Respuesta :
1. We have the formula for the volume of sphere: [tex]V= \frac{4 \pi }{3}r^3 [/tex]
where
[tex]V[/tex] is the volume
[tex]r[/tex] is the radius
We know from our problem that the volume of our spherical balloon is [tex]100in^3[/tex], so [tex]V=100in^3[/tex]. Lets replace that value in our formula and solve for [tex]r[/tex]:
[tex]100in^3= \frac{4 \pi }{3}r^3 [/tex]
[tex]3(100in^3)=4 \pi r^3[/tex]
[tex]300in^3=4 \pi r^3[/tex]
[tex]r^3= \frac{300in^3}{4 \pi } [/tex]
[tex]r= \sqrt[3]{\frac{300in^3}{4 \pi } } [/tex]
[tex]r=2.879in[/tex]
We can conclude that the radius of our spherical balloon is approximately 3 inches long. Therefore, the correct answer is A. 3 in.
2. Lets simplify each one of the expressions first:
F. [tex] \sqrt{x^2} [/tex] since the radicand is elevated to the same number as the index of the radical, we can cancel the radical and the exponent:
[tex] \sqrt{x^2} =x[/tex]
Since [tex]x\ \textgreater \ 0[/tex], this expression is equal to [tex]x[/tex].
G. [tex] \frac{1}{2} \sqrt[3]{8x^3} [/tex] 8 can be expressed as [tex]8=2*2*2=2^3[/tex], so we can rewrite our radicand:
[tex]\frac{1}{2} \sqrt[3]{8x^3} =\frac{1}{2} \sqrt[3]{2^3x^3} = \frac{1}{2} \sqrt[3]{(2x)^3} [/tex]
Since the radicand is elevated to the same number as the index of the radical, we can cancel the radical and the exponent:
[tex]\frac{1}{2} \sqrt[3]{(2x)^3}= \frac{1}{2} (2x)[/tex]
Now, we can cancel the 2 in the denominator with the one in the numerator:
[tex] \frac{1}{2} (2x)=x[/tex]
Since [tex]x\ \textgreater \ 0[/tex], this expression is equal to [tex]x[/tex].
H. [tex] \sqrt[3]{-x^3} [/tex] The radicand is elevated to the same number as the index of the radical, so we can cancel the radical and the exponent:
[tex]\sqrt[3]{-x^3}=-x[/tex]
Since [tex]x\ \textgreater \ 0[/tex], this expression is NOT equal to [tex]x[/tex].
We can conclude that the correct answer is H. [tex] \sqrt[3]{-x^3} [/tex].
3. The fourth root of [tex]- \frac{16}{81} [/tex] is [tex] \sqrt[4]{ -\frac{16}{81} } [/tex]. Remember that negative numbers don't have real even roots since a number raised to an even exponent is either positive or 0. Since 4 is even and [tex]- \frac{16}{81} [/tex] is negative, we can conclude that [tex]- \frac{16}{81} [/tex] has not a real fourth root.
The correct answer is D. no real root found.
4. This time we are going to take a different approach. We are going to simplify [tex] \sqrt{a^2(x+a^2)} [/tex] first, and then, we are going to compare the result with our given options:
[tex] \sqrt{a^2(x+a^2)} [/tex]
Lets apply the product rule for a radical [tex] \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b} [/tex]:
[tex]\sqrt{a^2(x+a^2)} = \sqrt{a^2} \sqrt{x+a^2} [/tex]
Notice that in our first product the radicand is raised to the same number as the index, so we can cancel the radical and the exponent:
[tex]\sqrt{a^2} \sqrt{x+a^2}=a \sqrt{x+a^2}[/tex]
We can conclude that the correct answer is F. [tex]a \sqrt{x+a^2} [/tex]
5. Just like before, we are going to simplify [tex] \sqrt{4x^2y^4} [/tex] firts, and then, we are going to compare the result with our given options. To simplify our radical expression we are going to use some laws of radicals:
[tex] \sqrt{4x^2y^4} [/tex]
Applying product rule for a radical [tex] \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b} [/tex]:
[tex]\sqrt{4x^2y^4} = \sqrt{4} \sqrt{x^2} \sqrt{y^4} [/tex]
Notice that [tex] \sqrt{4} =2[/tex], so:
[tex]\sqrt{4} \sqrt{x^2} \sqrt{y^4} =2 \sqrt{x^2} \sqrt{y^4} [/tex]
Notice that [tex]y^4=(y^2)^2[/tex], so we can rewrite our expression:
[tex]2 \sqrt{x^2} \sqrt{y^4}=2 \sqrt{x^2} \sqrt{(y^2)^2} [/tex]
Applying the radical rule [tex] \sqrt[n]{a^n} =a[/tex]:
[tex]2 \sqrt{x^2} \sqrt{(y^2)^2} =2xy^2[/tex]
We can conclude that the correct answer is A. [tex]2xy^2[/tex].
