Answer:Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of
0
.
x
+
1
4
x
2
-
2
x
-
5
Divide the highest order term in the dividend
4
x
2
by the highest order term in divisor
x
.
4
x
x
+
1
4
x
2
-
2
x
-
5
Multiply the new quotient term by the divisor.
4
x
x
+
1
4
x
2
-
2
x
-
5
+
4
x
2
+
4
x
The expression needs to be subtracted from the dividend, so change all the signs in
4
x
2
+
4
x
4
x
x
+
1
4
x
2
-
2
x
-
5
-
4
x
2
-
4
x
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
4
x
x
+
1
4
x
2
-
2
x
-
5
-
4
x
2
-
4
x
-
6
x
Pull the next terms from the original dividend down into the current dividend.
4
x
x
+
1
4
x
2
-
2
x
-
5
-
4
x
2
-
4
x
-
6
x
-
5
Divide the highest order term in the dividend
−
6
x
by the highest order term in divisor
x
.
4
x
-
6
x
+
1
4
x
2
-
2
x
-
5
-
4
x
2
-
4
x
-
6
x
-
5
Multiply the new quotient term by the divisor.
4
x
-
6
x
+
1
4
x
2
-
2
x
-
5
-
4
x
2
-
4
x
-
6
x
-
5
-
6
x
-
6
The expression needs to be subtracted from the dividend, so change all the signs in
−
6
x
−
6
4
x
-
6
x
+
1
4
x
2
-
2
x
-
5
-
4
x
2
-
4
x
-
6
x
-
5
+
6
x
+
6
After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.
4
x
-
6
x
+
1
4
x
2
-
2
x
-
5
-
4
x
2
-
4
x
-
6
x
-
5
+
6
x
+
6
+
1
The final answer is the quotient plus the remainder over the divisor.
4
x
−
6
+
1
x
+
1
Step-by-step explanation:
Lets call those two unknown numbers a, b and write the info in the problem as equations:
a*b = 30
a + b = 40
lets solve for a in the second equation and substitute in the first:
<span>a + b = 40
</span>a = 40 - b
therefore:
<span>a*b = 30
</span>(40 - b)b = 30
40b - b^2 = 30
b^2 - 40b + 30 = 0
if we apply the general quadratic equation to solve we have:
b = (40 +- √(1600 - 120))/2
b = (40 +- √(1480<span>))/2
</span>b = (40 +- 38.47)/2
There are two solutions:
<span>b1 = (40 + 38.47)/2
</span><span>b1 = 39.24
b2 = (40 - 38.47)/2
</span>b2 = 0.765
lets use the second solution <span>b2 = 0.765, and substitute in the first equation to find a:
</span><span>a*b = 30
</span>a*0.765 = 30
a = 30/0.765
a = 39.216
so the numbers are 39.216 and 0.765
Answer:
x=-12
Step-by-step explanation:
To remove the denominator we take LCD
LCD of 3 and 4 is 12
In the second step we multiply the whole equation by 12 not only the fractions
Now the denominator gets removed
1/3 * 12 = 4 , also 3/4 times 12 = 3*3 = 9
4x - 48 = 9x + 12
Subtract 9x on both sides
-5x - 48 = 12
Now add 48 on both sides
-5 x = 60
Divide both sides by -5
so x= -12
The value of x= -12
14n-8 is how you find the answer, and the next term is 62
Given function is
We know
Domain of a function is defined as set of those real values of x for which function is well defined.
So, y is defined when
![\rm\implies \:x \: \in \: ( - \infty , \: - 2] \: \cup \: [2, \: \infty )](https://tex.z-dn.net/?f=%5Crm%5Cimplies%20%5C%3Ax%20%5C%3A%20%20%5Cin%20%5C%3A%20%28%20-%20%20%5Cinfty%20%2C%20%20%5C%3A%20-%202%5D%20%5C%3A%20%20%5Ccup%20%5C%3A%20%5B2%2C%20%5C%3A%20%20%5Cinfty%20%29)
Now, <u>To find the range of function </u>
We know,
Range of a function is defined as set of those real values which is obtained by assigning the values to x.
So,
On squaring both sides, we get
Now, x is defined when
But,
<u>Hence, </u>
So, we have
![\rm\implies \:Domain \: of \: function : x \: \in \: ( - \infty , \: - 2] \: \cup \: [2, \: \infty )](https://tex.z-dn.net/?f=%5Crm%5Cimplies%20%5C%3ADomain%20%5C%3A%20of%20%5C%3A%20function%20%3A%20x%20%5C%3A%20%20%5Cin%20%5C%3A%20%28%20-%20%20%5Cinfty%20%2C%20%20%5C%3A%20-%202%5D%20%5C%3A%20%20%5Ccup%20%5C%3A%20%5B2%2C%20%5C%3A%20%20%5Cinfty%20%29)
and
