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shtirl [24]
3 years ago
7

HELP ME PLEASEEEEEEEEEEEEEEE

Mathematics
2 answers:
Murljashka [212]3 years ago
5 0
Well i'm gonna be honest : my answer probably is wrong, but here ya go :

so since the y intercept of f(x) is -2 translating that 6 units upwards will make it positive 4

so if six units upwards takes it to g(x) the answer should be 4

anygoal [31]3 years ago
4 0
Ok so the formula they gave you is in the form y=mx+b. What's different however is that they say the graph is translated up 6 units. So the formula f(x)=-4x-2 becomes f(x)=-4x+4 (since b represents how high the graph is). And to find any y intercept, you must make X zero, and -4(0) +4 is now equal to 4 and 4 is your answer
You might be interested in
Show that if the first 10 positive integers 1,2,3,···,10 are placed around a circle, in any order,there exists three integers in
zysi [14]

Answer:

Let A1=a1+a2+a3, A2=a2+a3+a4, and so on, A10=a10+a1+a2. Then A1+A2+⋯+A10=3(a1+a2+⋯+a10)=(3)(55)=165, so some Ai≥165/10=16.5, so some Ai≥17.

Step-by-step explanation:

7 0
3 years ago
How to find the smallest integer k such that 60k is a perfect square
suter [353]

9514 1404 393

Answer:

  k = 15

Step-by-step explanation:

Look for the missing factors that make all of the factors of 60k be squares.

  60 = 2² × 3 × 5

The factors 3 and 5 each have an odd exponent (1), so those two factors must be part of k.

  60k = 2² × 3 × 5 × k

is a perfect square when ...

  3 × 5 = k = 15

For k = 15, 60k = 900 = 30²

6 0
3 years ago
Find the length of UT.
otez555 [7]

Answer: A. 32

Concept:

Here, we need to know the idea of the intersecting chord theorem and segment addition postulate.

The<u> intersecting chord theorem </u>states that when two chords intersect at a point, P, the product of their respective partial segments is equal.

The<u> Segment Addition Postulate</u> states that given 2 points A and C, a third point B lies on the line segment AC if and only if the distances between the points satisfy the equation AB + BC = AC.

If you are still confused, please refer to the attachment below for a graphical explanation.

Solve:

<u>Given information</u>

CW = 12

TW = 14

VW = 2x + 5

UW = 2x + 2

<u>Given expression deducted from intersecting chord theorem</u>

CW · VW = TW · UW

<u>Substitute values into the expression</u>

(12) · (2x + 5) = (14) · (2x + 2)

<u>Expand parentheses and apply the distributive property</u>

24x + 60 = 28x + 28

<u>Subtract 14x on both sides</u>

24x + 60 - 24x = 28x + 28 - 24x

60 = 4x + 28

<u>Subtract 28 on both sides</u>

60 - 28 = 4x + 28 - 28

32 = 4x

<u>Divide 4 on both sides</u>

32 / 4 = 4x / 4

x = 8

<u>Given expression deducted from the segment addition postulate</u>

UT = UW + TW

<u>Substitute values into the expression</u>

UT = 2x + 2 + 14

<u>Substitute x value into the expression</u>

UT = 2 (8) + 2 + 14

UT = 16 + 2 + 14

<u>Combine like terms</u>

UT = 32

Hope this helps!! :)

Please let me know if you have any questions

5 0
2 years ago
Please read "The lost ship" on storyworks.
Degger [83]
1. it suggests the article could be about the tale of a lost ship, how it became lost, or how it is doing now.
2. could you please provide a picture
3.again, we cannot see a picture
4. we cannot see if the article has a map
7 0
3 years ago
g 1) The rate of growth of a certain type of plant is described by a logistic differential equation. Botanists have estimated th
alexira [117]

Answer:

a) The expression for the height, 'H', of the plant after 't' day is;

H = \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}

b) The height of the plant after 30 days is approximately 19.426 inches

Step-by-step explanation:

The given maximum theoretical height of the plant = 30 in.

The height of the plant at the beginning of the experiment = 5 in.

a) The logistic differential equation can be written as follows;

\dfrac{dH}{dt} = K \cdot H \cdot \left( M - {P} \right)

Using the solution for the logistic differential equation, we get;

H = \dfrac{M}{1 + A\cdot e^{-(M\cdot k) \cdot t}}

Where;

A = The condition of height at the beginning of the experiment

M = The maximum height = 30 in.

Therefore, we get;

5 = \dfrac{30}{1 + A\cdot e^{-(30\cdot k) \cdot 0}}

1 + A = \dfrac{30}{5} = 6

A = 5

When t = 20, H = 12

We get;

12 = \dfrac{30}{1 + 5\cdot e^{-(30\cdot k) \cdot 20}}

1 + 5\cdot e^{-(30\cdot k) \cdot 20} = \dfrac{30}{12} = 2.5

5\cdot e^{-(30\cdot k) \cdot 20} =  2.5 - 1 = 1.5

∴ -(30·k)·20 = ㏑(1.5)

k = ㏑(1.5)/(30 × 20) ≈ 6·7577518 × 10⁻⁴

k ≈ 6·7577518 × 10⁻⁴

Therefore, the expression for the height, 'H', of the plant after 't' day is given as follows

H = \dfrac{30}{1 + 5\cdot e^{-(30\times 6.7577518 \times 10^{-4}) \cdot t}} =  \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}

b) The height of the plant after 30 days is given as follows

H =  \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \cdot t}}

At t = 30, we have;

H =  \dfrac{30}{1 + 5\cdot e^{-(2.02732554 \times 10^{-3}) \times 30}} \approx 19.4258866473

The height of the plant after 30 days, H ≈ 19.426 in.

3 0
3 years ago
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