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3 years ago

Please helpanswer choices are:A. 46°B. 67°C. 115°D. 134°

Mathematics

1 answer:

bekas [8.4K]3 years ago

8 0

Answer:

134°

Step-by-step explanation:

∠BAC = ∠BCA because it is an isosceles triangle

∠XCA = ∠XAC . Congruent angles

∠XAC = 1/2(46) = 23° . because CE bisects ∠ C

∠XCA = 23°

180° - 23° - 23° = 134° . Sum of all angles in a triangle is 180°

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How does one integrate the following:

grin007 [14]

Consider substituting $u=4+xy$ , so that $\mathrm du=x\,\mathrm dy$ . Then

$\displaystyle\int_{x=a}^{x=b}\int_{y=a}^{y=b}\frac x{(4+xy)^2}\,\mathrm dy\,\mathrm dx=\int_{x=a}^{x=b}\int_{u=4+ax}^{u=4+bx}\frac1{u^2}\,\mathrm du\,\mathrm dx$
$=\displaystyle\int_{x=a}^{x=b}\left(-\frac1u\right)\bigg|_{u=4+ax}^{u=4+bx}\,\mathrm dx$
$=\displaystyle\int_{x=a}^{x=b}\left(\frac1{4+ax}-\frac1{4+bx}\right)\,\mathrm dx$

Then by similar substitutions, you can easily find that you end up with

$\dfrac1a\ln|4+ax|-\dfrac1b\ln|4+bx|\bigg|_{x=a}^{x=b}$
$=\dfrac1a\ln|4+ab|-\dfrac1b\ln|4+b^2|-\dfrac1a\ln|4+a^2|+\dfrac1b\ln|4+ab|$
$=\dfrac1a\ln\left|\dfrac{4+ab}{4+a^2}\right|+\dfrac1b\ln\left|\dfrac{4+ab}{4+b^2}\right|$

Of course, this all assumes that the integrand is continuous over the domain of integration, which would require that $a,b$ are chosen such that $xy\neq-4$ for any $(x,y)\in[a,b]^2$ . If in particular $ab>-4$ , then we can write

$=\dfrac1a\ln\dfrac{4+ab}{4+a^2}+\dfrac1b\ln\dfrac{4+ab}{4+b^2}$

and you can combine the logarithms if you like as

$=\ln\sqrt[a]{\dfrac{4+ab}{4+a^2}}+\ln\sqrt[b]{\dfrac{4+ab}{4+b^2}}$
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3 years ago

I need help ! Can somebody help please ?

crimeas [40]

Help me on my recent questions I’ll give BRAINLIEST even if they have a answer I just wanna make sure my problems are easy just need to make sure if they are correct I’ll give you BRAINLIEST

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3 years ago

For each of the following functions, find the maximum and minimum values of the function on the rectanglar region: −1≤x≤1,−2≤y≤2

Nesterboy [21]

Answer:

A) The maximum point is 6 at (1,2) while minimum is 0 at (-1,-2)

B) The maximum is 19 at (-1,2), (1,-2),(1,2) & minimum is 0 at (0 , 0)

C) The maximum and minimum is 0 at (+/- 1, +/- 2) and (0,0)

Step-by-step explanation:

From the question, the region is;

−1≤x≤1, −2≤y≤2

In order for us to find the maximum or minimum of a function, we need to find the critical points by finding the first partial derivative and then second derivative if needed;

Thus;

A) f(x,y) = x+y+3

f'(x) = 1 and f'(y)= 1

Therefore, we have no critical points and so we are going to check only the endpoints into the original equation as the follows;

x+y+3;

f(-1 , -2) = -1 + (-2) + 3 = -3 + 3 = 0

f(-1 , 2) = -1 + 2 + 3 = 3 + 1 = 4

f(1 , -2) = 1 + (-2) + 3 = -1 + 3 = 2

f(1 , 2) = 1 + 2 + 3 = 3 + 3 = 6

From the values gotten, we can see that, the maximum value is 6 while the minimum is 0. Thus the maximum point is (1,2) while minimum is at (-1,-2)

