Select all the correct answers,<br> Which relations represent functions?

an equilateral triangle and a regular pentagon have the same perimeter. Each side of the pentagon is 3 inches shorter than each

klemol [59]

The answer is 4.5 for the side length of the pentagon and 7.5 is the side length for the triangle because 7.5*3(3 sides) = 22.5 and 4.5*5(5 sides)= 22.5. The perimeter is 22.5

6 0

3 years ago

Evaluate z + 2+ z for x = 2, y=-3, z = -4.

Artyom0805 [142]

Answer:

-6

Step-by-step explanation:

Are you sure that z + 2+ z is correct? x and y do not appear here.

z + 2+ z simplifies to 2z + 2, and so, if z = -4, z + 2+ z has the value

2(-4) + 2, or -6.

7 0

3 years ago

The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p

Juliette [100K]

Check the picture below.

based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,

$\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)$

$\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}$

now, we can plug those values on A = (1/2)bh,

$\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5$

7 0

3 years ago

Can you guys help we with this question.

satela [25.4K]

Answer:

40

Step-by-step explanation:

(2x+1/(2x))^5 *(2x -1/(2x))^5

= ((2x)^2 -1/(2x)^2)^5 (a+b)*(a-b) =a2-b2

= (4x^2-1/4(x)^2)^5

now

x =4x^2. ,a = 1/4(x)^2 ,n =5

we have

general term = Cr *x^r *a^(n-r)

= Cr * (4x^2)^r * (1/4(x)^2)^(n-r)

= Cr *4^r * X^2r * 1/( 4^(n-r) *x^(2n-2r)

= Cr * 4^r/4^(n-r) * x^(2r)/x^(2n-2r)

= Cr * 4(2r-n) *x(4r-2n)

now for x^2

4r-2n = 2

4r -10=2

4r =12

r = 3

now for coeff

C(5,3) * 4^(2*3-5)

5!/(3!*(5-3)!) * 4

5*4/(2*1)*4

40

4 0

3 years ago

Read 2 more answers

Can anyone help me out please!!!

Yuliya22 [10]

Hope this helps you!!!!

5 0

4 years ago

Select all the correct answers, Which relations represent functions?