5. Lets check our statements:
F. [tex]x[/tex] is always greater than [tex] \sqrt{x} [/tex].
If [tex]x=0[/tex], [tex] \sqrt{x} =0[/tex]. Therefore, this statement is FALSE.
G. [tex]x= \sqrt{x} [/tex] only if [tex]x=0[/tex]. For a number bigger than zero the square root of that number will be always less than the number; therefore, this statement is TRUE.
H. Using a calculator we can check that [tex] \sqrt{ \frac{1}{2} } =0.7071[/tex] and [tex] \frac{1}{2} =0.5[/tex]. Since [tex]0.7071\ \textgreater \ 0.5[/tex], we can conclude that [tex] \sqrt{ \frac{1}{2} } \ \textgreater \ \frac{1}{2} [/tex].
We can conclude that this statement is TRUE.
J. [tex]x^4[/tex] is always bigger than [tex]x^2[/tex].
If [tex]x= \frac{a}{b} [/tex] and [tex]b\ \textgreater \ a[/tex], [tex]x^4[/tex] will always be smaller than [tex]x^2[/tex]. Therefore, we can conclude that this statement is FALSE.
6. We know that the formula for the volume of cube is [tex]V=s^3[/tex]
where
[tex]V[/tex] is the volume
[tex]s[/tex] is the side
We know for our problem that a cubical storage bin has a volume of 5832 cubic inches, so [tex]V=5832in^3[/tex]. Lets replace that value in our formula and solve for [tex]s[/tex]:
[tex]5832in^3=s^3[/tex]
[tex]s^3=5832in^3[/tex]
[tex]s= \sqrt[3]{5832in^3} [/tex]
[tex]s=18in[/tex]
We can conclude that the length of the side of the cubical storage bin is 18 inches.
where
[tex]V[/tex] is the volume
[tex]r[/tex] is the radius
We know from our problem that the volume of our spherical balloon is [tex]100in^3[/tex], so [tex]V=100in^3[/tex]. Lets replace that value in our formula and solve for [tex]r[/tex]:
[tex]100in^3= \frac{4 \pi }{3}r^3 [/tex]
[tex]3(100in^3)=4 \pi r^3[/tex]
[tex]300in^3=4 \pi r^3[/tex]
[tex]r^3= \frac{300in^3}{4 \pi } [/tex]
[tex]r= \sqrt[3]{\frac{300in^3}{4 \pi } } [/tex]
[tex]r=2.879in[/tex]
We can conclude that the radius of our spherical balloon is approximately 3 inches long. Therefore, the correct answer is A. 3 in.
2. Lets simplify each one of the expressions first:
F. [tex] \sqrt{x^2} [/tex] since the radicand is elevated to the same number as the index of the radical, we can cancel the radical and the exponent:
[tex] \sqrt{x^2} =x[/tex]
Since [tex]x\ \textgreater \ 0[/tex], this expression is equal to [tex]x[/tex].
G. [tex] \frac{1}{2} \sqrt[3]{8x^3} [/tex] 8 can be expressed as [tex]8=2*2*2=2^3[/tex], so we can rewrite our radicand:
[tex]\frac{1}{2} \sqrt[3]{8x^3} =\frac{1}{2} \sqrt[3]{2^3x^3} = \frac{1}{2} \sqrt[3]{(2x)^3} [/tex]
Since the radicand is elevated to the same number as the index of the radical, we can cancel the radical and the exponent:
[tex]\frac{1}{2} \sqrt[3]{(2x)^3}= \frac{1}{2} (2x)[/tex]
Now, we can cancel the 2 in the denominator with the one in the numerator:
[tex] \frac{1}{2} (2x)=x[/tex]
Since [tex]x\ \textgreater \ 0[/tex], this expression is equal to [tex]x[/tex].
