Which value is a solution to the inequality x < 4

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Answer 1
Answer:

Step-by-step explanation:

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a 12-foot piece of string is cut into two pieces so that the longer piece is 3 feet longer than twice the shorter piece. find the length of both pieces. what is the length of the shorter piece
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3+ 6 x{( 15 +9)=3-2}. Rajah got 49 while Obet got 39. Whose answer was correct? Prove your answer.asap i need it pls i need solution​
The area of triangle

Write a simplified expression for the area of the rectangle below

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Answer:

= 12x + 40

Step-by-step explanation:

area = l * b \n  = 20 *  ((3)/(5) x + 2)  \n  =  (60x)/(5)  + 40 \n  = 12x + 40

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Hellllllllllllllllllllllllllllllllllpppppppppppppppppppppppp whats the x

Answers

Easy buddy....

2 x + 68 = 180

Subtract the sides of the equation minus68

2x = 180 - 68

2x = 112

Divided the sides of the equation by2

x =  (112)/(2)   = 56  \n

So ;

x = 56 \: degreees

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And we're done.

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PLZ HELP ME GEOMETRY
question is below

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The missing justification is for the statement that three angles add to a particular angle. The appropriate choice is ...

... c. Angle Addition Postulate

Leo dug up 6 4/5 pounds oftreasure on his first dig. Anne dug
up 4 1/2 pounds of treasure. How
much more treasure did Leo dig
up than Anne?

Answers

6 4/5 = 6.8
4 1/2 = 4.5

6.8-4.5= 2.3

Answer: 2 3/10 more pounds of treasure

Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity​, where a is a positive real number. Find all points on the parabola at which r and bold r prime are orthogonal.

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Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=<2at,1>

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=<at^2+1\cdot\left(2at\right), t\cdot \left(1)>

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




The directions for a calculation problem are to subtract 7 from 16 and then multiply by 4. Which expression shows this calcu

Answers

Answer:

(16-7)x4

Step-by-step explanation:

You put 16-7 in parenthesis so it means to do it first, then multiply by 4.