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creativ13 [48]
3 years ago
12

CAN YOU GUYS HELP MY SISTER OUT FOR POINTS AND BRINLESET?!?!?!?

Mathematics
1 answer:
Pani-rosa [81]3 years ago
4 0

Answer:

1. 1/6

2. 1/9

3. 1/10

4.it would still be 9/10

-----------------------------------------------------------------------------------------------------------------

I cant answer that one sorry

-----------------------------------------------------------------------------------------------------------------

4 (im a little less confident in  this one :( )

-----------------------------------------------------------------------------------------------------------------

1/4

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Cost of 5 m ribbon = 7.5 and cost of 1 m ribbon is​
Umnica [9.8K]

Answer:

1.50

Step-by-step explanation:

7.5 is <em>1.5</em><em> </em><em>x</em><em> </em>5 so, 1 x 1.5 = 1.5

5 0
3 years ago
Does each function describe exponential growth or decay?
zhannawk [14.2K]

Hey Jinx :)

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Note that a decay has a rate between 0 and 1, and a growth as a rate of 1 or higher.

y = 1/2(1 + 0.03)^t → y = 1/2(1.03)^t (Rate is more then 0) GROWTH

y = 0.3(0.95)^t → (Rate is between 0 and 1) DECAY

y = ((1+0.03)^1/2)^2t → y = y = 1.014^2t (Rate is more then 0) GROWTH

y = 200(0.73)^t → (Rate is between 0 and 1) DECAY

y = 4(1/4)^t → (Rate is between 0 and 1) DECAY

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4 0
3 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

7 0
3 years ago
Please help me with these two question
BaLLatris [955]
I beleive 1 is C, and 2 is B!
6 0
3 years ago
Read 2 more answers
Find the value of x.
Tatiana [17]

\large \mathfrak{Solution : }

The given pair of angles are vertical opposite angle pair so, they are equal. i.e

  • (x + 90)\degree = 4x\degree

  • (4x - x) \degree= 90\degree

  • 3x\degree = 90\degree

  • x\degree =  \dfrac{90\degree}{3\degree}

  • x\degree = 30\degree
8 0
3 years ago
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