Jackson travels 2 km north, then 3 km east, and finally 2 km south. Which

The length of a rectangle is 20 units more than its width. The area of the rectangle is x4−100.

Mice21 [21]

Answer:

$1. x^2-10$ because the area expression can be rewritten as $(x^2-10)(x^2+10)$ which equals $(x^2-10)((x^2-10)+20).$

Step-by-step explanation:

Area of the rectangle $=(x^4-100)$

$x^4-100=(x^2)^2-10^2\\$Applying difference of two squares: a^2-b^2=(a-b)(a+b)\\(x^2)^2-10^2=(x^2-10)(x^2+10)$

Since the length of a rectangle is 20 units more than its width.

$Width: x^2-10\\Length=x^2+10=x^2-10+20$

The correct option is therefore 1.

4 0

2 years ago

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HELP FAST PLEASE

kramer

Answer:

y = 14

Step-by-step explanation:

proportional relation: y = kx

only y = 14x has the situation where constant is 14 (k=14)

5 0

2 years ago

1. Walking to improve health. In a study investigating a link between walking and improved health (Social Science & Medicine

lora16 [44]

Answer:

a) We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:

Null hypothesis: $\mu \geq 5.5$

Alternative hypothesis: $\mu < 5.5$

b) For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

c) For this case the significance level is given $\alpha=0.01$ and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

$z_{cric}= -2.326$

And the rejection zone would be: $(\infty ,-2.326)$

And the region is on the figure attached

Step-by-step explanation:

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true average number of days in the past month that adults walked for the purpose of health or recreation is lower than 5.5 days (a;ternative hypothesis) , the system of hypothesis would be:

Null hypothesis: $\mu \geq 5.5$

Alternative hypothesis: $\mu < 5.5$

Part b

For this case since we want to determine if the true mean is lower than an specified value is a one tailed test

Part c

For this case the significance level is given $\alpha=0.01$ and we want to find the critical value, so we need to find a critical value on the normal standard distribution who accumulate 0.01 of the area on the right tail and if we use excel or a table we got:

$z_{cric}= -2.326$

And the rejection zone would be: $(\infty ,-2.326)$

And the region is on the figure attached

3 0

2 years ago

The smaller rectangle was scaled by a factor of 2 to produce the larger figure. use the drop-down menus to show the missing dime

inessss [21]

If the rectangle is scaled by a factor of 2, it means that all of the sides are multiplied by 2. This means the length is changed from 6 to 12, and the width is changed from 2 to 4. The area however, is not 24, but 48, because BOTH the sides doubled.

5 0

2 years ago

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The center of a hyperbola is (−2,4) , and one vertex is (−2,7) . The slope of one of the asymptotes is 1/2 .

maxonik [38]

Answer:

<h2>The equation is $\frac{(y - 4)^{2} }{9 } - \frac{(x + 2)^{2} }{36 } = 1$ .</h2>

Step-by-step explanation:

The equation of a hyperbola is represented by $\frac{(y - k)^{2} }{b^{2} } - \frac{(x - h)^{2} }{a^{2} } = 1$ , where (h, k) is the center of the hyperbola.

As per the given condition, h = -2 and k = 4.

Thus, the equation becomes $\frac{(y - 4)^{2} }{b^{2} } - \frac{(x + 2)^{2} }{a^{2} } = 1$ .

One vertex is (-2, 7).

Hence, putting x = -2 and y = 7, we get $b^{2} = 9$ .

The slope of the asymptote is $\frac{b}{a}$ .

Hence, $\frac{b^{2} }{a^{2} } = \frac{1}{4} \\a^{2} = 36$ .

Thus, the equation is $\frac{(y - 4)^{2} }{9 } - \frac{(x + 2)^{2} }{36 } = 1$ .

7 0

2 years ago