Apply division by 10 when one tenth of a number is required and apply multiplication by 10 when 10 times of a number is required.
<u>Solution:</u>
Need to determine what operation is required to get one-tenth of a number and 10 times of a number
To get one tenth of a number, divide the number by 10.
For example to get one – tenth of 100, divide it by 10, we get 10 as a result.
To get ten times of a number, multiply the number by 10
For example 10 times of 10 = 10 x 10 = 100
Hence apply division by 10 when one tenth of a number is required and apply multiplication by 10 when 10 times of a number is required.
The remainder theorem says that dividing a polynomial <em>f(x)</em> by a 1st-degree polynomial <em>g(x)</em> = <em>x</em> - <em>c</em> leaves a remainder of exactly <em>f(c)</em>.
(a) With <em>f(x)</em> = <em>px</em> ³ + 4<em>x</em> - 10 and <em>d(x)</em> = <em>x</em> + 3, we have a remainder of 5, so
<em>f</em> (-3) = <em>p</em> (-3)³ + 4(-3) - 10 = 5
Solve for <em>p</em> :
-27<em>p</em> - 12 - 10 = 5
-27<em>p</em> = 27
<em>p</em> = -1
(b) With <em>f(x)</em> = <em>x</em> + 3<em>x</em> ² - <em>px</em> + 4 and <em>d(x)</em> = <em>x</em> - 2, we have remainder 8, so
<em>f</em> (2) = 2 + 3(2)² - 2<em>p</em> + 4 = 8
-2<em>p</em> = -10
<em>p</em> = 5
(you should make sure that <em>f(x)</em> was written correctly, it's a bit odd that there are two <em>x</em> terms)
(c) <em>f(x)</em> = 2<em>x</em> ³ - 4<em>x</em> ² + 6<em>x</em> - <em>p</em>, <em>d(x)</em> = <em>x</em> - 2, <em>R</em> = <em>f</em> (2) = 18
<em>f</em> (2) = 2(2)³ - 4(2)² + 6(2) - <em>p</em> = 18
12 - <em>p</em> = 18
<em>p</em> = -6
The others are done in the same fashion. You would find
(d) <em>p</em> = 14
(e) <em>p</em> = -4359
(f) <em>p</em> = 10
(g) <em>p</em> = -13/2 … … assuming you meant <em>f(x)</em> = <em>x</em> ⁴ + <em>x</em> ³ + <em>px</em> ² + <em>x</em> + 20
4+9g I can’t find an exact answer for g unless it’s an equation with an equal sign for example if it’s an equation like 10(.4 + .5g) + 4g = g the exact answer for g would be -1/2
Answer:
26. t<-70 see description below.
Step-by-step explanation:
To solve an inequality in one variable, solve using inverse operations like any other equations. However, if dividing or multiplying by a negative at any points, flip the sign of the inequality.
Example:
-7 was multiplied to as an inverse operation to division by -7. Because it was negative, the inequality sign was flipped. To graph, plot -70 on the number line with an open circle. Then draw an arrow from -70 to the left.
