All categories

3 years ago

Help me asap please !!

Mathematics

1 answer:

pav-90 [236]3 years ago

4 0

I’m not positive but I think all you have to do is minus 62 from 180 which would be 119

You might be interested in

A total of 504tickets were sold for the school play. They were either adult tickets or student tickets. There were 54more studen

LuckyWell [14K]

Answer:

225 adult tickets

Step-by-step explanation:

S: no. of students

A: no. of adults

S = A + 54

S + A = 504

A + A + 54 = 504

2A = 450

A = 225

S = 225 + 54

S = 279

3 0

3 years ago

Which properties could you use to solve the following expression?

lesantik [10]

Answer:

associative only!!!!!!!!!!!!!

7 0

3 years ago

Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

5 0

3 years ago

Read 2 more answers

Someone please help me ASAP!!!! I really need it

MatroZZZ [7]

^ this guy is a scammer don’t click the link

7 0

3 years ago

A number increased by 4 is the same as 19 - 2 times the number

masha68 [24]

The number is 5

x + 4 = 19 - 2x
+2x +2x
3x + 4 = 19
-4 -4
3x = 15
/3 /3
x = 5

6 0

3 years ago