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3 years ago

PPPPPPlzzz helpppppppp

Mathematics

1 answer:

avanturin [10]3 years ago

7 0

the only 2 possible answers are 2 and -2 .But the answer is y=2 because x is positive

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Correlation coefficients can test ______ variable(s) at a time. a. two or more b. only one c. one or more d. only two

Vika [28.1K]

Correlation coeffiencients is used to determine the relationships between data, Correlation coefficients can test two or more variable(s) at a time

<h3 /><h3>What are correlation coefficients?</h3>

This is a linear correlation between sets of data. This coefficient can only be between -1 and 1.

Positive values show a positive correlation while a negative value shows negative correlation. A correlation coefficient of zero shows no relationship between variables.

Based on the explanation, we can conclude that Correlation coefficients can test two or more variable(s) at a time

Learn more on correlation here: brainly.com/question/4219149

4 0

2 years ago

Which is the equation of the graphed line written in standard form? y = x x – 1/2y = 0 x – y = 0 y =1/2x

AysviL [449]

The standard for of this equation is x – y = 0

Firstly, options A and D are not in standard form, so they are not options.

To decided between B and C, we choose a point on the line and see if it works. For example, we'll use (2, 2).

x – y = 0

2 - 2 = 0

0 = 0

This is true for option C. Therefore, it is correct.

5 0

3 years ago

Find the arc length of the given curve between the specified points. x = y^4/16 + 1/2y^2 from (9/16), 1) to (9/8, 2).

lutik1710 [3]

Answer:

The arc length is $\dfrac{21}{16}$

Step-by-step explanation:

Given that,

The given curve between the specified points is

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

The points from $(\dfrac{9}{16},1)$ to $(\dfrac{9}{8},2)$

We need to calculate the value of $\dfrac{dx}{dy}$

Using given equation

$x=\dfrac{y^4}{16}+\dfrac{1}{2y^2}$

On differentiating w.r.to y

$\dfrac{dx}{dy}=\dfrac{d}{dy}(\dfrac{y^2}{16}+\dfrac{1}{2y^2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}\dfrac{d}{dy}(y^4)+\dfrac{1}{2}\dfrac{d}{dy}(y^{-2})$

$\dfrac{dx}{dy}=\dfrac{1}{16}(4y^{3})+\dfrac{1}{2}(-2y^{-3})$

$\dfrac{dx}{dy}=\dfrac{y^3}{4}-y^{-3}$

We need to calculate the arc length

Using formula of arc length

$L=\int_{a}^{b}{\sqrt{1+(\dfrac{dx}{dy})^2}dy}$

Put the value into the formula

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4}-y^{-3})^2}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-2\times\dfrac{y^3}{4}\times y^{-3}}dy}$

$L=\int_{1}^{2}{\sqrt{1+(\dfrac{y^3}{4})^2+(y^{-3})^2-\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4})^2+(y^{-3})^2+\dfrac{1}{2}}dy}$

$L=\int_{1}^{2}{\sqrt{(\dfrac{y^3}{4}+y^{-3})^2}dy}$

$L= \int_{1}^{2}{(\dfrac{y^3}{4}+y^{-3})dy}$

$L=(\dfrac{y^{3+1}}{4\times4}+\dfrac{y^{-3+1}}{-3+1})_{1}^{2}$

$L=(\dfrac{y^4}{16}+\dfrac{y^{-2}}{-2})_{1}^{2}$

Put the limits

$L=(\dfrac{2^4}{16}+\dfrac{2^{-2}}{-2}-\dfrac{1^4}{16}-\dfrac{(1)^{-2}}{-2})$

$L=\dfrac{21}{16}$

Hence, The arc length is $\dfrac{21}{16}$

6 0

3 years ago

My friend sets out walking at a speed of 3 miles per hour. I set out behind her 5 minutes later at 4 miles per hour.

MissTica

3 m/h=0.05m/m
.05•5=0.25 m/h I believe?

7 0

3 years ago

If the common difference of an ap is 3/2 and its 20 th term 35×1/2 find first term and 15 th term

Lady bird [3.3K]

Answer:

Step-by-step explanation:

d = 3/2

a₂₀ = a₁+19d

35/2 = a₁ + 19×3/2

a₁ = 35/2 - 19×3/2 = -11

a₁₅ = a₁+14d = -11 + 14×3/2 = 10

4 0

3 years ago