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3 years ago

Can someone help please

Mathematics

2 answers:

bonufazy [111]3 years ago

5 0

It’s negative 2 because it quadrants the exhale

Ilya [14]3 years ago

4 0

Y=1.5x+2 because the rate of change is 1.5 and the y intercept is 2

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Use the area of the rectangle to find the area of the triangle.

tatyana61 [14]

Answer:

Answer is A

Step-by-step explanation:

Area of triangle = $\frac{bxh}{2}$

= $\frac{10x5}{2}$

=25

Area of rectangle = bxh

=10x5

= 50

6 0

2 years ago

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What happens to the value of the expression \dfrac5x+5 x 5 +5start fraction, 5, divided by, x, end fraction, plus, 5 as xxx de

aleksandr82 [10.1K]

Answer:

It increases

Step-by-step explanation:

(5/x) + 5

As x decreases, 5/x increases. So the expression increases.

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3 years ago

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

djyliett [7]

Answer: use the website desmos and plug in the cordinate it gives u the line automaticaly.

Step-by-step explanation:

7 0

2 years ago

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The volume of a shampoo filled into a container is uniformly distributed between 370 and 385 milliliters. (a) What are the mean

Olenka [21]

Answer:

(a) mean = 377.5ml. standard deviation = 4.33

(b) 0.333

Step-by-step explanation:

(a)The mean:

$\mu = \frac{385 + 370}{2} = 377.5 ml$

The standard deviation:

$\sigma = \frac{385 - 370}{\sqrt{12}} = 4.33$

(b)

$P(x < 375) = \frac{375 - 370}{385 - 370} = \frac{1}{3} \approx 0.333$

(c)P(x > m) = 0.95

$\frac{m - 370}{385 - 370} = 0.95$

$m - 370 = 14.25$

$m = 384.25ml$

So the volumn of shampoo should be 384.25ml to exceed 95% of containers.

7 0

3 years ago

Need help with Calculus 1 inverse trig functions

lidiya [134]

Answer:

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}$

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

$\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

$\displaystyle (tan^{-1}u)'=\frac{u'}{1+u^2}$

Where u is a function of x as provided:

$y=3tan^{-1}(x+\sqrt{1+x^2})$

If we set

$u=(x+\sqrt{1+x^2})$

Then

$\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}$

$\displaystyle u'=1+\frac{x}{\sqrt{1+x^2}}$

Taking the derivative of y

$y'=3[tan^{-1}(x+\sqrt{1+x^2})]'$

Using the change of variables

$\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}$

$\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}$

Operating

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}$

$\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}$

8 0

3 years ago