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3 years ago

HELP ASAP PLEASE

Mathematics

1 answer:

miss Akunina [59]3 years ago

4 0

X > 0 refers to the right half of the coordinate plane.
y < 0 refers to the bottom half of the coordinate plane.

Both these conditions are met in the lower right quadrant of the coordinate plane. Quadrants are numbered counterclockwise from upper right, so the lower right quadrant is number IV.

The name of the quadrant in which (x >0, y < 0) lies is
IV

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Please help! (would really appreciate it)

love history [14]

Divide:

( 8x^4 - 3x^2 + x - 10) by ( x - 1 ) ============================> 8x^5 − 8x^4 − 3x^3 + 4x^2 − 11x +10

Simplify:

(8x^4 − 3x^2 + x − 10)(x − 1)

(8x^4 + −3x^2 + x + −10)(x + −1)

(8x^4)(x) + (8x^4)(−1) +(−3x^2)(x) + (−3x^2)(−1) + (x)(x) + (x)(−1) + (−10)(x) +(−10)(−1)

8x^5 − 8x^4 −3x^3 + 3x^2 + x^2 − x − 10x +10

Hence, Your Answer, =====> 8x^5 − 8x^4 − 3x^3 + 4x^2 − 11x + 10

Hope that helps!!!! : )

6 0

3 years ago

Order from greatest to least: 2 6/8, 1 7/12, 2 5/6

anastassius [24]

2 5/6, 2 6/8 1 7/12 is the way you would order it.

6 0

4 years ago

Read 2 more answers

Find the third order maclaurin polynomial. Use it to estimate the value of sqrt1.3

vodka [1.7K]

$\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$ is the maclaurin polynomial and estimate value of $\sqrt{1.3}$ is 1.14. This can be obtained by using the formula to find the maclaurin polynomial.

<h3>Find the third order maclaurin polynomial:</h3>

Given the polynomial,

$f(x)=\sqrt{1+3x}=(1+3x)^{\frac{1}{2} }$

The formula to find the maclaurin polynomial,

$f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}$

Next we have to find f'(x), f''(x) and f'''(x),

$f'(x) = \frac{3}{2}(1+3x)^{-\frac{1}{2} }$

$f''(x) =-\frac{9}{4}(1+3x)^{-\frac{3}{2} }$

$f'''(x) = \frac{81}{8}(1+3x)^{-\frac{5}{2} }$

By putting x = 0 , we get,

$f(0)=(1+3(0))^{\frac{1}{2} }=1$

$f'(0) = \frac{3}{2}(1+3(0))^{-\frac{1}{2} }=\frac{3}{2}$

$f''(0) =-\frac{9}{4}(1+3(0))^{-\frac{3}{2} }=-\frac{9}{4}$

$f'''(0) = \frac{81}{8}(1+3(0))^{-\frac{5}{2} }=\frac{81}{8}$

Therefore the maclaurin polynomial by using the formula will be,

$\sqrt{1+3x}=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}$

$\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$

To find the value of $\sqrt{1.3}$ we can use the maclaurin polynomial,

$\sqrt{1.3}$ is $\sqrt{1+3x}$ with x = 1/10,

$\sqrt{1+3(1/10)}=1+\frac{3}{2} (1/10)-\frac{9}{8} (1/10)^{2} + \frac{81}{8}(1/10)^{3}$

$\sqrt{1+3(1/10)}=\frac{18247}{16000} = 1.14$

Hence $\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$ is the maclaurin polynomial and estimate value of $\sqrt{1.3}$ is 1.14.

Learn more about maclaurin polynomial here:

brainly.com/question/24188694

#SPJ1

6 0

2 years ago

Read 2 more answers

Iteru [2.4K]

Think it’s option 1.3

6 0

3 years ago

Is this a function please help I’m failing

igomit [66]

Answer:

Yes

Step-by-step explanation:

The relation is a function. For a relation to be a function there must be a unique x value for each y value. So this means x's can not repeat, and in this relation, the x-values never repeat. Therefore this is a function.

7 0

3 years ago