Answer:
its x=1/3 :)))
X^9/7 convert from exponent form to radical form
Answer:
[tex]x^{\frac{9}{7}} = \sqrt[7]{x^9}[/tex]
Step-by-step explanation:
You can solve this by realising that the denominator of a fractional exponent can be expressed as the base of a radical.
Note also that the order does not matter. You could also express it as
[tex]\sqrt[7]{x}^9[/tex]
The reason this works is that you're effectively breaking the exponent into fractions. The first answer is the equivalent of:
[tex](x^9)^{1/7}[/tex]
and the second would be:
[tex](x^{1/7})^9[/tex]
In both cases, the exponents would be multiplied, giving the same result.
Luis solves the following system of equations by elimination.
5s+ 3t = 30
2s+3t=-3
What is the value of s in the solution of the system?
The circumference of a circle is 6π ft. What is the area, in square feet? Express your answer in terms of
π
Please help
Answer:
Circumference of the circle 2π(radius)=6π ft
Therefore, radius=3 ft
Area=π(radius)²=π×3²=9π ft²
9π ft² is the right answer.
The area of circle is 9π square feet.
What is area?Area is the amount of area occupied by an object's flat (2-D) surface or shape.
The circumference of circle = 2πr
6 = 2πr
r = 3π
And the area of circle is πr².
The area of the circle = 9π
Therefore, the area of the circle is 9π square feet.
To learn more about the area;
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Asphere has a radius of 27 inches. A horizontal plane passes through the center of the sphere.
Part 1 out od 2
Describe the cross section formed by the plane and the sphere.
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Answer:
circle of radius 27 inches
Step-by-step explanation:
Anywhere a plane cuts a sphere, the cross section is a circle. When the plane includes the center of the sphere, the circle has the same radius the sphere has.
The cross section is a circle of radius 27 inches.
20 less than 5 times a number is 85. What is the number
Answer: I don’t understand the question??
Step-by-step explanation: Can you say the question again like the same way the homework does
If -xy – 5+ y^2+ x^2= 0 and it is known that dy/dx= y-2x/-x+2y, find all
coordinate points on the curve where x = -1 and the line tangent to the
curve is horizontal, or state that no such points exist.
Answer:
There is no point of the form (-1, y) on the curve where the tangent is horizontal
Step-by-step explanation:
Notice that when x = - 1. then dy/dx becomes:
dy/dx= (y+2) / (2y+1)
therefore, to request that the tangent is horizontal we ask for the y values that make dy/dx equal to ZERO:
0 = ( y + 2) / (2 y + 1)
And we obtain y = -2 as the answer.
But if we try the point (-1, -2) in the original equation, we find that it DOESN'T belong to the curve because it doesn't satisfy the equation as shown below:
(-1)^2 + (-2)^2 - (-1)*(-2) - 5 = 1 + 4 + 2 - 5 = 2 (instead of zero)
Then, we conclude that there is no horizontal tangent to the curve for x = -1.
Solve x4 - 11x2 + 18 = 0.
Answer:
x=√2, x=-√2, x= 3 and x=-3
Step-by-step explanation:
We need to solve the equation x^4 - 11x^2+18=0
We can replace x^4 = u^2 and x^2 = u
So, the equation will become
u^2 -11u+18 = 0
Factorizing the above equation:
u^2 -9u-2u+18 =0
u(u-9)-2(u-9)=0
(u-2)(u-9)=0
u=2, u=9
As, u = x^2, Putting back the value:
x^2 =2 , x^2 =9
taking square roots:
√x^2 =√2 ,√x^2=√9
x=±√2 , x = ±3
so, x=√2, x=-√2, x= 3 and x=-3
Answer:
The second answer is correct.
Step-by-step explanation:
Mark brainliest if correct please