help help help please

its actually y=>3x-1 because it's shaded below the line.

below the line: < or <=

above the line: > or>=

To convert between meters and gigameters, use the equation 1 Gm = 10^9 m. This relationship is important for converting units in scientific and mathematical contexts.

To show the equivalency between meters and gigameters, we can use the following equation:

1 gigameter (Gm) = 10⁹ meters (m).

This equation means that one gigameter is equal to one billion meters. We can use this equivalency for converting measurements from meters to gigameters or vice versa. For example, to convert 2 gigameters to meters, simply multiply 2 by 10⁹ to get 2 x 10⁹ meters.

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Equations may be true or false, just as word sentences may be true or false. The equation:

3 + x = 7

will be false if any number except 4 is substituted for the variable. The value of the variable for which the equation is true (4 in this example) is called the solution of the equation. We can determine whether or not a given number is a solution of a given equation by substituting the number in place of the variable and determining the truth or falsity of the result.

Example 1 Determine if the value 3 is a solution of the equation

4x - 2 = 3x + 1

Solution We substitute the value 3 for x in the equation and see if the left-hand member equals the right-hand member.

4(3) - 2 = 3(3) + 1

12 - 2 = 9 + 1

10 = 10

Ans. 3 is a solution.

The first-degree equations that we consider in this chapter have at most one solution. The solutions to many such equations can be determined by inspection.

Example 2 Find the solution of each equation by inspection.

a. x + 5 = 12
b. 4 · x = -20

Solutions a. 7 is the solution since 7 + 5 = 12.
b. -5 is the solution since 4(-5) = -20.

The constraints for this problem are represented by the inequalities x > 3y and 15x + 25y ≥ 700. The combinations that satisfy the constraints is (45,10) as it meets both conditions.

The constraints for the art store situation are represented by the inequalities x > 3y and 15x + 25y ≥ 700. The first inequality, x > 3y, represents the condition that the number of small prints made must be more than three times the number of large prints sold. The second inequality, 15x + 25y ≥ 700, characterizes the fact the store needs to make a daily profit of at least $700 in order to cover costs.

With these inequalities, we can determine which pairs of small print and large print quantities satisfy this system of constraints. Checking each pair, we find that the combination (45,10) satisfies both inequalities as 45 is more than three times 10, and 15*45 + 25*10 = $775, which is greater than $700. Thus, selling 45 small prints and 10 large prints will satisfy the constraints.

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Answer:  12

Step-by-step explanation:

Given : You are making identical gift bags using 24 candles and 36 bottles of lotion .

Then, the greatest number of gift bags you can make with no items left over= greatest common factor of 24 and 36

Prime factorization of 24 and 36 :_

[tex]24=2\times2\times2\times3\\\\ 36=2\times2\times3\times3[/tex]

GCF(24 , 36)= [tex]2\times2\times3=12[/tex]

Hence, the greatest number of gift bags you can make with no items left over =12

Midpoint, as the word suggests, means the point which lies in the middle of something. If m is the midpoint of PQ then PM will be half of PQ.

Midpoint, as the word suggests, means the point which lies in the middle of something. The midpoint of a line segment means a point that lies in the mid of the given line segment.

Given that m is the midpoint of PQ. Therefore, it will divide PQ into two equal parts. Therefore,  we can write,

PQ = PM + MQ

But since PM = MQ,

PQ = PM + PM

PQ = 2 PM

PM = PQ/2

Hence, if m is the midpoint of PQ then the relationship between PM and PQ will be half of PQ.

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Answer:

200 mi.

Step-by-step explanation:

200 mi. is more or greater than 200 km. Because, 200 mi. in km will be 321 km so that is greater than 200 km.  

I hope this helps you.

1) There are no prime numbers whose sum is divisible by 7 and is a perfect square.

2) 1 pipe will take 2 hours to fill 1 pool, as it fills at the same rate as all 5 pipes combined.

3) Ron's score is 20%.

4) There are 20 students in the class.

5) Katie is 14 years old.

Let's evaluate each of the statements you provided:

1) The sum of 2 prime numbers is ALWAYS even, so if x and y are prime numbers, then (x+y)/7 = z².

In this case either x = 2 or y = 5 (or vice versa). This is the only possibility to be divisible by 7.

Then (5+2)/7 = 1² = 1.

The first part is correct; the sum of two prime numbers is always even. However, the statement about (x+y)/7 = z² doesn't hold true for all prime numbers. It might work for the specific values of x and y you mentioned, but it's not a general rule.

5 pipes → 2 hours → 2 pools, then 1 pipe takes (5x2) 10 hours to fill 2 pools, or 1 pipe takes (10/2) 5 hours to fill 1 pool.

This is a correct interpretation of the information provided. If 5 pipes can fill 2 pools in 2 hours, then each pipe alone takes 10 hours to fill 2 pools, or 5 hours to fill 1 pool.

It's 6/30 = 0.2 or 20%.

Correct, 6/30 simplifies to 0.2, which is equivalent to 20%.

35% represents 7 students, and 100% of the class represents x students. Therefore, 35%/7 = 100%/x. Cross-multiplying, we get 0.35x = 7, and x = 7/0.35, which simplifies to x = 20 students.

Let K be the age of Katie, and B her brother's age. K + 2 = B (Today). In 12 years, B+12 will be twice K (remember K = B-2):

B+12 = 2(K) OR B+12 = 2(B-2)

B+12 = 2(B-2)

B+12 = 2B - 4

12 + 4 = B, and B = 16. If the brother is 16, then Kate is 14.

