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3 years ago

Find AM in the parallelogram

Mathematics

1 answer:

n200080 [17]3 years ago

5 0

Answer:

AM = 6

Step-by-step explanation:

Using the property of a parallelogram

• The diagonals bisect each other

MO is a diagonal, hence

AM = AO = 6

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Solve the following equation:<br> 1/3(y - 2) -5/6 (y + 1) =3/ 4(0 - 3) - 2

Evgesh-ka [11]

Answer:

11/5

Step-by-step explanation:

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2 years ago

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sergiy2304 [10]

Answer:

Step-by-step explanation:

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2 years ago

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Can anyone help with math models?? Please…?

Solnce55 [7]

Answer:

12*1/4= 3

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1 year ago

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Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95

Virty [35]

Answer:

$L(x)=1+\dfrac{1}{3}x$

$\sqrt[3]{0.95} \approx 0.9833$

$\sqrt[3]{1.1} \approx 1.0333$

Step-by-step explanation:

Given the function: $g(x)=\sqrt[3]{1+x}$

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial, $L(x)=f(a)+f'(a)(x-a)$

a=0

$g(0)=\sqrt[3]{1+0}=1$

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

Therefore:

$L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x$

(b) $\sqrt[3]{0.95}= \sqrt[3]{1-0.05}$

When x = - 0.05

$L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833$

$\sqrt[3]{0.95} \approx 0.9833$

(c)

(b) $\sqrt[3]{1.1}= \sqrt[3]{1+0.1}$

When x = 0.1

$L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333$

$\sqrt[3]{1.1} \approx 1.0333$

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2 years ago

WHOEVER ANSWERS THE FASTEST WINS BRAINIEST ANSWER AND GETS 98 POINTS FOR THIS QUESTION!!!!!! WHO CAN DO IT?

olga nikolaevna [1]

coordinates of b'a'c are

A:(-5,6)

B:(-5,10)

C:(-3,8)

ANswers of ' are

A'(-6,5)

B'(-10,5)

C'(-8,3)

hope this helped

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2 years ago

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