I need help! Please. Justify answer using Intermediate value theorem (IVT)

Mathematics

1 answer:

STatiana [176]3 years ago

8 0

Step-by-step explanation:

N(t) is continuous at t = 6, so f(6) = 25(6) + 150 = 300.

f(0) = 80.

f(t) is continuous, so by Intermediate Value Theorem, f(t) must equal 250 at some point between t=0 and t=6.

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What is the cost price of 600 pens

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3 years ago

(2a^2b + 3ab^2 - b^3) - (4b^3 - 3ab^2 -6a^2b)

skelet666 [1.2K]

Hi there!

~

$(2a^2b+3ab^2-b^3)-(4b^3-3ab^2-6a^2b)$

Distribute the Negative Sign:

$= 2a^2 b + 3ab^2 - b^3 + -1(4b^3 - 3ab^2 - 6a^2b)\\= 2a^2 b + 3ab^2 + - b^3 + -1 (4b^3) + -1 (-3ab^2) + -1 (-6a^2b)\\= 2a^2 b + 3ab^2 + -b^3 + -4b^3 + 3ab^2 + 6a^2 b$

Combine Like Terms:

$= 2a^2 b + 3ab^2 - b^3 + -4ab^3 + 3ab^2 + 6a^2 b\\= (2a^2 b + 6a^2 b) + ( 3ab^2 + 3ab^2) + (-b^3 + -4b^3) \\= 8a^2 b + 6ab^2 + -5b^3$

Answer : $8a^2 b + 6ab^2 + -5b^3$

Hope this helped you!

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3 years ago

LAST ATTEMPT MARKING AS BRAINLIEST!! ( graph the image of the figure using the dilation given)

Natali5045456 [20]

<h3>Answers:</h3>

$W(-2,-1) \to \\W \ ' \ (-4,4)\\\\X(0,2) \to \\X \ ' \ (0,4)\\\\Y(3,-1) \to \\Y \ ' \ (6,-2)\\\\V(-2,-1) \to \\V \ ' \ (-4,-2)\\\\$

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Explanation:

The scale factor is 2, which means we double each coordinate of each point. The general rule is $(x,y) \to (kx,ky)$ with k = 2. So we can say the more specific dilation rule is $(x,y) \to (2x,2y)$

Something like W(-2,2) moves to W ' (-4, 4) after multiplying each coordinate by 2. Do the same for the other points as well.

The given preimage points

V = (-2, -1)
W = (-2, 2)
X = (0, 2)
Y = (3, -1)

will dilate to the corresponding image points

V ' = (-4, -2)
W ' = (-4, 4)
X ' = (0, 4)
Y ' = (6, -2)

as shown below. This causes the image to be larger compared to the preimage (since the scale factor is larger than 1). Any given point is twice as far from the origin as compared to before, which in turn means the distance between any two points is twice as much.