The calculations and logic are correct. If Katie's brother is 16, then Katie is indeed 14 years old.

for such more question on prime numbers

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Answer:

quotient=2,731

Remainder=0

Step-by-step explanation:

893) 2,438,783 (2,731

      - 1 786

      ---------

          6527

        - 6251

        ---------

            2768

          - 2679

         -----------

                893

             -  893

             ---------

                 0

1. The algebraic property of equality that is represented is the multiplication property. The answer is letter P.

2. The algebraic property of equality that is represented is the subtraction property. The answer is letter E.


3. The algebraic property of equality that is represented is the substitution property. The answer is letter M.


4. The algebraic property of equality that is represented is the division property. The answer is letter A.


5. The algebraic property of equality that is represented is the addition property. The answer is letter A.


6. The algebraic property of equality that is represented is the distributive property. The answer is Letter E.


7. The algebraic property of equality that are represented is the transitive property. The answer is letter is E.


8. The algebraic property of equality that are represented is the symmetric property. The answer is Letter A.


9. The algebraic property of equality that are represented is the reflexive property. The answer is letter C.


10. The algebraic property of equality that are represented by is the multiplication property. The answer is letter P.


11. The algebraic property of equality that are represented is the subtraction property. The answer is letter L.

12. The algebraic property of equality that are represented is the reflexive property. The answer is letter S.

13. The algebraic property of equality that are represented is the transitive property. The answer is letter U.


14. The algebraic property of equality that are represented is the symmetric property. The answer is letter K.

The given is 3 5 14 2  4 1 10 11 6 12 8 13 9 7

Plugging in the letters we got for each number, gives us MAKE APPLESAUCE.

As the number of apples is less than the total number of friends among which the apples need to be distributed so, no one will get a complete piece of apple so, 5 apples can be shared in between 7 friends as [tex]\frac{5}{7}[/tex].

According to the question, the total number of apples are 5 and the total number of friends are 7.

So, each friend will get (bu unitary method)-

[tex]\rm{Each\;friend\;will\;get}=\dfrac{5}{7}\;\rm apples[/tex]

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Answer:

y= -2x-2

Step-by-step explanation:

point slope- y=mx+b

y-y1=m(x-x1)

you have the point   (-1,0) and you have to use (x1, y1)

your slope is -2      

Substitute

y-0= -2(x +1)        

Distribute

y-0=-2x-2

+0         +0

y= -2x-2

The y-intercept of the function g(x) = (1/4)^x is (0,1), and the horizontal asymptote is at y=0.

The correct option is (c).

Given the function g(x) = (1/4)^x, we need to find the y-intercept and the equation of the horizontal asymptote.

The y-intercept occurs where x = 0. Substituting 0 for x in g(x), we get:

g(0) = (1/4)^0 = 1

So the y-intercept is at the point (0,1).

For horizontal asymptotes, we look at the behavior of the function as x goes to positive or negative infinity. In this case, as x goes to infinity, (1/4)^x approaches 0. Therefore, the horizontal asymptote of g(x) is at y = 0.

The correct answer is c. The y-intercept is (0,1) and the asymptote is at y=0.

In this case, we are simply asked to find for the value of:

xbya or abxy

Since we know that given values of ab and xy, therefore abxy is:

abxy = 37 * (1 / 37)

abxy = 37 / 37

abxy = 1

12 months per year

3200/12 = 266.66 per month

jan/feb = 2 months

266.666 x 2 = 533.33 round off to 533 cars sold

If 3,200 cars were sold all year, then approximately 533 cars were sold in the Jan/Feb period.

We need to calculate the proportion of the year that these two months represent and then apply that proportion to the total number of cars sold in the year.

Since there are 12 months in a year, the Jan/Feb period represents [tex]\(\frac{2}{12}\)[/tex] of the year. We can simplify this fraction to [tex]\(\frac{1}{6}\)[/tex] by dividing both the numerator and the denominator by 2.

To find the number of cars sold in the Jan/Feb period, we multiply the total number of cars sold in the year by the fraction representing Jan/Feb:

[tex]\[ \text{Number of cars sold in Jan/Feb} = \text{Total cars sold} \times \frac{2}{12} \][/tex]

Substituting the given total number of cars sold:

[tex]\[ \text{Number of cars sold in Jan/Feb} = 3,200 \times \frac{2}{12} \][/tex]

[tex]\[ \text{Number of cars sold in Jan/Feb} = 3,200 \times \frac{1}{6} \][/tex]

[tex]\[ \text{Number of cars sold in Jan/Feb} = \frac{3,200}{6} \][/tex]

Performing the division:

[tex]\[ \text{Number of cars sold in Jan/Feb} = 533.\overline{3} \][/tex]

[tex]\[ \text{Number of cars sold in Jan/Feb} \approx 533[/tex]

132 + 56 + 79 = 267

360 - 267 = 93

answer is 93 degrees

An equation with all real numbers as solutions is one that, regardless of what value a variable might have, both sides will always be equal, such as 2x+3 = 2x+3. An equation with no real solutions is a statement where the two sides cannot ever be equal, like 2x+5 = 2x+7.

An example of an equation with variables on both sides that has all real numbers as the solution is a statement where both sides are identical. For instance, 2x+3 = 2x+3. All real numbers can satisfy this equation for x because both sides are always equal, regardless of what your value of x is.

On the other hand, an equation with no real solutions is one where it's impossible to find any real number that satisfies the equation. An example can be 2x+5 = 2x+7. No matter what value you put in place of x, the two sides of the equation will never be equal. Hence, this equation has no real solution.

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