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Dafna11 [192]
3 years ago
13

I need to turn this in by 10:00 plz help thank you :)

Mathematics
1 answer:
kati45 [8]3 years ago
4 0

Answer:

11 is one 0f them

Step-by-step explanation:

11x11

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Find the minimum value if f(x) = xe^x over [-2,0]
SashulF [63]

Given function:

f(x)=xe^x

The minimum value of the function can be found by setting the first derivative of the function to zero.

f^{\prime}(x)=xe^x+e^x\begin{gathered} xe^x+e^x\text{ = 0} \\ e^x(x\text{ + 1)  = 0} \end{gathered}

Solving for x:

\begin{gathered} x\text{ + 1 = 0} \\ x\text{ = -1} \end{gathered}\begin{gathered} e^x\text{ = 0} \\ \text{Does not exist} \end{gathered}

Substituting the value of x into the original function:

\begin{gathered} f(x=1)=-1\times e^{^{-1}}_{} \\ =\text{ -}0.368 \end{gathered}

Hence, the minimum value in the given range is (-1, -0.368)

6 0
1 year ago
What is (5x-2)(4x^2 -3x-2)
Ronch [10]
Answer: 20x^3 - 23x^2 - 4x + 4

Explanation:

Use distributive property:

(5x-2)(4x^2 - 3x-2)
= 20x^3 - 15x^2 - 10x - 8x^2 + 6x + 4
= 20x^3 - 23x^2 - 4x + 4
8 0
3 years ago
Read 2 more answers
use 3.14 n for to estimate the area of a circle if the diameter is given on your answer to the nearest hundredth if necessary​
s2008m [1.1K]

Answer: The Answer is pie

Step-by-step explanation:

8 0
3 years ago
The diagram shows a logo​
Charra [1.4K]

Answer/Step-by-step explanation:

✔️Find EC using Cosine Rule:

EC² = DC² + DE² - 2*DC*DE*cos(D)

EC² = 27² + 14² - 2*27*14*cos(32)

EC² = 925 - 756*cos(32)

EC² = 283.875639

EC = √283.875639

EC = 16.85 cm

✔️Find the area of ∆DCE:

Area = ½*14*27*sin(32)

Area of ∆DCE = 100.15 cm²

✔️Since ∆DCE and ∆ABE are congruent, therefore,

Area of ∆ABE = 100.15 cm²

✔️Find the area of the sector:

Area of sector = 105/360*π*16.85²

Area = 260.16 cm² (nearest tenth)

✔️Therefore,

Area of the logo = 100.15 + 100.15 + 260.16 = 460.46 ≈ 460 cm² (to 2 S.F)

5 0
3 years ago
(will mark as brainliest)
Helga [31]

Find the x and y-intercepts of the equation. Then calculate the points at constant intervals and plot and connect those points on a makeshift graph. A quadratic equation can have 2 solutions, 1 solution, or none. The corresponding graphs(if two solutions) would most probably cross over the x-axis twice.

Hope this helps.

4 0
3 years ago
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