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CaHeK987 [17]
3 years ago
7

Can someone please help me with this?

Mathematics
1 answer:
Marta_Voda [28]3 years ago
4 0

Answer:

yeet

Step-by-step explanation:

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The graph of the continuous function g, the derivative of the function f, is shown above. The function g is piecewise linear for
4vir4ik [10]
A) g=f' is continuous, so f is also continuous. This means if we were to integrate g, the same constant of integration would apply across its entire domain. Over 0, we have g(x)=2x. This means that


f_{0


For f to be continuous, we need the limit as x\to1^- to match f(1)=3. This means we must have


\displaystyle\lim_{x\to1}x^2+C=1+C=3\implies C=2


Now, over x, we have g(x)=-3, so f_{x, which means f(-5)=17.


b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,


\displaystyle\int_1^6g(x)=4+\int_3^62(x-4)^2\,\mathrm dx=4+6=10


c) f is increasing when f'=g>0, and concave upward when f''=g'>0, i.e. when g is also increasing.

We have g>0 over the intervals 0 and x>4. We can additionally see that g'>0 only on 0 and x>4.


d) Inflection points occur when f''=g'=0, and at such a point, to either side the sign of the second derivative f''=g' changes. We see this happening at x=4, for which g'=0, and to the left of x=4 we have g decreasing, then increasing along the other side.


We also have g'=0 along the interval -1, but even if we were to allow an entire interval as a "site of inflection", we can see that g'>0 to either side, so concavity would not change.
5 0
2 years ago
Given that lines a and b are parallel, what angles formed on line b when cut by the transversal are congruent with ∠2?
Kruka [31]

Answer:

The angles formed on line b when cut by the transversal are congruent with ∠2 are \angle{6}\text{ and }\angle{7}

Step-by-step explanation:

Consider the provided information.

If transversal line crossed by two parallel lines, then, the corresponding angles and alternate angles are equal .

The angles on the same corners are called corresponding angle.

Alternate Angles:  Angles that are in opposite positions relative to a transversal intersecting two lines.

∠2 and ∠6 are corresponding angles

Therefore, ∠2 = ∠6

∠2 and ∠7 are alternate exterior angles

Therefore, ∠2 = ∠7

Hence, the angles formed on line b when cut by the transversal are congruent with ∠2 are \angle{6}\text{ and }\angle{7}

3 0
2 years ago
Y=(4x-5)(x-?) what does x equal
Whitepunk [10]
Set the factor '(4 + -1x)' equal to zero and attempt to solve: Simplifying 4 + -1x = 0 Solving 4 + -1x = 0 Move all terms containing x to the left, all other terms to the right. Add '-4' to each side of the equation. 4 + -4 + -1x = 0 + -4 Combine like terms: 4 + -4 = 0 0 + -1x = 0 + -4 -1x = 0 + -4 Combine like terms: 0 + -4 = -4 -1x = -4 Divide each side by '-1'. x = 4 Simplifying x = 4

Easy answer
Its 0

I hope it's correct Good luck...
8 0
3 years ago
Find the area.<br> 2 cm<br> 6 cm<br> 4 cm
hammer [34]

Answer:

A=1/2(a+b)h

A=1/2(2cm+4cm)6cm

A=1/2×6cm×6cm

A=18cm²

3 0
2 years ago
(1 point) A fish tank initially contains 15 liters of pure water. Brine of constant, but unknown, concentration of salt is flowi
Drupady [299]

Answer:

a. \dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

b. x(t) = 6\frac{2}{3} \cdot c \cdot t

c. c  = \dfrac{3}{8}  \ g/L

Step-by-step explanation:

a. The volume of water initially in the fish tank = 15 liters

The volume of brine added per minute = 5 liters per minute

The rate at which the mixture is drained = 5 liters per minute

The amount of salt in the fish tank after t minutes = x

Where the volume of water with x grams of salt = 15 liters

dx =  (5·c - 5·c/3)×dt = 20/3·c = 6\frac{2}{3} \cdot c \cdot dt

\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

b. The amount of salt, x after t minutes is given by the relation

\dfrac{dx}{dt} = 6\frac{2}{3} \cdot c

dx = 6\frac{2}{3} \cdot c \cdot dt

x(t) = \int\limits \, dx  = \int\limits \left ( 6\frac{2}{3} \cdot c \right) \cdot dt

x(t) = 6\frac{2}{3} \cdot c \cdot t

c. Given that in 10 minutes, the amount of salt in the tank = 25 grams, and the volume is 15 liters, we have;

x(10) = 25 \ grams(15 \ in \ liters) = 6\frac{2}{3} \times c \times 10

6\frac{2}{3} \times c  =\dfrac{25 \ grams }{10}

c  =\dfrac{25 \ g/L }{10 \times 6\frac{2}{3} }  = \dfrac{25 \ g/L}{10 \times \dfrac{20}{3} } =\dfrac{3}{200} \times 25 \ g/L= \dfrac{75}{200}  \ g/L = \dfrac{3}{8}  \ g/L

c  = \dfrac{3}{8}  \ g/L

4 0
2 years ago
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