqrt{x}=2\implies \sqrt{x+8}=2-\sqrt{x}\impliedby \textit{we raise both at }^2
\\\\\\
(\sqrt{x+8})^2=(2-\sqrt{x})^2\implies x+8=(2-\sqrt{x})(2-\sqrt{x})
\\\\\\
x+8=\stackrel{\textit{and we FOIL here}}{4-4\sqrt{x}+(\sqrt{x})^2}\implies x+8=4-4\sqrt{x}+x
\\\\\\
4=-4\sqrt{x}\implies \cfrac{4}{-4}=\sqrt{x}\implies -1=\sqrt{x}\impliedby \stackrel{again}{\textit{we raise both at }^2}
\\\\\\
(-1)^2=(\sqrt{x})^2\implies 1=x$

3 0

4 years ago