The graph represents the piecewise function: f(x) = What is the domain and range of the function? Domain: Range:

A music competition on television had five elimination rounds. After each elimination, only half of the contestants were sent to

ikadub [295]

Its c, average rate, im sure of it

6 0

1 year ago

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A rectangular portrait measures 50cm by 70cm. It is surrounded by a rectangular frame of uniform width. If the area of the frame

snow_tiger [21]

Let us assume uniform width = x cm wide.

Length of rectangular portrait = 50cm and width of rectangular portrait = 70cm.

Therefore, length of rectangle made by frame = 50 + x+x = (2x+50) cm.

And width of rectangle made by frame = 70+x+x = (2x+70) cm.

We know, the area of rectangular portrait = 50 × 70 = 3500 cm^2.

Total area of the rectangle made by frame would be = (2x+50) * (2x+70)

We know,

Actual area of frame = Area of rectangle made by frame - area of rectangular portrait.

We also given "the area of the frame is the same as the area of the portrait."

We can setup an equation now,

3500 = (2x+50) * (2x+70) - 3500.

Subtracting 3500 from both sides, we get

3500-3500 = (2x+50) * (2x+70) - 3500-3500.

0 = (2x+50) * (2x+70) -7000.

FOIL (2x+50) * (2x+70), we get

0 = 2x*2x +2x*70 + 50*2x +50*70 - 7000.

0 = 4x^2 +140x +100x +3500 -7000.

4x^2 +240x -3500 = 0.

Dividing whole equation by 4, we get

x^2 +60x - 875 =0

Applying quadratic formula $=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ , we get

$=\frac{-60\pm \sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}$

$x=\frac{-60+\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}=5\left(\sqrt{71}-6\right)$

$x=\frac{-60-\sqrt{60^2-4\cdot \:1\left(-875\right)}}{2\cdot \:1}:\quad -5\left(6+\sqrt{71}\right)$

We cant take negative value.

So, $x=5\left(\sqrt{71}-6\right)=12.13$

We could take approximately 12 cm.

7 0

2 years ago

20 POINTS!!!! PLZ HELP IN A HURRY!!!!!!

MakcuM [25]

Answer:

Volume is how much you can put in something like a box. Surface area is how much wrapping paper is needed to cover the box. Words like fill often designated volume while words like cover mean surface area.

Step-by-step explanation:

Volume is how much you can put in something like a box. Surface area is how much wrapping paper is needed to cover the box. Words like fill often designated volume while words like cover mean surface area.

8 0

1 year ago

Read 2 more answers

t Technodynamics, Inc., a randomly-selected hiring committee of 3 people is formed from a group of 4 employees in marketing an

atroni [7]

Answer: a) 0.25, b) 0.78, c) 0.71

Step-by-step explanation:

Since we have given that

Number of employees in marketing = 4

Number of employees in management = 7

We need to hire committee of 3 people.

a) Find the probability that the committee has exactly 2 employees from marketing.

So, Probability becomes

$\dfrac{^4C_2\times ^7C_1}{^{11}C_3}\\\\=\dfrac{42}{165}\\\\=0.25$

b) Find the probability that the committee has at least one employee from marketing.

$P(x\geq 1)=1-P(x=0)\\\\P(x\geq 1)=1-\dfrac{^7C_3}{^{11}C_3}=1-0.212=0.78$

c) Find the probability that the committee has at most one employee from management.

$P(x\leq 1)=P(x=0)+P(x=1)\\\\P(x\leq 1)=\dfrac{^7C_3}{^{11}C_3}+\dfrac{^7C_2\times ^4C_1}{^{11}C_3}\\\\P(x\leq 1)=0.21+0.50=0.71$

Hence, a) 0.25, b) 0.78, c) 0.71

8 0

1 year ago

Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.

xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form $p(t)=at^2$ , where a is in ℝ.

2.Let W2 be the set of all polynomials of the form $p(t)=t^2+a$ , where a is in ℝ.

3. Let W3 be the set of all polynomials of the form $p(t)=at^2+at$ , where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

$p_1 (t) = at^2, p_2(t)=bt^2$ .