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2 years ago

The length of a rectangle is 6 times its width. If the perimeter of the rectangle is 98in , find its area.

Mathematics

1 answer:

blagie [28]2 years ago

4 0

Answer:

294in²

Step-by-step explanation:

Let the width be x

Perimeter = 2(length+width)

98 = 2(6x + x)

49 = 6x + x

49 = 7x

7 = x

The width is 7 and the length is 7 x 6 = 42

Area = length × width

42 × 7 = 294in²

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A line crosses the coordinates (-3,5) and (4,-2). What is the slope of the line?

ICE Princess25 [194]

Answer:

Slope = -1

Step-by-step explanation:

$\frac{-2-5}{4-(-3)}=\frac{-7}{7} =-1$

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3 years ago

Each drink cost d in dollars

zhenek [66]

There is no picture what do you mean

6 0

2 years ago

A driving exam consists of 30 multiple-choice questions. Each of the 30 answers is either right or wrong. Suppose the probabili

Butoxors [25]

Answer:

a. P(x>20)=0.19

b. P(x≥6)=0.72

c. P(x≤20)=0.81

d. A and C

Step-by-step explanation:

We know that:

1) the probability that a student makes fewer than 6 mistakes is 0.28

$P(x$

2) The probaiblity that a student makes between 6 to 20 mistakes is 0.53.

$P(6\leq x\leq20)=0.53$

We will express the proabilibities in function of the information we have.

a. Probability that a student makes more than 20 mistakes.

$P(x>20)=1-P(x\leq20)=1-(P(x20)=1-(0.28+0.53)=1-0.81=0.19$

b. Probability that the student make 6 or more mistakes

$P(x\geq6)=1-P(Px$

c. Probability that a student makes 20 mistakes at most

$P(x\leq20)=P(x$

d. A and C, because A takes a event of more than 20 mistakes and C takes the event of 20 or less mistakes. Both events cover a probability of 1.

7 0

2 years ago

Can someone plz help me with this!

asambeis [7]

Answer:

(arranged from top to bottom)

System #3, where x=6

System #1, where x=4

System #7, where x=3

System #5, where x=2

System #2, where x=1

Step-by-step explanation:

System #1: x=4

$2x+y=10\\x-3y=-2$

To solve, start by isolating your first equation for y.

$2x+y=10\\y=-2x+10$

Now, plug this value of y into your second equation.

$x-3(-2x+10)=-2\\x+6x-30=-2\\7x=28\\x=4$

System #2: x=1

$x+2y=5\\2x+y=4$

Isolate your second equation for y.

$2x+y=4\\y=-2x+4$

Plug this value of y into your first equation.

$x+2(-2x+4)=5\\x+(-4x)+8=5\\x-4x+8=5\\-3x=-3\\x=1$

System #3: x=6

$5x+y=33\\x=18-4y$

Isolate your first equation for y.

$5x+y=33\\y=-5x+33$

Plug this value of y into your second equation.

$x=18-4(-5x+33)\\x=18+20x-132\\-19x=-114\\x=6$

System #4: all real numbers (not included in your diagram)

$y=13-2x\\8x+4y=52$

Plug your value of y into your second equation.

$8x+4(13-2x)=52\\8x+52-8x=52\\0=0$

<em>all real numbers are solutions</em>

System #5: x=2

$x+3y=5\\6x-y=11$

Isolate your second equation for y.

$6x-y=11\\-y=-6x+11\\y=6x-11$

Plug in your value of y to your first equation.

$x+3(6x-11)=5\\x+18x-33=5\\19x=38\\x=2$

System #6: no solution (not included in your diagram)

$2x+y=10\\-6x-3y=-2$

Isolate your first equation for y.

$2x+y=10\\y=-2x+10$

Plug your value of y into your second equation.

$-6x-3(-2x+10)=-2\\-6x+6x-30=-2\\-30=-2$

<em>no solution</em>

System #7: x=3

$y=10+x\\2x+3y=45$

Plug your value of y into your second equation.

$2x+3(10+x)=45\\2x+30+3x=45\\5x=15\\x=3$

3 0

2 years ago

4. The reflecting dish of a parabolic microphone has a cross-section in the shape of a parabola. The microphone itself is placed

elixir [45]

Assume the parabola is placed on a graph where the x-axis is the top of the dish.
The vertex is then at (0,-30) The x-intercepts or zeros are at (-30,0) and (30,0)

The equation of such parabola would be:
$y = a(x+30)(x-30)$
Plug in vertex to find value of 'a'
$-30 = a(0+30)(0-30) \\ \\ a = \frac{-30}{(-30)(30)} = \frac{1}{30}$

Now find the focus given that $p = \frac{1}{4a}$
$p = \frac{1}{4(1/30)} = \frac{30}{4} = 7.5$

Answer: the microphone should be placed 7.5 inches from vertex.

8 0

3 years ago