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coldgirl [10]
3 years ago
13

Pls help being timed

Mathematics
2 answers:
kotykmax [81]3 years ago
6 0
I did the equation, however as for my final answer, i got 13/105 but the fraction doesn't match with any of the choices...

Alex_Xolod [135]3 years ago
4 0
11/70•30/30=770/2100
11/30•70/70=330/2100
330/2100-770/2100=-440
-440/2100=-22/105

The answer is B







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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
3 years ago
Solve the following quadratic equation by completing the square ✓3x^2 + 10x + 7✓3 = 0​
solmaris [256]

\sqrt{3}x^2+10x+7\sqrt{3}=0\\\\\sqrt3(x^2+\dfrac{10x}{\sqrt{3}}+7)=0\\\\x^2+\dfrac{10x}{\sqrt{3}}+7=0\\\\x^2+\dfrac{10x}{\sqrt{3}}+\dfrac{25}{3}-\dfrac{25}{3} +7=0\\\\(x+\dfrac{5}{\sqrt{3}})^2 = \dfrac{4}{3}\\\\|x+\dfrac{5}{\sqrt{3}}| = \dfrac{2}{\sqrt{3}}\\\\x_1 =   \dfrac{2}{\sqrt{3}}-\dfrac{5}{\sqrt{3}} = -\dfrac{3}{\sqrt{3}} = -\sqrt{3}\\\\x_2 =    -\dfrac{2}{\sqrt{3}}-\dfrac{5}{\sqrt{3}} = \dfrac{-7}{\sqrt{3}} = -\dfrac{-7\sqrt{3}}{3}

5 0
3 years ago
What is the interquartile range of the data ? 0,2,4,0,2,3,8,6
Alexeev081 [22]

(arrange the data set from least to greatest)

0, 0, 2, 2, 3, 4, 6, 8

(find the median: *the middle number*)

Median: 2.5

Lower quartile: 1

Upper quartile: 5

Interquartile range : upper quartile - lower quartile = answer

Interquartile range: 5 - 1 = 4

So the IQR or interquartile for the following data set is 4.

6 0
3 years ago
I need help understanding how to solve this problem.
LUCKY_DIMON [66]

Answer:

  • <u><em>D. (-5/13, -12/13)</em></u>

Explanation:

It is indicated to <em>find the coordinates</em> of the <em>point T (5/13, 12/13) </em>at a distance <em>π + x.</em>

  • π is an arc of half a circle, i.e. 180º.

Then, the distance π+ x means that the <em>point T</em> is rotated 180º.

The rule for the rotation of a point 180º around the origin is:

  • (x, y) → (-x, - y)

Applying that rule to the point T:

  • (5/13, 12/13) → (-5/13, - 12/13) ← answer
6 0
4 years ago
Read 2 more answers
Please help me out with this please
ki77a [65]

Answer:

P = 9 is the max value

Step-by-step explanation:

Sketch

2x + 4y = 10

with x- intercept = (5, 0) and y- intercept (0, 2.5)

x + 9y = 12

with x- intercept = (12, 0) and y- intercept = (0, \frac{4}{3} )

Solve

2x + 4y = 10 and x + 9y = 12 to find the point of intersection at (3, 1)

The region corresponding to the solution of the system of constraints

Has vertices at (0, \frac{4}{3}), (0, 0) , (5, 0) and (3, 1)

Now evaluate the objective function at each vertex.

(0, 0) can be excluded as it will not give a maximum

(5, 0) → P = 5 + 0 = 5

(0, \frac{4}{3}) → 0 + 8 = 8

(3, 1) → 3 + 6(1) = 3 + 6 = 9 ← maximum value

Thus the maximum value is 9 when x = 3 and y = 1

7 0
3 years ago
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