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3 years ago

I need help please it would mean alot

Mathematics

2 answers:

REY [17]3 years ago

6 0

Answer:

I got 20 for the answer

Andreyy893 years ago

6 0

Answer:

Step-by-step explanation:

I typed this into the calculator however, the steps are listed below.

-6×10(-1)=-0.6

[10(-1)=0.1]

-3×10(-2)=-0.03

[10(-2)=0.01]

-0.6÷-0.03=20

(If you do not know how to divide fractions please google. there are multiple ways.)

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jeka94

Answer:

Step-by-step explanation:

I hope this helps

3 0

3 years ago

The monthly rent of Marilyn's house went from $500 to $440, If p is the

Mila [183]

Answer:

Step-by-step explanation:

5 0

3 years ago

The distance between Eagle Lake and Swallow Lake on a map is 9 cm. The scale of the map is 4 cm = 2 km.

Sedbober [7]

Answer:

4½ km

Step-by-step explanation:

2 cm= 1 km

1 cm= ½ km

4cm×2km = 8cm

8 cm (4 km) + 1 cm (½ km) = 9 cm

9 cm/ 4 ½ km

4 0

3 years ago

Solve attachment 30 points!!!!!

Effectus [21]

Answer:

$A=\frac{B^3}{5g+m}$

Step-by-step explanation:

Step 1: Write out equation

5gA + mA = B³

Step 2: Factor out A

A(5g + m) = B³

Step 3: Divide both sides by 5g + m

$A=\frac{B^3}{5g+m}$

4 0

3 years ago

1. Approximate the given quantity using a Taylor polynomial with n3.

Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

$f(x) = x^{1/4}$

The n-th order Taylor polynomial for function f with its center at a is:

$p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}$

As n = 3 So,

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}$

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}$

$p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}$

$p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}$

$p_{3} (x)$ = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6

(0.0000018522752) (x-81)³

$p_{3} (x)$ = 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

(x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

$p_{3} (94)$ = 0.0092592593 (94) - 0.000042866941 (94 - 81)² +

0.00000030871254 (94-81)³ + 2.25

= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +

2.25

= 0.87037 03742 - 0.000042866941 (169) +

0.00000030871254(2197) + 2.25

= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

$p_{3} (94)$ = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute $\sqrt[4]{94}$ as $94^{1/4}$ using calculator

Exact value:

$E_{a}$ (94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94) = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94) = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94) = 0.0001

4 0

3 years ago