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2 years ago

Uma peça teatral teve, ao todo, 860 expectadores. Desses expectadores, 90% pagaram

Mathematics

2 answers:

Vika [28.1K]2 years ago

5 0

A 1634 hope this helps

alina1380 [7]2 years ago

4 0

A 1634 source trust me bro

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(57x2:3)+32x450-[39-(4+2):2[+8=

tigry1 [53]

Answer:

57×2/3+32×450-[39-4+2]/2+8

57×2/3+32×450-(39-6/2)+8

57×2/3+32×450-36+8

57×2/3+14400-28

57×2/3+14372

38+14372

14410

8 0

2 years ago

Can anyone help me solve this Algebra 2 Problem, i literally cant figure it out

Andrews [41]

Answer:

Zeros : 1 , -1, 3

Degree : 4

End Behaviour : At x-> ∞ f(x) -> ∞ and x->-∞ f(x) -> ∞

Y - intercept : -3

Extra Points: (0,-3), (2,-3)

Step-by-step explanation:

f(x) = 0 to find the zeros

$Therefore (x+1)(x-1)^{2} (x-3) = 0$

Clearly x = -1,1,3

Here 1 is a repeating root as it is (x-1)²

Degree is highest power of x in f(x)

Clearly it is x*x²*x = x⁴ is the maximum power of x

Thus degree is 4

Looking at end behavior we substitute x->∞ and x-> -∞

Clearly f(x)>0 as all terms are positive and f(x)->∞

Similarly when x->-∞

f(x)>0 as 2 terms are -ve and their product is positive thus f(x)-> ∞

Y-Intercept is f(0)

f(0) = (0+1)(0-1)²(0-3) = 1*1*-3 = -3

Thus Y-Intercept is -3

Substitute x = 0 , 2 for extra points

Thus f(0) = -3

and f(2) = -3

Thus points on the graph (0,-3), (0,2)

We can use all this information to draw a graph remember that 1 is a repeating root so that will be a point of minima. The graph is a parabola that passes through x-axis at x = -1, 3.

5 0

3 years ago

The midpoint of GH is M( 6,-4). One endpoint is G(8,-2). Find the coordinates of endpoint H.

Finger [1]

Using the midpoint formula, the coordinates of endpoint H are (4, -6).

<h3>The Midpoint Formula</h3>

The midpoint formula is given as: $(x_m, y_m) = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} )$

Where,

$(x_m, y_m)$ = coordinates of the midpoint

$(x_1, y_1)$ = coordinates of the first point

$(x_2, y_2)$ = coordinates of the second point

Given the following:

$(x_m, y_m)$ = M( 6,-4)

$(x_1, y_1)$ = G(8,-2)

$(x_2, y_2)$ = H(?, ?)

Plug in the values into the midpoint formula

$M(6, -4) = (\frac{8 + x_2}{2}, \frac{-2 + y_2}{2} )$

Solve for the x-coordinate and y-coordinate separately

$6 = \frac{8 + x_2}{2}$

Multiply both sides

$6 \times 2 = 8 + x_2\\\\12 = 8 + x_2\\\\12 - 8 = x_2\\\\4 = x_2\\\\\mathbf{x_2 = 4}$

$-4 = \frac{-2 + y_2}{2}\\\\-8 = -2 + y_2\\\\-8 + 2 = y_2\\\\-6 = y_2\\\\\mathbf{y_2 = -6}$

Therefore, using the midpoint formula, the coordinates of endpoint H are (4, -6).

Learn more about midpoint formula on:

brainly.com/question/13115533

3 0

2 years ago

If x=0. Find the value of 8x-2(6-x)

Harrizon [31]

Answer:

-12

Step-by-step explanation:

8x - 2(6 - x)

Substitute 0 for each x

8*0 - 2(6 - 0)

Solve.

0 - 2*6

0 - 12

-12

7 0

3 years ago

Read 2 more answers

Drag the tiles to the boxes to form correct pairs. Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

Solving the expression

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

Solving the expression

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

Solving the expression

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

Solving the expression

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

Learn more about algebraic expression from brainly.com/question/11336599

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5 0

3 years ago

Read 2 more answers