H. [tex] \sqrt[3]{-x^3} [/tex] The radicand is elevated to the same number as the index of the radical, so we can cancel the radical and the exponent:
[tex]\sqrt[3]{-x^3}=-x[/tex]
Since [tex]x\ \textgreater \ 0[/tex], this expression is NOT equal to [tex]x[/tex].
We can conclude that the correct answer is H. [tex] \sqrt[3]{-x^3} [/tex].
3. The fourth root of [tex]- \frac{16}{81} [/tex] is [tex] \sqrt[4]{ -\frac{16}{81} } [/tex]. Remember that negative numbers don't have real even roots since a number raised to an even exponent is either positive or 0. Since 4 is even and [tex]- \frac{16}{81} [/tex] is negative, we can conclude that [tex]- \frac{16}{81} [/tex] has not a real fourth root.
The correct answer is D. no real root found.
4. This time we are going to take a different approach. We are going to simplify [tex] \sqrt{a^2(x+a^2)} [/tex] first, and then, we are going to compare the result with our given options:
[tex] \sqrt{a^2(x+a^2)} [/tex]
Lets apply the product rule for a radical [tex] \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b} [/tex]:
[tex]\sqrt{a^2(x+a^2)} = \sqrt{a^2} \sqrt{x+a^2} [/tex]
Notice that in our first product the radicand is raised to the same number as the index, so we can cancel the radical and the exponent:
[tex]\sqrt{a^2} \sqrt{x+a^2}=a \sqrt{x+a^2}[/tex]
We can conclude that the correct answer is F. [tex]a \sqrt{x+a^2} [/tex]
5. Just like before, we are going to simplify [tex] \sqrt{4x^2y^4} [/tex] firts, and then, we are going to compare the result with our given options. To simplify our radical expression we are going to use some laws of radicals:
[tex] \sqrt{4x^2y^4} [/tex]
Applying product rule for a radical [tex] \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b} [/tex]:
[tex]\sqrt{4x^2y^4} = \sqrt{4} \sqrt{x^2} \sqrt{y^4} [/tex]
Notice that [tex] \sqrt{4} =2[/tex], so:
[tex]\sqrt{4} \sqrt{x^2} \sqrt{y^4} =2 \sqrt{x^2} \sqrt{y^4} [/tex]
Notice that [tex]y^4=(y^2)^2[/tex], so we can rewrite our expression:
[tex]2 \sqrt{x^2} \sqrt{y^4}=2 \sqrt{x^2} \sqrt{(y^2)^2} [/tex]
Applying the radical rule [tex] \sqrt[n]{a^n} =a[/tex]:
[tex]2 \sqrt{x^2} \sqrt{(y^2)^2} =2xy^2[/tex]
We can conclude that the correct answer is A. [tex]2xy^2[/tex].
5. Lets check our statements:
F. [tex]x[/tex] is always greater than [tex] \sqrt{x} [/tex].
If [tex]x=0[/tex], [tex] \sqrt{x} =0[/tex]. Therefore, this statement is FALSE.
G. [tex]x= \sqrt{x} [/tex] only if [tex]x=0[/tex]. For a number bigger than zero the square root of that number will be always less than the number; therefore, this statement is TRUE.
H. Using a calculator we can check that [tex] \sqrt{ \frac{1}{2} } =0.7071[/tex] and [tex] \frac{1}{2} =0.5[/tex]. Since [tex]0.7071\ \textgreater \ 0.5[/tex], we can conclude that [tex] \sqrt{ \frac{1}{2} } \ \textgreater \ \frac{1}{2} [/tex].
We can conclude that this statement is TRUE.
J. [tex]x^4[/tex] is always bigger than [tex]x^2[/tex].
If [tex]x= \frac{a}{b} [/tex] and [tex]b\ \textgreater \ a[/tex], [tex]x^4[/tex] will always be smaller than [tex]x^2[/tex]. Therefore, we can conclude that this statement is FALSE.
6. We know that the formula for the volume of cube is [tex]V=s^3[/tex]
where
[tex]V[/tex] is the volume
[tex]s[/tex] is the side
We know for our problem that a cubical storage bin has a volume of 5832 cubic inches, so [tex]V=5832in^3[/tex]. Lets replace that value in our formula and solve for [tex]s[/tex]:
[tex]5832in^3=s^3[/tex]
[tex]s^3=5832in^3[/tex]
[tex]s= \sqrt[3]{5832in^3} [/tex]
[tex]s=18in[/tex]
We can conclude that the length of the side of the cubical storage bin is 18 inches.
