F(x)

There are 450 people at a play, the total receipt for $600 and the admission was $2 for adults and $0.75 for children, how many

sdas [7]

Answer:

340 children, 210 adults.

Step-by-step explanation:

$Let \ a \ - adults, \ c - children;\\c + a = 450\\0.75c + 2a = 600|\cdot 4\\3c + 8a = 2400;\\\\c + a = 450|\cdot(-3)\\3c + 8a = 2400\\-3c -3a = -1350\\5a = 1050 => a = 210, c = 450 -210 = 340.$

8 0

3 years ago

The coordinates of △ABC are A(12,8), B(10,18), C(4,16) . After a dilation, the coordinates are A'(6,4), B'(5,9), C'(2,8) A′(6,4)

gogolik [260]

Take note of that every single coordinate had their values divided by 2 after the dilation. Therefore, the scaling factor is 1/2, or .5

7 0

3 years ago

Read 2 more answers

How do you find the missing height of a triangle?

g100num [7]

Answer:

Use the Pythagorean theorem

Step-by-step explanation:

The pythagorean theorem is $a^{2} +b^{2} =c^{2}$ where a is a height, b is a height, and c is the hypotenuse, which is the longest side of a triangle. For example, if the triangle has two side with lengths of three and four we would put them into the equation to get:

$3^{2} +4^{2} =c^{2}$

Simplify the equation to get:

$9+16=c^2$

Add 16 and 9 to get 25 and take the square root of both sides:

$\sqrt{25} =c$

The sqrt of 25 is 5. Therefore, the length of the hypotenuse, or c, is 5.

8 0

3 years ago

Read 2 more answers

Can you answer these please many thanks

givi [52]

Given: (a) $2(x+3)=2x+6$ and (b) $4(y-3)=4y-12$ .

To find: The expanded form of the given expressions.

(c) $4(m+n)=4m+4n$ (using $a(b+c)=ab+ac$ )

(d) $3(5-q)=15-3q$ (using $a(b-c)=ab-ac$ )

(e) $5(2c+1)=10c+5$ (using $a(b+c)=ab+ac$ )

(f) $3(2x-5)=6x-15$ (using $a(b-c)=ab-ac$ )

(g) $7(4b-1)=28b-7$ (using $a(b-c)=ab-ac$ )

(h) $3(2x+y-5)=3((2x+y)-5)$

⇒ $3(2x+y-5)=3(2x+y)-15$ (using $a(b-c)=ab-ac$ )

⇒ $3(2x+y-5)=6x+3y-15$ (using $a(b+c)=ab+ac$ )

(i) $2(6a-4b+3)=2((6a-4b)+3)$

⇒ $2(6a-4b+3)=2(6a-4b)+6$ (using $a(b+c)=ab+ac$ )

⇒ $2(6a-4b+3)=12a-8b+6$ (using $a(b-c)=ab-ac$ )

(j) $6(m+n+p)=6((m+n)+p)$

⇒ $6(m+n+p)=6(m+n)+6p$ (using $a(b+c)=ab+ac$ )

⇒ $6(m+n+p)=6m+6n+6p$ (using $a(b+c)=ab+ac$ )

(k) $y(y+2)=y^2+2y$ (using $a(b+c)=ab+ac$ )

(l) $g(g-3)=g^2-3g$ (using $a(b-c)=ab-ac$ )

(m) $n(4-n)=4n-n^2$ (using $a(b-c)=ab-ac$ )

(n) $a(b+c)=ab+ac$ (using $a(b+c)=ab+ac$ )

(o) $s(3s-4)=3s^2-4s$ (using $a(b-c)=ab-ac$ )

(p) $2x(x+5)=2x^2+10x$ (using $a(b+c)=ab+ac$ )

(q) $4y(x-3)=4xy-12y$ (using $a(b-c)=ab-ac$ )

(r) $5a(2b-5)=10ab-25a$ (using $a(b-c)=ab-ac$ )

(s) $4a(3b+2c)=12ab+8ac$ (using $a(b+c)=ab+ac$ )

(t) $5p(4p-5q)=20p^2-25pq$ (using $a(b-c)=ab-ac$ )

6 0

3 years ago

If f(x) is a fifth degree polynomial, is it possible that it has exactly 5 complex roots?

Zolol [24]

No.

A fifth degree polynomial, having a graph that increases and starts from below x-axis.

Therefore, no matter what equation it is. The fifth degree polynomial will intercept x-axis AT LEAST one.

The fifth degree polynomial can have only at maximum, 4 complex roots.

<em>You can try drawing or seeing the graph of fifth-degree polynomial function. No matter what equations, they still intercept at least one x-value.</em>

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4 0

3 years ago