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denis23 [38]
3 years ago
9

Plz help will mark you brainliest if correct

Mathematics
2 answers:
katrin [286]3 years ago
7 0

Answer:

The answer would be B instead of D

inn [45]3 years ago
5 0

Answer:

B

Step-by-step explanation:

5/1 is not equal to 6/2 or 7/3 or 8/4

You might be interested in
Solve for y.<br> r/3-2/y=s/5
nordsb [41]

Answer:

y = 2 / (r/3 - s/5)

Step-by-step explanation:

r/3 - 2/y = s/5

add 2/y to both sides

r/3 = s/5 + 2/y

Subtract s/5 from both sides

r/3 - s/5 = 2/y

multiply both sides by y

y(r/3 - s/5) = 2

Divide both sides by r/3 - s/5

y = 2 / (r/3 - s/5)

8 0
3 years ago
The volume of a cone is 3πx3 cubic units and its height is x units. Which expression represents the radius of the cone’s base, i
dalvyx [7]

Answer:

<em>r=3x</em>

Step-by-step explanation:

<u>The Volume of a Cone</u>

The volume of a cone of radius r and height h is:

\displaystyle V=\frac{1}{3}\pi hr^2

We are given the volume of a cone is

V=3\pi x^3

Equating:

\displaystyle \frac{1}{3}\pi hr^2=3\pi x^3

Multiplying by 3:

\pi hr^2=9\pi x^3

Simplifying by pi:

hr^2=9 x^3

Since h=x:

xr^2=9 x^3

Dividing by x:

r^2=9 x^2

Taking square roots:

r=\sqrt{9 x^2}=3x

Thus: r=3x

5 0
3 years ago
Read 2 more answers
The points -4. -4 -4, 4
Ilia_Sergeevich [38]

Answer:

whats the question?

Step-by-step explanation:

I don't get the question so what is the question?

3 0
3 years ago
Let f(x)=x^3-x-1
alexira [117]
f(x) = x^3 - x - 1

To find the gradient of the tangent, we must first differentiate the function.

f'(x) = \frac{d}{dx}(x^3 - x - 1) = 3x^2 - 1

The gradient at x = 0 is given by evaluating f'(0).

f'(0) = 3(0)^2 - 1 = -1

The derivative of the function at this point is negative, which tells us <em>the function is decreasing at that point</em>.

The tangent to the line is a straight line, so we will have a linear equation of the form y = mx + c. We know the gradient, m, is equal to -1, so

y = -x + c

Now we need to substitute a point on the tangent into this equation to find c. We know a point when x = 0 lies on here. To find the y-coordinate of this point we need to evaluate f(0).

f(0) = (0)^3 - (0) - 1 = -1

So the point (0, -1) lies on the tangent. Substituting into the tangent equation:

y = -x + c \\\\ -1 = -(0) + c \\\\ -1 = c \\\\ \text{Equation of tangent is } \boxed{y = -x - 1}
6 0
3 years ago
PLEASE HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
jekas [21]

Answer:

smal and blurry

Step-by-step explanation:

4 0
3 years ago
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