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andrezito [222]
3 years ago
15

Grandpa Ernie is shrinking! Over the past 4 years his height decreased by a total of 2.4 cm.It decreased by the same amount each

year. What was the change in Grandpa Ernie's height each year? AllNewsImagesMapsMore SettingsTools
Mathematics
1 answer:
Nuetrik [128]3 years ago
6 0

Your answer would be -0.6 cm.

just take -2.4/4

= -0.6

It is -2.4 because his height decreased each year.

<em>Hope helps!-Aparri</em>

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When is g(x) = 0 for the function g(x) = 5.2 3x + 4?
kicyunya [14]
G(x) = 0 can never happen.

When x is 0, the answer is 4
When x is 1, the answer is 19.6
When x is 4, the answer is 66.4

According to Wolfram Alpha though, x is equal to -(10/39). If that's not a choice for you, don't worry about it. :)
6 0
3 years ago
box with a square base and open top must have a volume of 4,000 cm3. Find the dimensions of the box (in cm) that minimize the am
Naya [18.7K]

Answer:

Hence the width, length is 20 cm and height is 10 cm

Step-by-step explanation:

Since the box has a square base, let length = width = x. Also, let the height = y, therefore:

The volume of box = width * length * height

4000 = x * x * y

4000 = x²y

y=4000/x²

The surface area (SA) = area of the base + sum of the area of each side

SA = x² + xy + xy + xy + xy

SA = x² + 4xy

substitute y = 4000/x²

SA = x² + 4x(4000/x²)

SA = x² + 16000/x

Taking the derivative:

SA' = 2x - 16000/x²

making SA' = 0:

0 = 2x - 16000/x²

2x = 16000/x²

2x³ = 16000

x³ = 8000

x = 20 cm

y = 4000 / x² = 4000 / 20² = 10 cm

Hence the width, length is 20 cm and height is 10 cm

6 0
3 years ago
Choose two angles that are each separately alternate exterior angles with 412.
Alborosie

Answer:

∠2 and ∠5

Step-by-step explanation:

we know that

<u>Alternate Exterior Angles</u> are a pair of angles on the outer side of each of those two lines but on opposite sides of the transversal

In this problem

∠12 and ∠2 are alternate exterior angles

∠12 and ∠5 are alternate exterior angles

therefore

∠2 and ∠5 are each separately alternate exterior angles with ∠12

6 0
3 years ago
Use the function below to find f(-4)
Anastaziya [24]

Answer:

B. 1/16

Step-by-step explanation:

f(-4) = 2^(-4) = 1/(2^4) = 1/16

3 0
3 years ago
Apply The Remainder Theorem, Fundamental Theorem, Rational Root Theorem, Descartes Rule, and Factor Theorem to find the remainde
Over [174]

9514 1404 393

Answer:

  possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}

  actual roots: -1, (2 ±4i√2)/3

  no turning points; no local extrema

  end behavior is same-sign as x-value end-behavior

Step-by-step explanation:

The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.

The Rational Root Theorem tells us any rational roots will be of the form ...

  ±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}

  = ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots

Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.

The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.

The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):

  f(x) = (x +1)(3x^2 -4x +12)

The complex roots of the quadratic can be found using the quadratic formula:

  x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6

  x = (2 ± 4i√2)/3 . . . . complex roots

__

The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.

4 0
3 years ago
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