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2 years ago

Help me please!!!!!!!

Mathematics

1 answer:

weqwewe [10]2 years ago

4 0

Answer: 31

Step-by-step explanation:

The sum of all triangles is 180 degrees. If you add 116+33 you get 149. If you subtract 149 from 180 you get 31 degrees.

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Please help to answer ASAP

miv72 [106K]

143.17 times 1.35 = 193.28

5 0

2 years ago

HELP PLEASEEE. I DONT UNDERSTAND

Shtirlitz [24]

Answer:

Step-by-step explanation:

So we have the equation:

$x^2-6x+58=0$

And we want to solve for x.

We can solve it by completing the square.

First, subtract 58 from both sides:

$x^2-6x=-58$

Divide the b term by 2 and square it:

$(-6)/2=-3\\(-3)^2=9$

So, add 9 to both sides:

$(x^2-6x+9)=-58+9$

On the left, the perfect square trinomial pattern. Add on the right. So:

$(x-3)^2=-49$

Take the square root of both sides:

$x-3=\pm \sqrt{-49}$

The square root of -49 is 7i:

$x-3=\pm 7i$

Add 3 to both sides:

$x=3\pm 7i$

So, our answer is D.

And we're done!

4 0

2 years ago

The vertex form of the equation of a parabola is y = 2(x + 4)2 - 21.

Ronch [10]

Answer:

y = 2x^2 + 16x + 11

Step-by-step explanation:

2(x + 4)^2 - 21

2(x + 4)(x + 4) - 21

2(x^2 + 8x + 16) - 21

2x^2 + 16x + 32 - 21

2x^2 + 16x + 11

6 0

2 years ago

Ana,beto y Carla son confeccionará de polos,por cada 12 polos que confecciona Ana,Beto confecciona 7,mientras que por cada 28 po

deff fn [24]

Answer:

caca

Step-by-step explanation:

caca con caca e caca en bolas

6 0

2 years ago

An adult ticket to a museum costs 3$ more than a children’s ticket. When 200 adult tickets and 100 children tickets are sold, th

antiseptic1488 [7]

The cost of children’s ticket is $ 5

<h3><u>Solution:</u></h3>

Let "c" be the cost of one children ticket

Let "a" be the cost of one adult ticket

Given that adult ticket to a museum costs 3$ more than a children’s ticket

<em>Cost of one adult ticket = 3 + cost of one children ticket</em>

a = 3 + c ------ eqn 1

<em><u>Given that 200 adult tickets and 100 children tickets are sold, the total revenue is $2100</u></em>

200 adult tickets x cost of one adult ticket + 100 children tickets x cost of one children ticket = 2100

$200 \times a + 100 \times c = 2100$

200a + 100c = 2100 ------ eqn 2

<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "c"</u></em>

Substitute eqn 1 in eqn 2

200(3 + c) + 100c = 2100

600 + 200c + 100c = 2100

600 + 300c = 2100

300c = 1500

Thus the cost of children’s ticket is $ 5

6 0

3 years ago