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Svetradugi [14.3K]
3 years ago
14

which is steeper a hill that rises 2 feet for every 10 feet of run, or a hill that rises 2 feet for every 15 feet of run? Explai

n
Mathematics
2 answers:
spayn [35]3 years ago
4 0
In order to figure this out you need to use slope intercept for for each of these hills. Let's say the first hill is HA and the second hill is B

slope intercept: y = mx + b

HA: y = (2/10)x
HB: y = (2/15)x

As you can see when you plot this on a graph, HB is steeper because the line rises more sharply.
LuckyWell [14K]3 years ago
4 0

Answer: Ok, the first hill rises 2 feet for every 10 feet of run.

So if both hills start in the ground where y = 0, we can write the height as a function of x in the next way: y = (2/10)*X so for every 10 feet you advance in x, you rise 2 feet in Y, where here 2/10 is the slope.

The second hill is described by y = (2/15)*X, so for every 15 feet in x, you rise 2 feet in y, and the slope is 2/15

Is easy to see that 2/10 > 2/15, so the slope of the first hill is bigger than the slope of the second hill, this means that the first hill is steeper than the second.

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2.1(2.3 + 2.1x) = 11.65 + x. Step by step equation please help
DedPeter [7]

Answer:

x=2

Step-by-step explanation:

2.1(2.3 + 2.1x) = 11.65 + x

Distribute

4.83+4.41x=11.65+x

bring 4.83 over (-)

4.41x=6.82+x

bring x over

4.41x-1(x)=6.82

4.41-1=3.41 so

3.41x=6.82

divide 3.41

x=2

3 0
3 years ago
What is the domain of the function y = RootIndex 3 StartRoot x minus 1 EndRoot?
finlep [7]

By definition of cubic roots and power properties, we conclude that the domain of the cubic root function is the set of all real numbers.

<h3>What is the domain of the function?</h3>

The domain of the function is the set of all values of x such that the function exists.

In this problem we find a cubic root function, whose domain comprise the set of all real numbers based on the properties of power with negative bases, which shows that a power up to an odd exponent always brings out a negative result.

<h3>Remark</h3>

The statement is poorly formatted. Correct form is shown below:

<em>¿What is the domain of the function </em>y = \sqrt[3]{x - 1}<em>?</em>

<em />

To learn more on domain and range of functions: brainly.com/question/28135761

#SPJ1

4 0
1 year ago
A kitchen sink faucet streams 0.5 gallon of water in 10 seconds. A bathroom sink faucet stream 0.75 gallon of water in 18 second
Masteriza [31]

Answer:

kitchen sink faucet stream

Step-by-step explanation:

with kitchen sink faucet you fill 0.05gallons of water per second you get that by dividing 0.5 by 10

and the bathroom sink faucet does 0.041gallons of water per second you get that by dividing 0.75 by 18

7 0
3 years ago
The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each
Fynjy0 [20]

Answer:

\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

Step-by-step explanation:

Given equation of ellipsoids,

u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}

The vector normal to the given equation of ellipsoid will be given by

\vec{n}\ =\textrm{gradient of u}

            =\bigtriangledown u

           

=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})

           

=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}

           

=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}

Hence, the unit normal vector can be given by,

\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}

             =\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}

             

=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

Hence, the unit vector normal to each point of the given ellipsoid surface is

\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}

3 0
3 years ago
Consider the hypotheses below. Upper H 0​: mu equals 50 Upper H 1​: mu not equals 50 Given that x overbar equals 60​, s equals 1
Neko [114]

Answer:

We conclude that  \mu\neq 50.

Step-by-step explanation:

We are given that x bar = 60​, s = 12​, n = 30​, and alpha = 0.10

Also, Null Hypothesis, H_0 : \mu = 50

Alternate Hypothesis, H_1 : \mu \neq 50

The test statistics we will use here is;

                    T.S. = \frac{Xbar-\mu}{\frac{s}{\sqrt{n} } } ~  t_n_-_1

where, X bar = sample mean = 60

              s = sample standard deviation = 12

              n = sample size = 30

So, Test statistics = \frac{60-50}{\frac{12}{\sqrt{30} } } ~ t_2_9

                             = 4.564

At 10% significance level, t table gives critical value of 1.311 at 29 degree of freedom. Since our test statistics is more than the critical value as 4.564 > 1.311 so we have sufficient evidence to reject null hypothesis as our test statistics will fall in the rejection region.

P-value is given by, P(t_2_9 > 4.564) = less than 0.05% {using t table}

Here also, P-value is less than the significance level as 0.05% < 10% , so we will reject null hypothesis.

Therefore, we conclude that \mu\neq 50.

7 0
3 years ago
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