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Romashka-Z-Leto [24]
3 years ago
14

Find the missing geometric means in this sequence.

Mathematics
1 answer:
amid [387]3 years ago
3 0
What's the problem?
You might be interested in
(08.07 HC)
andreev551 [17]

Answer:

\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}

\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)

C)  See attachment.

Step-by-step explanation:

Given function:

f(x)=2x^2-x-10

<h3><u>Part A</u></h3>

To factor a <u>quadratic</u> in the form  ax^2+bx+c<em> , </em>find two numbers that multiply to ac and sum to b :

\implies ac=2 \cdot -10=-20

\implies b=-1

Therefore, the two numbers are -5 and 4.

Rewrite b as the sum of these two numbers:

\implies f(x)=2x^2-5x+4x-10

Factor the first two terms and the last two terms separately:

\implies f(x)=x(2x-5)+2(2x-5)

Factor out the common term  (2x - 5):

\implies f(x)=(x+2)(2x-5)

The x-intercepts are when the curve crosses the x-axis, so when y = 0:

\implies (x+2)(2x-5)=0

Therefore:

\implies (x+2)=0 \implies x=-2

\implies (2x-5)=0 \implies x=\dfrac{5}{2}

So the x-intercepts are:

x=-2, \:\:x=\dfrac{5}{2}

<h3><u>Part B</u></h3>

The x-value of the vertex is:

\implies x=\dfrac{-b}{2a}

Therefore, the x-value of the vertex of the given function is:

\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}

To find the y-value of the vertex, substitute the found value of x into the function:

\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}

Therefore, the vertex of the function is:

\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)

<h3><u>Part C</u></h3>

Plot the x-intercepts found in Part A.

Plot the vertex found in Part B.

As the <u>leading coefficient</u> of the function is positive, the parabola will open upwards.  This is confirmed as the vertex is a minimum point.

The axis of symmetry is the <u>x-value</u> of the <u>vertex</u>.  Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is <u>symmetrical</u>.

Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).

5 0
2 years ago
10 points! What is the slope of the given line?
aleksklad [387]

Answer:

-5/2

Step-by-step explanation:

8 0
3 years ago
HELP ME PLEASEEEEEEEEEEEEEEEEEEE
aniked [119]

Answer:

the last one :)

Step-by-step explanation:

3 0
3 years ago
The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=
Leno4ka [110]
1 step: S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}, S_{3}=T_{1}+T_{2}+T_{3}, then
 S_{5}-S_{3}=T_{4}+T_{5}=5.

2 step: T_{n}=T_{1}*q^{n-1}, then 
T_{6}=T_{1}*q^{5}
T_{5}=T_{1}*q^{4}
T_{4}=T_{1}*q^{3}
T_{3}=T_{1}*q^{2}
and \left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right. will have form \left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right..

3 step: Solve this system  \left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right. and dividing first equation on second we obtain \frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}. So, \frac{(q-1)(q+1)}{q+1} = \frac{1}{2} and q-1= \frac{1}{2}, q= \frac{3}{2} - the common ratio.

4 step: Insert q= \frac{3}{2}into equation T_{1}*q^{3}*(q+1)=5 and obtain T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5, from where T_{1}= \frac{16}{27}.




5 0
3 years ago
A line with a slope of 3 passes through the points (9,7) and (10,c) what is the value of c?
Zina [86]

Answer:

The answer is 4. The line passes through the point (10,4).

Step-by-step explanation:

7 0
3 years ago
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