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3 years ago

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A rectangular garden has length twice as great as its width. A second rectangular garden has the same length as the first garden

and width that is 4 meters greater than the width of the first garden. The second garden has area of 120 square meters. What is the length of the two gardens?

1 answer:

Slav-nsk [51]3 years ago

4 0

I say 12 because that would mean the first rectangle is 6x12 and the second is 10x12

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let x1,x2, and x3 be linearly independent vectors in R^(n) and let y1=x2+x1; y2=x3+x2; y3=x3+x1. are y1,y2,and y3 linearly indep

Nutka1998 [239]

Answer with Step-by-step explanation:

We are given that

$x_1,x_2$ and $x_3$ are linearly independent.

By definition of linear independent there exits three scalar $a_1,a_2$ and $a_3$ such that

$a_1x_1+a_2x_2+a_3x_3=0$

Where $a_1=a_2=a_3=0$

$y_1=x_2+x_1,y_2=x_3+x_2,y_3=x_3+x_1$

We have to prove that $y_1,y_2$ and $y_3$ are linearly independent.

Let $b_1,b_2$ and $b_3$ such that

$b_1y_1+b_2y_2+b_3y_3=0$

$b_1(x_2+x_1)+b_2(x_3+x_2)+b_3(x_3+x_1)=0$

$b_1x_2+b_1x_1+b_2x_3+b_2x_2+b_3x_3+b_3x_1=0$

$(b_1+b_3)x_1+(b_2+b_1)x_2+(b_2+b_3)x_3=0$

$b_1+b_3=0$

$b_1=-b_3$ ...(1)

$b_1+b_2=0$

$b_1=-b_2$ ..(2)

$b_2+b_3=0$

$b_2=-b_3$ ..(3)

Because $x_1,x_2$ and $x_3$ are linearly independent.

From equation (1) and (3)

$b_1=b_2$ ...(4)

Adding equation (2) and (4)

$2b_1==0$

$b_1=0$

From equation (1) and (2)

$b_3=0,b_2=0,b_3=0$

Hence, $y_1,y_2$ and $y_3$ area linearly independent.

5 0

3 years ago

Find the height of the building below by using your knowledge of trigonometry.

Lilit [14]

Answer:

50 $\sqrt{3}$

Step-by-step explanation:

The figure represents a 30-60-90 special triangle

in this special right triangle the measure of sides lengths is represented by

a, a $\sqrt{3}$ , and 2a

The side length that sees angle measure 30 is represented by a and here we see a = 50m

The height of the building, which sees angle measure 60, is equal to

50 $\sqrt{3}$

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2 years ago

Help, please? Pleeeeeeeaaaaaaasssssssseeeeeee??

Lilit [14]

Answer:

Graph A passes THrough the Center making it proportional when graph B is not passing

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

After a college football team once again lost a game to their archrival, the alumni association conducted a survey to see if alu

valentina_108 [34]

Answer:

P-value for this hypothesis test is 0.00175.

Step-by-step explanation:

We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.

A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.

<u><em>Let p = proportion of all living alumni who favored firing the coach</em></u>

SO, Null Hypothesis, $H_0$ : p = 0.50 {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, $H_A$ : p > 0.50 {means that the majority of alumni are in favor of firing the coach}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of the alumni in the sample who were in favor of firing the coach = $\frac{64}{100}$ = 0.64

n = sample of alumni = 100

So, <em><u>test statistics</u></em> = $\frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }$

= 2.92

<u>Now, P-value of the hypothesis test is given by ;</u>

P-value = P(Z > 2.92) = 1 - P(Z $\leq$ 2.92)

= 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

4 0

3 years ago

Rollie was succesful in losing weight he had a goal weight in mind he went on a diet for three months each month he would lose o

Llana [10]

Let's say that in the beginning he weighted x and at the end he weighted x-y, y being the number of kg he wanted to loose.

first month he lost
y/3

then he lost:
(y-y/3)/3
this is
(2/3y)/3=2/9y
explanation: ((y-y/3) is what he still needed to loose: y minus what he lost already

and then he lost
(y-2/9y-1/3y)/3+3 (the +3 is his additional 3 pounts)
(y-2/9y-1/3y)/3-3=(7/9y-3/9y)/3+3=4/27y+3

it's not just y/3 because each month he lost one third of what the needed to loose at the current time, not in totatl

and the weight at the end of the 3 months was still x-y+3, 3 pounds over his goal weight!

so: x -y/3-2/9y-4/27y-3=x-y+3

we can subtract x from both sides:
-y/3-2/9y-4/27y-3=-y+3
add everything up:

-19/27y=-y+6

which means
-19/27y=-y+6

y-6=19/27y

8/27y=6
4/27y=3
y=20.25

so... that's how much he wanted to loose, but he lost 3 less than that, so 23.25

ps. i hope I didn't make a mistake in counting, let me know if i did. In any case you know HOW to solve it now, try to do the calculations yourself to see if they're correct!

5 0

3 years ago