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3 years ago

Name each term of the polynomial : (–9x^4)+(y^3)–7y+19

Mathematics

1 answer:

zvonat [6]3 years ago

5 0

Answer:

$9x^4, y^3, -7x, and 19$ are the terms of the polynomial.

Step-by-step explanation:

A polynomials is an expression with many terms. Terms are each piece separated by addition or subtraction. They include variables, coefficients and constants.

This polynomial has a quartic term with degree 4, a cubic term with degree 3, a linear term with degree 1 and a constant.

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If logb2=x and logb3=y, evaluate the following in terms of x and y:

Alja [10]

$log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x$

Solution:

Given that,

$log_b2 = x\\\\log_b3 = y --------(i)$

Use the following log rules

Rule 1: $log_b(ac) = log_ba + log_bc$

Rule 2: $log_b\frac{a}{c} = log_ba - log_bc$

Rule 3: $log_ba^c = clog_ba$

$a) log_b{162}$

Break 162 down to primes:

$162 = 2^1 \times 3^4$

$log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y$

Thus we get,

$log_b162 = x + 4y$

$b) log_b 324$

Break 324 down to primes:

$324 = 2^2 \times 3^4$

$log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y$

$c) log_b\frac{8}{9}$

By rule 2

$log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} = 3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} = 3x - 2y$

$d) \frac{log_b27}{log_b16}$

By rule 2

$\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} = 3y - 4x$

Thus the given are evaluated in terms of x and y

3 0

2 years ago

How to solve for 6x-12+2x=3+8x-15

Alecsey [184]

6x−12+2x=3+8x−15

Simplify:

6x+−12+2x=3+8x+−15

(6x+2x)+(−12)=(8x)+(3+−15)(Combine Like Terms)

8x+−12=8x+−12

8x−12=8x−12

Subtract 8x from both sides.

8x−12−8x=8x−12−8x

−12=−12

Add 12 to both sides.

−12+12=−12+12

0=0

The real numbers are the only solution we can have.

4 0

3 years ago

Write the standard equation of a circle with center (0,0) and radius 7

lana66690 [7]

Answer:

x^2 + y^2 = 49

Step-by-step explanation:

For center (h,k) and radius r, eqution of the respective circle is

(x-h)^2 + (y-k)^2 = r^2

7 0

3 years ago

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