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2 years ago

Hello, i need help with this question please! solve x/2 +3 > 4

Mathematics

2 answers:

kirill [66]2 years ago

7 0

Answer:

x > 2

Step-by-step explanation:

A linear inequality is given to us and we need to find the interval of x. The given inequality is

⇒ x/2 + 3 > 4

Subtract 3 to both sides,

⇒ x/2 + 3 - 3 > 4 -3

⇒ x/2 > 1

Multiply both sides by 2 ,

⇒ x > 2

Hence x can have all values more than 2 .

jeka57 [31]2 years ago

3 0

Answer:

x >2

Step-by-step explanation:

x/2 +3 > 4

Subtract 3 from each side

x/2 +3-3 > 4-3

x/2 >1

Multiply each side by 2

x/2*2 >1*2

x >2

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Find p(2) and p(4) for the function p(x) = 6x4 + 4x3 – 3x2 + 8x + 15.

gulaghasi [49]

Answer:

p(2) =147 and p(4) = 1791

Step-by-step explanation:

We are given p(x)= 6x^4 + 4x^3 – 3x^2 + 8x + 15.

Now we need to find value of p(2) and p(4)

Put x =2,

p(2) = 6(2)^4 + 4(2)^3 – 3(2)^2 + 8(2) + 15

p(2) = 6(16)+4(8)-3(4)+8(2)+15

p(2) = 96+32-12+16+15

p(2) = 147

Now put x = 4

p(4) = 6(4)^4 + 4(4)^3 – 3(4)^2 + 8(4) + 15

p(4) = 6(256)+4(64)-3(16)+8(4)+15

p(4) = 1536+256-48+32+15

p(4) = 1791

3 0

3 years ago

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Help jabsjsjsjsj thank

vodomira [7]

The answer is probably c if I’m correct

5 0

3 years ago

Is 1/2x-4 a function?

lakkis [162]

Answer:

Yes it is

Step-by-step explanation:

Each x value is represented by exactly one y value.

7 0

3 years ago

If Tiva babysits for hours, she will earn $32.

Viktor [21]

Answer:

true i think not 100% sure tho

Step-by-step explanation:

6 0

2 years ago

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I really need help! This is last minute and its 1 am. I'm tired so I'm going to leave it to you guys to solve my problems and pr

Illusion [34]

Answer:

<h2>question 1</h2>

$n^{2} - 20n -96 = 0$

use product and sum method

product = -96

sum = -20

numbers needed = ( -24 , 4)

n - 24 = 0

n + 4 = 0

hence n = 24 and n = -4

<h2>Question 2 </h2>

$x^{2} + 12 x = 48$

in the form $ax^{2} +bx +c = 0$

= $x^{2} +12x - 48$

make use of the formula :

$\frac{-b+-\sqrt{b^{2} -4ac} }{2a}$

replace values to make 2 equations :

1. $\frac{-12+\sqrt{12^{2} -4*1*-48} }{2*1}$ = 3.17

2. $\frac{-12-\sqrt{12^{2} -4*1*-48} }{2*1}$ = -15.2

hence x = 3.17 and x = -15.2

<h2>Question 3 </h2>

$x^{2} -14x+40=0$

use product and sum method

product = 40

sum = -14

numbers needed = (-10 , -4)

x - 10 = 0

x - 4 = 0

hence x = 10 and x = 4

<h2>Question 4 </h2>

$5b^{2} -20b-18 = 7$

in the form $ax^{2} +bx +c = 0$

this becomes $5b^{2} -20b-18-7$

= $5b^{2} -20b-25$

can simplify by 5

= $b^{2} -4b-5 =0\\$

use product and sum method

product = -5

sum = -4

numbers needed (-5 , 1)

b-5 = 0

b + 1 = 0

hence b = 5 and b = -1

5 0

3 years ago

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