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3 years ago

Name the corresponding angle or side. a, b, c, or D

Mathematics

1 answer:

AleksandrR [38]3 years ago

4 0

Answer:

I think it is B. LM

Step-by-step explanation:

i think this because the shape just looks like it is rotated.

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ANY MATH EXPERTS PLS HELP RN I GOT 30 MIN LEFT I PUT 100 POINTS

PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

x y

0 6

24 0

Points are (0,6) and (24,0)

6 0

3 years ago

Read 2 more answers

Which equation can be solved using the expression StartFraction negative 3 plus-or-minus StartRoot (3) squared + 4 (10) (2) EndR

katen-ka-za [31]

Answer: B. 2 = 3x + 10x2

Step-by-step explanation:

This is the concept of quadratic equations; We required to find the type of equation that can be solved using the model that has been used to solve the equation such that the answer is:

[-3+-sqrt(3^2+4(10)(2))]/(2(10))

The formual that was applied here was a quadratic formula given by:

x=[-b+\-sqrt(b^2-4ac)]/2a

whereby from the our substituted values above,

a=10,b=3 and c=-2

such that the quadratic equation will be:

10x^2+3x-2

3 0

3 years ago

Read 2 more answers

Is the square root of infinity still infinity?.

hram777 [196]

The square root of infinity is still infinity, or rather, undefined or indeterminate. For example, if you take the square root of a big number, you will still have a big number as a result, now if you make the number even bigger, the square root will also get bigger, now imagine what will happen if this number is infinite.

6 0

3 years ago

Solve only if you know the solution and show work.

SashulF [63]

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx$

For the remaining integral, let $t=\tan\dfrac x2$ . Then

$\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}$
$\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}$

and

$\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt$

Now the integral is

$\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt$

The first integral is trivial, so we'll focus on the latter one. You have

$\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt$

Decompose the integrand into partial fractions:

$\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}$

so you have

$\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$

which are all standard integrals. You end up with

$\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt$
$=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C$
$=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C$
$=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C$

To try to get the terms to match up with the available answers, let's add and subtract $\ln\left|1+\tan\dfrac x2\right|$ to get

$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C$
$2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C$

which suggests A may be the answer. To make sure this is the case, show that

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1$

You have

$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}$
$\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}$

So in the corresponding term of the antiderivative, you get

$\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|=\ln\left|\dfrac2{\cos x+\sin x+1}\right|$
$=\ln2-\ln|\cos x+\sin x+1|$

The $\ln2$ term gets absorbed into the general constant, and so the antiderivative is indeed given by A,

$\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=2x+5\ln\left|1+\tan\dfrac x2\right|-\ln|\cos x+\sin x+1|+C$

5 0

2 years ago

I need help with these 2

Slav-nsk [51]

Answer:

Step-by-step explanation: Hey for the second one I got y = -3/4x -0.5 I'm not sure if its correct though sorry if not
ill try my best to explain my solution though
1. From the parallel equation (3x + 4y = 12) all we need to do is find the slope

So the easiest way to do so is to put the said equation in <u>y-intercept </u>form

y=mx +b
m= slope

b= y intercept

so 1. 3x + 4y = 12

4y = 12-3x

divide that by 4 to get only y

y=3-3/4x

-3/4 is our slope

y=-3/4x+b

than we have a point -2, -2

if we put -2 for y

-2=-3/4x+b

and then we put our -2 for x
-2 = -3/4 * -2 + b

-2 = -1.5 +b
b=-0.5

Answer : y=-3/4x-0.5

4 0

1 year ago