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2 years ago

10

a sports team donates 157 tickets to a sporting event,with tickets to be shared equally among 9 classrooms.How many tickets does

each classroom receive? How many tickets are left over?

1 answer:

Valentin [98]2 years ago

5 0

Answer:

Each class should get 17 tickets. 4 tickets will be left over.

Step-by-step explanation:

157/9=17.444... -> 17 per class

17 x 9 = 153

157-153= 4 left

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Given right triangle ABC with altitude BD drawn to hypotenuse AC. If AD = 4

Vinil7 [7]

Answer:

BD = 6

Step-by-step explanation:

Theorem: The altitude to the hypotenuse of a right triangle is the mean proportional between the segments of the hypotenuse.

So, $\frac{4}{BD} = \frac{BD}{9}$

$BD^{2} = 36$

BD = 6

See how easy that is IF you know the theorem.

4 0

2 years ago

Which expression shows 18 + 15 written as a product of two factors?

____ [38]

1st one is ur answer

6 0

2 years ago

Read 2 more answers

Solve the system 2x+4y=16 5x +3y=19 then find the value of y/x

AURORKA [14]

Answer:

I would change 2x+4y=24 into x=12–2y

To do that, divide both sides by 2 and then subtract 2y on each side.

After that, substitute for x. 3x+2y=19 would become 3(12–2y)+2y=19.

Then solve.

3(12–2y)+2y=19

36–6y+2y=19

-4y+36=19

-4y=-17

y=4.25

Then, substitute y in either equation.

Either this:

3x+2y=19

3x+2(4.25)=19

3x+8.5=19

3x=10.5

x=3.5

Or:

2x+4y=24

2x+4(4.25)=24

2x+17=24

2x=7

x=3.5

Or you could solve it in the equation you created in the beginning:

x=12–2y

x=12–2(4.25)

x=12–8.5

x=3.5

The coordinates where the lines intercept are (3.5, 4.25).

Sorry for the long answer!

4 0

2 years ago

Help I am very bad at math I need answers by tomorrow if you can help please do

d1i1m1o1n [39]

Answer:

Step-by-step explanation:

(a - b)(a +b) = a² - b²

1 - Sin² A = Cos² A

$LHS = \frac{1}{1- Sin A} + \frac{1}{1 + Sin A}\\\\= \frac{1*(1 + Sin A)}{(1- Sin A)(1 + Sin A)} + \frac{1*(1- Sin A)}{(1 + Sin A)(1- Sin A)}\\\\= \frac{1 + Sin A+ 1 - Sin A}{1^{2}- Sin^{2} A}\\\\= \frac{2}{1 - Sin^{2} A}\\\\= \frac{2}{Cos^{2} A}\\\\= 2 Sec^{2} A$

2) Sec² A - Tan² A = 1

$LHS = \frac{1}{Sec A - Tan A}\\\\=\frac{1*(Sec A + Tan A)}{(Sec A - Tan A)(Sec A + Tan A)}\\\\=\frac{Sec A + Tan A}{Sec^{2} A - Tan^{2} A}\\\\=\frac{Sec A + Tan A }{1}\\\\= Sec A + Tan A = RHS\\\\\\$

3) LHS = Cosec² A + Cot² A

= Cosec² A + Cosec² A - 1

= 2Cosec² A - 1 = RHS

$4) LHS = \frac{Sec A}{Cos A}- \frac{Tan A}{Cot A}\\\\ = Sec A*\frac{1}{Cos A}-Tan A*\frac{1}{Cot A}\\\\ = Sec A * Sec A - Tan A * Tan A\\\\= Sec^{2} A - Tan^{2} A \\\\= 1$

3 0

2 years ago

2(3x+6) = 3(2x-6) <br> solve equations with variables on both sides

Serjik [45]

Answer:

2(3x+6) = 6x+12
3(2x-6) = 6x-18

6x+12=6x-18

Subtract 12 from 18, which gives you:

6x=6x-6

Subtract the 6x on the opposite side of the equal side, and your answer is: x = -6

6 0

2 years ago