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guapka [62]
2 years ago
15

PLEASE ANSWER ASAP WILL GIVE BRAINLEIST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Step2247 [10]2 years ago
5 0

Answer:

r = 16

Step-by-step explanation:

Cross multiply:

4 x 20 = r x 5

Simplifying

4 x 20 = r x 5

Multiply 4 x 20

80 = r x 5

Reorder the terms for easier multiplication:

80 = 5r

Solve

80 = 5r

Solve for variable 'r'.

Move all terms containing r to the left, all other terms to the right.

Add '-5r' to each side of the equation.

80 + -5r = 5r + -5r

Combine like terms: 5r + -5r = 0

80 + -5r = 0

Add '-80' to each side of the equation.

80 + -80 + -5r = 0 + -80

Combine like terms: 80 + -80 = 0

0 + -5r = 0 + -80

-5r = 0 + -80

Combine like terms: 0 + -80 = -80

-5r = -80

Divide each side by '-5'.

r = 16

Simplify

r = 16

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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
Kira has $15 in her savings account. She will get $5 per week for her allowance. Write an expression to represent the amount of
Fudgin [204]

Answer:

3??

Step-by-step explanation:

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6 0
2 years ago
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