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Phantasy [73]
3 years ago
5

How do you find the value of c that satisfy the equation:

Mathematics
1 answer:
victus00 [196]3 years ago
8 0

Answer:

The value of c = -0.5∈ (-1,0)

Step-by-step explanation:

<u>Step(i)</u>:-

Given function f(x) = 4x² +4x -3 on the interval [-1 ,0]

<u> Mean Value theorem</u>

Let 'f' be continuous on [a ,b] and differentiable on (a ,b). The there exists a Point 'c' in (a ,b) such that

f^{l} (c)  = \frac{f(b) -f(a)}{b-a}

<u>Step(ii):</u>-

Given  f(x) = 4x² +4x -3 …(i)

Differentiating equation (i) with respective to 'x'

          f¹(x) = 4(2x) +4(1) = 8x+4

<u>Step(iii)</u>:-

By using mean value theorem

f^{l} (c)  = \frac{f(0) -f(-1)}{0-(-1)}

8c+4 = \frac{-3-(4(-1)^2+4(-1)-3)}{0-(-1)}

8c+4 = -3-(-3)

8c+4 = 0

8c = -4

c = \frac{-4}{8} = \frac{-1}{2} = -0.5

c ∈ (-1,0)

<u>Conclusion</u>:-

The value of c = -0.5∈ (-1,0)

         

<u></u>

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Novosadov [1.4K]
I assume you mean 9+52,
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Hope this helps! :)
4 0
3 years ago
Describe the graph of the function f(x) = 2x^3 +14x^2 +13x +6. Include the y-intercept, x-intercepts, and the shape of the graph
vovikov84 [41]

f(x) = 2x^{3} + 14x^{2} + 13x + 6

The graph is attached below.

In order to find the y-intercept, check the graph and see where it hits the y-axis, so in this case the y-intercept would be  at (0,6).

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3 years ago
David made a triangle with sides 6 cm, 9cm , 12 cm. George wanted to make a triangle bigger in size but similar to that drawn by
SashulF [63]

Step-by-step explanation:

Given :

\triangle ABC \sim \triangle XYZ .△ABC∼△XYZ.

AB = 6\:cm, \:BC = 9 \:cm, CA = 12 \:cmAB=6cm,BC=9cm,CA=12cm

Let \: XY = 10\:cm, \:YZ = x \:cm, \: XZ = y /:cmLetXY=10cm,YZ=xcm,XZ=y/:cm

\frac{AB}{XY} = \frac{BC}{x} = \frac{CA}{y}XYAB=xBC=yCA

\blue {( The \: lengths \:of \: corresponding }(Thelengthsofcorresponding

\blue {sides \:are \: proportional .)}sidesareproportional.)

\implies \frac{6}{10} = \frac{9}{x} = \frac{12}{y}⟹106=x9=y12

i) \implies \frac{6}{10} = \frac{9}{x}i)⟹106=x9

\implies x = \frac{9 \times 10}{6}⟹x=69×10

\implies x = 15\:cm⟹x=15cm

ii) \implies \frac{6}{10} = \frac{12}{y}ii)⟹106=y12

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\red { Value \:of \:x } \green { = 15\:cm }Valueofx=15cm

\red { Value \:of \:y } \green { = 20\:cm }Valueofy=20cm

•••♪

8 0
3 years ago
MN is a diameter of a circle with centre ' O ' . If BD = CD , prove that ∠OAD = ∠OCD
Varvara68 [4.7K]

Answer:

\sf \: Proved \: \angle \: OAD  \:  =  \angle \: OCD

Step-by-step explanation:

<em>Given:</em>

Mn is diameter of circle having centre O

and BD = OD,

<em><u>To prove that:</u></em>

<u>\angle \: OAD  \:  =  \angle \: OCD</u>

<em>Solution:</em>

Join the points O and B and draw OB,

On joining the line,

in ∆OCD and ∆OBD,

OC =OB → (Radius of same circle)

BD =CD → (from given)

OD =OD → (Common side in both the triangles)

Hence ∆OCD and ∆OBD are congruent from SSS property.

so we can say that,

\angle \: OBD  \:  =  \angle \: OCD

Consider above prove as statement A

Corresponding angles of congruent traingle.

in ∆ OAB,

OA = OB (radius of same circle)

hence ∆OAB is an isosceles traingle.

We know that opposite angle of isosceles traingle are always equal. hence,

\angle \: OBD  \:  =  \angle \: OAB \\ \angle \: OAB  \:  =  \angle \: OAD (same \: angles) \\ \angle \: OBD  \:  =  \angle \: OAD

Consider above prove as statement B

From Statement A & B we can say that

\angle \: OAD  \:  =  \angle \: OCD

<em>Thanks for joining brainly community!</em>

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