I need help with this guys

Mathematics

2 answers:

ss7ja [257]3 years ago

5 0

$HEY THERE!!$

According to the question.

For the fair the organiser ordered 32 rolles of tickets .

though each role has 100 tickets .

so we have to find the total no. of tickets.

So we will multiply 32 with 100.

=> 32× 100= $\boxed{3200\: Tickets }$

HOPE IT HELPED YOU..

sergeinik [125]3 years ago

4 0

Answer:

3200

32 * 100 = 3200

You might be interested in

Indicate the method you would use

FinnZ [79.3K]

Answer:

correct choice is 1st option

Step-by-step explanation:

Two given triangles have two pairs of congruent sides: one pair of length 7 units and second pair of length 8 units. The third side is common, i.e the lengths of third sides are equal too.

Use SSS theorem that states that if three sides of one triangle are congruent to three sides of another triangle, then these two triangles are congruent.

Thus, given triangles are congruent by SSS theorem.

7 0

2 years ago

Solve.

jenyasd209 [6]

Answer:

no solution

Step-by-step explanation:

(you do the exact same you'd do in an equation problem. Get x by itself)

3x+1<2x-2<5x+7

1 < -x-2 < 5x+7

3< -x < 5x+9

3 < -6x < 9

-(1/2) > x > -(3/2)

x can't be bigger than -3 halves and be smaller than -1 half, so the answer is no solution

6 0

3 years ago

The sum of twice a first number and five times a second number is 78. If the second number is subtracted from five times the fir

ss7ja [257]

The numbers are: "9" and "12" .
___________________________________
Explanation:
___________________________________
Let: "x" be the "first number" ; AND:

Let: "y" be the "second number" .
___________________________________
From the question/problem, we are given:
___________________________________
2x + 5y = 78 ; → "the first equation" ; AND:

5x − y = 33 ; → "the second equation" .
____________________________________
From "the second equation" ; which is:

" 5x − y = 33" ;

→ Add "y" to EACH side of the equation;

   5x − y + y = 33 + y ;

to get: 5x = 33 + y ;

Now, subtract: "33" from each side of the equation; to isolate "y" on one side of the equation ; and to solve for "y" (in term of "x");

5x − 33 = 33 + y − 33 ;

to get: " 5x − 33 = y " ; ↔ " y = 5x − 33 " .
_____________________________________________
Note: We choose "the second equation"; because "the second equation"; that is; "5x − y = 33" ; already has a "y" value with no "coefficient" ; & it is easier to solve for one of our numbers (variables); that is, "x" or "y"; in terms of the other one; & then substitute that value into "the first equation".
____________________________________________________
Now, let us take "the first equation" ; which is:
"  2x + 5y = 78 " ;
_______________________________________
We have our obtained value; " y = 5x − 33 " .
_______________________________________
We shall take our obtained value for "y" ; which is: "(5x− 33") ; and plug this value into the "y" value in the "first equation"; and solve for "x" ;
________________________________________________
Take the "first equation":
________________________________________________
→ " 2x + 5y = 78 " ; and write as:
________________________________________________
→ " 2x + 5(5x − 33) = 78 " ;
________________________________________________
Note the "distributive property of multiplication" :
________________________________________________
a(b + c) = ab + ac ; AND:

a(b − c) = ab − ac .
________________________________________________
So; using the "distributive property of multiplication:

→ +5(5x − 33) = (5*5x) − (5*33) = +25x − 165 .
___________________________________________________
So we can rewrite our equation:

→ " 2x + 5(5x − 33) = 78 " ;

by substituting the: "+ 5(5x − 33) " ; with: "+25x − 165" ; as follows:
_____________________________________________________

  → " 2x + 25x − 165 = 78 " ;
_____________________________________________________
→ Now, combine the "like terms" on the "left-hand side" of the equation:

+2x + 25x = +27x ;

Note: There are no "like terms" on the "right-hand side" of the equation.
_____________________________________________________
→ Rewrite the equation as:
_____________________________________________________
→ " 27x − 165 = 78 " ;

Now, add "165" to EACH SIDE of the equation; as follows:

→ 27x − 165 + 165 = 78 + 165 ;

→ to get: 27x = 243 ;
_____________________________________________________
Now, divide EACH SIDE of the equation by "27" ; to isolate "x" on one side of the equation ; and to solve for "x" ;
_____________________________________________________
27x / 27 = 243 / 27 ;

→ to get: x = 9 ; which is "the first number" .
_____________________________________________________
Now; Let's go back to our "first equation" and "second equation" to solve for "y" (our "second number"):

2x + 5y = 78 ; (first equation);

5x − y = 33 ; (second equation);
______________________________
Start with our "second equation"; to solve for "y"; plug in "9" for "x" ;

→ 5(9) − y = 33 ;

45 − y = 33;

Add "y" to each side of the equation:

45 − y + y = 33 + y ; to get:

45 = 33 + y ;

↔ y + 33 = 45 ; Subtract "33" from each side of the equation; to isolate "y" on one side of the equation ; & to solve for "y" ;

→ y + 33 − 33 = 45 − 33 ;

to get: y = 12 ;

So; x = 9 ; and y = 12 . The numbers are: "9" and "12" .
____________________________________________
To check our work:
_______________________
1) Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;

→ 5x − y = 33 ; → 5(9) − 12 =? 33 ?? ; → 45 − 12 =? 33 ?? ; Yes!
________________________
2) Let us plug these values into the original "second equation" ; to see if the equation holds true (with "x = 9" ; and "y = 12") ;

→ 2x + 5y = 78 ; → 2(9) + 5(12) =? 78?? ; → 18 + 60 =? 78?? ; Yes!
_____________________________________
So, these answers do make sense!
______________________________________

3 0

3 years ago

A line passes through the points (2,4) and (5,6) .

Alenkinab [10]

Answer:

Option B and D are correct.

Step-by-step explanation:

Given: A line passes through the points (2,4) and (5,6).

* Case 1:

If a line passes through the points (2, 4) and (5, 6)

Point slope intercept form:

for any two points $(x_1,y_1)$ and $(x_2, y_2)$