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BaLLatris [955]
4 years ago
14

D is a point on segment CE. If CD = 9.8 and CE = 15.2, what is DE?​

Mathematics
1 answer:
Brrunno [24]4 years ago
8 0

Answer:

5.4

Step-by-step explanation:

imagine C-------------D---------------E

So CD = 9.8. and CE is 15.2

Subtract CD from CE.

15.2-9.8=5.4

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#43 please help have mid terms Tomorrow
nirvana33 [79]

Answer:

Job A is more profitable for nearly 49 months (or 50 months including the first month)

Job B is more profitable after 49 months (or 50 months including the first month).

Step-by-step explanation:

Let x be the number of months passed after first month

<u>Job A:</u>

$2,000 for the first month with a $300 raise every month thereafter

Function describing this situation:

f(x)=2,000+300x

<u>Job B:</u>

$1,500 for the first month with a 5% raise every month thereafter

Function describing this situation:

g(x)=1,500\cdot (1+0.05)^x\\ \\g(x)=1,500\cdot 1.05^x

Plot both graphs (see attached diagram). The diagram shows that the job A is more profitable for nearly 49 months (or 50 months including the first month) and the job B is more profitable after 49 months (or 50 months including the first month).

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The correct awnser would be D
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X over 2 -8=3<br> How do I find the answer if I do not get it plz help
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Find the axis of symmetry and the vertex of the graph of y = 3x2 + 2x.
hichkok12 [17]

Step-by-step explanation:

<u>Finding Vertex</u>

Given

y\:=\:3x^2\:+\:2x

The vertex of an up-down facing parabola of the form

y=ax^2+bx+c\:\mathrm{is}\:x_v=-\frac{b}{2a}

The Parabola params are:

a=3,\:b=2,\:c=0

x_v=-\frac{b}{2a}

x_v=-\frac{2}{2\cdot \:3}

x_v=-\frac{1}{3}

\mathrm{Plug\:in}\:\:x_v=-\frac{1}{3}\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=3\left(-\frac{1}{3}\right)^2+2\left(-\frac{1}{3}\right)

    =3\left(-\frac{1}{3}\right)^2-2\cdot \frac{1}{3}

    =\frac{1}{3}-\frac{2}{3}         ∵  3\left(-\frac{1}{3}\right)^2=\frac{1}{3}

    =\frac{1-2}{3}

    =\frac{-1}{3}

\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{b}=-\frac{a}{b}

   =-\frac{1}{3}

y_v=-\frac{1}{3}

Therefore the parabola vertex is:

\left(-\frac{1}{3},\:-\frac{1}{3}\right)

\mathrm{If}\:a

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

a=3

\mathrm{Minimum}\space\left(-\frac{1}{3},\:-\frac{1}{3}\right)

<u>Finding symmetry</u>

For a parabola in standard form y=ax^2+bx+c

the axis of symmetry is the vertical line that goes through the vertex x=\frac{-b}{2a}

\mathrm{Axis\:of\:Symmetry\:for}\:y=ax^2+bx+c\:\mathrm{is}\:x=\frac{-b}{2a}

a=3,\:b=2

x=\frac{-2}{2\cdot \:3}

x=-\frac{1}{3}

Therefore,

\mathrm{Axis\:of\:Symmetry\:for}\:y=3x^2+2x:\quad x=-\frac{1}{3}

3 0
3 years ago
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