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3 years ago

What is (4x+-7)(2x+-9)

Mathematics

1 answer:

uysha [10]3 years ago

8 0

Answer:x=1

Step-by-step explanation:(4x+-7)=(2x+-9)

We move all terms to the left:

(4x+-7)-((2x+-9))=0

We add all the numbers together, and all the variables

(4x-7)-((2x-9))=0

We get rid of parentheses

4x-((2x-9))-7=0

We calculate terms in parentheses: -((2x-9)), so:

(2x-9)

We get rid of parentheses

2x-9

Back to the equation:

-(2x-9)

We get rid of parentheses

4x-2x+9-7=0

We add all the numbers together, and all the variables

2x+2=0

We move all terms containing x to the left, all other terms to the right

2x=-2

x=-2/2

x=-1

(4x+-7)=(2x+-9)

We move all terms to the left:

(4x+-7)-((2x+-9))=0

We add all the numbers together, and all the variables

(4x-7)-((2x-9))=0

We get rid of parentheses

4x-((2x-9))-7=0

We calculate terms in parentheses: -((2x-9)), so:

(2x-9)

We get rid of parentheses

2x-9

Back to the equation:

-(2x-9)

We get rid of parentheses

4x-2x+9-7=0

We add all the numbers together, and all the variables

2x+2=0

We move all terms containing x to the left, all other terms to the right

2x=-2

x=-2/2

x=-1

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(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?

DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

$U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500$

There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

$U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111$

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

$P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611$

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3 years ago

What is an equation of the line that passes through the points (2,-3) and (-3,7)

sveticcg [70]

Answer:

-1 + 4

Step-by-step explanation:

i did the math

5 0

3 years ago

15. Suppose a box of 30 light bulbs contains 4 defective ones. If 5 bulbs are to be removed out of the box.

Naddika [18.5K]

Answer:

1) Probability that all five are good = 0.46

2) P(at most 2 defective) = 0.99

3) Pr(at least 1 defective) = 0.54

Step-by-step explanation:

The total number of bulbs = 30

Number of defective bulbs = 4

Number of good bulbs = 30 - 4 = 26

Number of ways of selecting 5 bulbs from 30 bulbs = $30C5 = \frac{30 !}{(30-5)!5!} \\$

$30C5 = 142506$ ways

Number of ways of selecting 5 good bulbs from 26 bulbs = $26C5 = \frac{26 !}{(26-5)!5!} \\$

$26C5 = 65780$ ways

Probability that all five are good = 65780/142506

Probability that all five are good = 0.46

2) probability that at most two bulbs are defective = Pr(no defective) + Pr(1 defective) + Pr(2 defective)

Pr(no defective) has been calculated above = 0.46

Pr(1 defective) = $\frac{26C4 * 4C1}{30C5}$

Pr(1 defective) = (14950*4)/142506

Pr(1 defective) =0.42

Pr(2 defective) = $\frac{26C3 * 4C2}{30C5}$

Pr(2 defective) = (2600 *6)/142506

Pr(2 defective) = 0.11

P(at most 2 defective) = 0.46 + 0.42 + 0.11

P(at most 2 defective) = 0.99

3) Probability that at least one bulb is defective = Pr(1 defective) + Pr(2 defective) + Pr(3 defective) + Pr(4 defective)

Pr(1 defective) =0.42

Pr(2 defective) = 0.11

Pr(3 defective) = $\frac{26C2 * 4C3}{30C5}$

Pr(3 defective) = 0.009

Pr(4 defective) = $\frac{26C1 * 4C4}{30C5}$

Pr(4 defective) = 0.00018

Pr(at least 1 defective) = 0.42 + 0.11 + 0.009 + 0.00018

Pr(at least 1 defective) = 0.54

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3 years ago

Compared to its 'parent' function f(x)=x^2, describe the changes if f(x)=-x^2-3?

zimovet [89]

Answer:

it opens down meaning it has a maximum instead of a minimum. the graph has a horizontal shift of 3 spaces to the right

Step-by-step explanation:

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3 years ago

A cylinder has radius r and height h. A. How many times greater is the surface area of a cylinder when both dimensions are multi

Ierofanga [76]

Answer: A. Factor 2 => 4x greater

Factor 3 => 9x greater

Factor 5 => 25x greater

Step-by-step explanation: A. A cylinder is formed by 2 circles and a rectangle in the middle. That's why surface area is given by circumference of a circle, which is the length of the rectangle times height of the rectangle, i.e.:

A = 2.π.r.h

A cylinder of radius r and height h has area:

$A_{1}$ = 2πrh

If multiply both dimensions by a factor of 2:

$A_{2}$ = 2.π.2r.2h

$A_{2}$ = 8πrh

Comparing $A_{1}$ to $A_{2}$ :

$\frac{A_{2}}{A_{1}}$ = $\frac{8.\pi.rh}{2.\pi.rh}$ = 4

Doubling radius and height creates a surface area of a cylinder 4 times greater.

By factor 3:

$A_{3} = 2.\pi.3r.3h$

$A_{3} = 18.\pi.r.h$

Comparing areas:

$\frac{A_{3}}{A_{1}}$ = $\frac{18.\pi.r.h}{2.\pi.r.h}$ = 9

Multiplying by 3, gives an area 9 times bigger.

By factor 5:

$A_{5} = 2.\pi.5r.5h$

$A_{5} = 50.\pi.r.h$

Comparing:

$\frac{A_{5}}{A_{1}}$ = $\frac{50.\pi.r.h}{2.\pi.r.h}$ = 25

The new area is 25 times greater.

B. By analysing how many times greater and the factor that the dimensions are multiplied, you can notice the increase in area is factor². For example, when multiplied by a factor of 2, the new area is 4 times greater.

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3 years ago