B) f(x,y) = 3x² + 4y²

f'(x) = 6x and f'(y)= 8y

If we equate each of them to zero to find the critical point, we'll obtain:

6x = 0 and so, x = 0

8y = 0 and so, y = 0

So, critical point is at (0 , 0)

Now, let's find the second derivative to check ;

f''x = 6, f''xy = 0, f''yy = 8

D = (f''x)(f''y) - (f''xy)² = (6 x 8) - (0)² = 48

Since D(0, 0) = 48 > 0 and f''x (0, 0) = 0, therefore it might be maximum or minimum and we can't say which it is.

So, let's check the critical point with the endpoints into the original equation to obtain:

f(0 , 0) = 3(0²) + 4(0²) = 0

f(-1 , -2) = 3(-1²) + 4(-2)² = 3 + 16 = 19

f(-1 , 2) = 3(-1²) + 4(2)² = 3 + 16 = 19

f(1 , -2) = 3(1²) + 4(-2)² = 3 + 16 = 19

f(1 , 2) = 3(1²) + 4(2)² = 3 + 16 = 19

Thus maximum occurs at (-1,2), (1,-2),(1,2) & minimum occurs at (0 , 0)

C)f(x , y) = 4x² - y²

Repeating the same process as B above;

f'x = 8x

f'y = -2y

If we equate each of them to zero to find the critical point, we'll obtain:

8x = 0 and so, x = 0

-2y = 0 and so, y = 0

So, critical point is at (0 , 0)

Now, let's find the second derivative to check ;

f''x = 8, f''xy = 0, f''y = -2

D = (f''x)(f''y) - (f''xy)² = (8 x 2) - (0)²= 16

Since D(0, 0) = 16 > 0 and f''x (0, 0) = 0, therefore it might be maximum or minimum and we can't say which it is.

So, let's check the critical point with the endpoints into the original equation to obtain:

f(0 , 0) = 4(0)² - 0² = 0

f(-1 , -2) = 4(-1)² - (-2)² = 4 - 4 = 0

f(-1 , 2) = 4(-1)² - 2² = 4 - 4 = 0

f(1 , -2) = 4(1)² - (-2)² = 4 - 4 = 0

f(1 , 2) = 4(1)² - (2)² = 4 - 4 = 0

Looking at the values, maximum and minimum is 0 at (+/- 1, +/- 2) and (0,0)

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3 years ago

Which of these strategies would eliminate a variable in the system of equations?

Mashcka [7]

Answer:

Option A: Multiply the top equation by $\frac{\textbf{1}}{\textbf{2}}$ , then add the equations.

Step-by-step explanation:

OPTION A:

When we multiply the top equation by $\frac{1}{2}$ we get:

$\frac{1}{2}10x + \frac{1}{2}4y = \frac{1}{2}(-2)$

$= 5x + 2y = -1$

Now, we add the second equation to this, we get:

5x + 2y + 5x - 2y = -1 + 2

$\implies 10x = 1$

The 'y' variable is eliminated.

OPTION B: Note that multiplying the second equation by 2 would result in:

10x - 4y = 4. To eliminate 'y' we should add this equation to the top equation not subtract it. So, this option is wrong.

OPTION C:

Adding the equations also will result in a equation of two variables, viz:

15x + 2y = 0 which does not eliminate any variable at all.

So, OPTION C is also wrong.

Hence, OPTION A is the answer.

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3 years ago

What is the effective annual yield. 360 days in a year. 9.5%compounded monthly.

Elodia [21]

Answer:

Step-by-step explanation:

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3 years ago

Please helpanswer choices are:A. 46°B. 67°C. 115°D. 134°​

Please helpanswer choices are:A. 46°B. 67°C. 115°D. 134°