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3 years ago

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There were 19,748 adults at the baseball game. There were 5,136 fewer children at the baseball gamr than adults. How many childr

en were at the baseball game

1 answer:

Harman [31]3 years ago

6 0

There were 13,612 children at the baseball game

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Select all pairs of numbers that have LCM of 18?

spayn [35]

Only C and D....................

8 0

4 years ago

Here is another if you answer first you get a brainlist SIMPLe

tigry1 [53]

8x - 2x = -17 + 35
6x = 18
divide by 6
x = 3

5 0

3 years ago

Tell whether the following equations are true or false. Then, explain your reasoning.

lorasvet [3.4K]

Answer and explanation:

Given : Equations

1) $x + 6g-6g = x$

2) $2f-4e + 4e = 2f$

To find : Tell whether the following equations are true or false. Then, explain your reasoning.

Solution :

Equation means when left hand side and right hand side of the equation is same.

1) $x + 6g-6g = x$

Take LHS,

$LHS=x + 6g-6g$

Same term with opposite sign cancel each other or sum is zero,

$LHS=x$

$LHS=RHS$

Therefore, Yes, it is true as it satisfy definition of equation.

2) $2f-4e + 4e = 2f$

Take LHS,

$LHS=2f-4e + 4e$

Same term with opposite sign cancel each other or sum is zero,

$LHS=2f$

$LHS=RHS$

Therefore, Yes, it is true as it satisfy definition of equation.

4 0

4 years ago

Hamburger Hut sells regular hamburgers as well as a larger burger. Either type can include cheese, relish, lettuce, tomato, must

Studentka2010 [4]

Answer:

a) 40 different hamburgers can be ordered with exactly three extras

b) 20 different regular hamburgers can be ordered with exactly three extras

c) 7 different regular hamburgers can be ordered with at least five extras

Step-by-step explanation:

The order in which the extras are ordered is not important. So we use the combinations formula to solve this question.

Combinations formula:

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

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

In this problem:

2 options of hamburger(regular or larger)

6 options of extras(cheese, relish, lettuce, tomato, mustard, or catsup.).

(a) How many different hamburgers can be ordered with exactly three extras?

1 hamburger type, from a set of 2.

3 extras, from a set of 6. So

$C_{2,1}*C_{6,3} = \frac{2!}{1!(2-1)!}*\frac{6!}{3!(6-3)!} = 2*20 = 40$

40 different hamburgers can be ordered with exactly three extras

(b) How many different regular hamburgers can be ordered with exactly three extras?

3 extras, from a set of 6. So

$C_{6,3} = \frac{6!}{3!(6-3)!} = 20$

20 different regular hamburgers can be ordered with exactly three extras

(c) How many different regular hamburgers can be ordered with at least five extras?

Five extras:

5 extras, from a set of 6. So

$C_{6,5} = \frac{6!}{5!(6-5)!} = 6$

Six extras:

6 extras, from a set of 6. So

$C_{6,6} = \frac{6!}{6!(6-6)!} = 1$

6 + 1 = 7

7 different regular hamburgers can be ordered with at least five extras

8 0

3 years ago

Match the graph shown with the following equation<br> Answer a and b please

julia-pushkina [17]

Answer:

a) D b)B

Step-by-step explanation:

a) the graph is of negative slope as it slants downwards hence slop/gradient is negative. y intercept is positive so only option d is right

b) sub each option point into 3y = 2x - 1 if they equate to be the same, it is correct.

3(2) = 2(6) - 1

6 = 11 (not equal)

3(1) = 2(2) - 1

3 = 3 (equal) this is the answer

of course you can sub in the rest to confirm but I'm lazy so yeah I'll stop here

Topic: Linear Graphs and its equations

If you would like to venture further, you can check out my Instagram page (learntionary) where I post notes and mathematics tips. Thanks!

8 0

3 years ago