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4 years ago

In standard form the y intercept is

Mathematics

1 answer:

labwork [276]4 years ago

4 0

Answer:

the y intercept is the c term in standard form.

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What is 766 divided by 21 with fractions

Olin [163]

36 and 12/25 hope this helps

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3 years ago

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If sin0=2/3 find the values of the other five trigonometric functions of 0.

mamaluj [8]

Answer:

See image

Step-by-step explanation:

It helps a lot to draw a triangle. Sin is OPPOSITE/ HYPOTENUSE. You can use Pythagorean thm to find the third side. Then use cos = ADJ/HYP

tan = OPP/HYP

Then, find reciprocal functions. Reciprocal just means you are flipping the ratios over.

csc = HYP/OPP

sec = HYP/ADJ

tan = ADJ/OPP

The only tricky remaining part is that you cant leave a radical (the square root symbol) on the bottom of a ratio. So you have to fix those. See image.

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3 years ago

What is the area of a rectangle that is 3.1 cm wide and 4.4 cm long? Enter the full-precision answer first to see the correspond

Brrunno [24]

Answer: Without rounding : 13.64 (square cm)

Rounded answer : 13.6(square cm)

Step-by-step explanation:

We know that the area of a rectangle is given by :-

$A=l\times w$ , where l is the length and w is the width of the rectangle.

Given : The dimensions of a rectangle are 3.1 cm wide and 4.4 cm long.

Then, the area of rectangle will be:-

$A=4.4\times 3.1=\dfrac{44}{10}\times\dfrac{31}{10}\\\\=\dfrac{1364}{100}=13.64\text{ cm}^2\approx13.6\ cm^2 \text{Rounded to the nearest tenth}$

Therefore, the area of rectangle = $13.6\text{ cm}^2$

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4 years ago

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snow_lady [41]

Answer: 98

Step-by-step explanation:

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4 years ago

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Find the absolute maximum and absolute minimum values of f on the given interval.

anyanavicka [17]

The question is missing parts. Here is the complete question.

Find the absolute maximum and absolute minimum values of f on the given interval.

$f(x)=xe^{-\frac{x^{2}}{32} }$ , [ -2,8]

Answer: Absolute maximum: f(4) = 2.42;

Absolute minimum: f(-2) = -1.76;

Step-by-step explanation: Some functions have absolute extrema: maxima and/or minima.

Absolute maximum is a point where the function has its greatest possible value.

Absolute minimum is a point where the function has its least possible value.

The method for finding absolute extrema points is

1) Derivate the function;

2) Find the values of x that makes f'(x) = 0;

3) Using the interval boundary values and the x found above, determine the function value of each of those points;

4) The highest value is maximum, while the lowest value is minimum;

For the function given, absolute maximum and minimum points are:

$f(x)=xe^{-\frac{x^{2}}{32} }$

Using the product rule, first derivative will be:

$f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )$

$f'(x)=e^{-\frac{x^{2}}{32} }(1-\frac{x^{2}}{16} )$ = 0

$1-\frac{x^{2}}{16}=0$

$\frac{x^{2}}{16}=1$

$x^{2}=16$

x = ±4

x can't be -4 because it is not in the interval [-2,8].

$f(-2)=-2e^{-\frac{(-2)^{2}}{32} }=-1.76$

$f(4)=4e^{-\frac{4^{2}}{32} }=2.42$

$f(8)=8e^{-\frac{8^{2}}{32} }=1.08$

Analysing each f(x), we noted when x = -2, f(-2) is minimum and when x = 4, f(4) is maximum.

Therefore, absolute maximum is f(4) = 2.42 and

absolute minimum is f(-2) = -1.76

8 0

3 years ago