In triangle ABC, if b is the longest side, what can you conclude about angle A?

Mathematics

2 answers:

Yuri [45]3 years ago

5 0

Answer:

It is less than 90degrees

Step-by-step explanation:

malfutka [58]3 years ago

5 0

It is less than 90 degrees

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PLS ANSWER THE BELOW QUESTION AND I'LL MARK U AS BRAINLIST

MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

$a) \dfrac{3x}{2}-5=4$

Add 5 to both sides

$\dfrac{3x}{2}-5 +5 = 4 + 5\\\\\dfrac{3x}{2} = 9 \\\\Now \ multiply \ both \ sides \ by \ 2 \\\\\dfrac{3x}{2}*2 = 9*2\\\\3x = 18\\Now \ divide \ both \ sides \ by \ 3\\\\x =\dfrac{18}{3}\\\\$

x = 6

$b) \dfrac{x-3}{5} + 1 = -2\\\\\\Subtract \ 1 \ from \ both \ sides\\\\\dfrac{x-3}{5} = -2 -1\\\\\dfrac{x-3}{5} = -3\\\\Multiply \ both \ side \ by \ 5\\\\x - 3 = -3*5\\\\x -3 = -15\\\\$

Add 3 to both sides

x = -15 +3

x = -12

7 0

3 years ago

Read 2 more answers

Solve dis attachment and show all work ( I got it all wrong and I want to know how to solve it )

DedPeter [7]

(a) First find the intersections of $y=e^{2x-x^2}$ and $y=2$ :

$2=e^{2x-x^2}\implies \ln2=2x-x^2\implies x=1\pm\sqrt{1-\ln2}$

So the area of $R$ is given by

$\displaystyle\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\left(e^{2x-x^2}-2\right)\,\mathrm dx$

If you're not familiar with the error function $\mathrm{erf}(x)$ , then you will not be able to find an exact answer. Fortunately, I see this is a question on a calculator based exam, so you can use whatever built-in function you have on your calculator to evaluate the integral. You should get something around 0.5141.

(b) Find the intersections of the line $y=1$ with $y=e^{2x-x^2}$ .

$1=e^{2x-x^2}\implies 0=2x-x^2\implies x=0,x=2$

So the area of $S$ is given by

$\displaystyle\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}(2-1)\,\mathrm dx+\int_{1+\sqrt{1-\ln2}}^2\left(e^{2x-x^2}-1\right)\,\mathrm dx$
$\displaystyle=2\int_0^{1-\sqrt{1-\ln2}}\left(e^{2x-x^2}-1\right)\,\mathrm dx+\int_{1-\sqrt{1-\ln2}}^{1+\sqrt{1-\ln2}}\mathrm dx$

which is approximately 1.546.

(c) The easiest method for finding the volume of the solid of revolution is via the disk method. Each cross-section of the solid is a circle with radius perpendicular to the x-axis, determined by the vertical distance from the curve $y=e^{2x-x^2}$ and the line $y=1$ , or $e^{2x-x^2}-1$ . The area of any such circle is $\pi$ times the square of its radius. Since the curve intersects the axis of revolution at $x=0$ and $x=2$ , the volume would be given by

$\displaystyle\pi\int_0^2\left(e^{2x-x^2}-1\right)^2\,\mathrm dx$