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3 years ago

Rewrite the expression by factoring out (w-5)

Mathematics

1 answer:

lara [203]3 years ago

5 0

Answer:

(w - 5)(2w² + 7)

Step-by-step explanation:

Given

2w²(w - 5) + 7(w - 5) ← factor out (w - 5) from each term

= (w - 5)(2w² + 7)

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Scott took out a 72 month loan for $35,000 to purchase a new boat. If Scott paid $8,925 in simple interest, what was the interes

diamong [38]

Answer:

Option 4.7% = 3,500 x 4.7% =$164.50 simple annual interest.

82.25 this is what Scott will pay in 6 months at simple interest.

Option 4.2% =3,500 x (1 +0.042/12)^6 =3,500 x 1.0035^6=$3,574.15.

3,500 =$74.15 this is what Scott will pay in 6 months at compounded interest.

The compound option is cheaper by: 74.15 =$8.10.

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2 years ago

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There are 446 highlighters in a carton there are 38 cartons. What is the total amount of highlighters in all Carton. Please show

Kisachek [45]

You would do 38 times 446 which equals 16056

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3 years ago

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Please help me on this question

Alina [70]

Answer:

DE = 13.4 cm (to 1 decimal place)

Step-by-step explanation:

Given: ABCD is a square

BC = AC = 12 cm (opposite sides of a square are congruent)

E is midpoint of BC (given)

BE = EC = 12/2 = 6 cm

CD = AB = 12 cm (opposite sides of a square are congruent)

angle ECD is a right angle (interior angles of a square are 90 deg.)

Consider right triangle ECD

DE = sqrt(EC^2+CD^2) ............. pythagorean theorem

= sqrt(6^2+12^2)

= sqrt ( 36+144 )

= sqrt (180)

= 2 sqrt(45)

= 13.416 (to three dec. places)

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2 years ago

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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

Find the solutions to the equation<br><br> -3y(2y+5)=0

noname [10]

Answer:

The solutions are y=0, -5/2

Step-by-step explanation:

-3y(2y+5)=0

We can use the zero product property

-3y =0 2y+5 =0

y =0 2y = -5

y = -5/2

The solutions are y=0, -5/2

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2 years ago