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3 years ago

Please help! S is the segment bisector of RT. Find RT.

Mathematics

1 answer:

kap26 [50]3 years ago

5 0

Answer:

Step-by-step explanation:

algebra and bisector means its half

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1. Simplify: (6-√5)(5+√5)

aniked [119]

150 because 6 times 5 is 30 then you multiply whats inside the square root thing so you'd get 25, 25 is a perfect square which simplifies to 5 then 5 times 30 is 150.

7 0

3 years ago

Read 2 more answers

Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

3 0

3 years ago

12 boys collected 16 aluminum cans each 15 girls collected 14 aluminum cans how meany cans did the girls collected than the boys

ANEK [815]

Answer:

Step-by-step explanation:

Since the boys collected 16 cans each and there are 12 of them it means they collected 16*12= 192 cans in total

there are 15 girls and each collected 14 cans. Therefore they collected 15*14=210 cans in total

210-192=18 cans the girls collected more than the boys

5 0

3 years ago

The area of the right triangle shown is 24 square feet. Which equations can be used to find the lengths of the legs of the trian

Gwar [14]

The equations to calculate the legs are 0.5(x)(x + 2) = 24, x^2 + 2x - 48 = 0 and x^2 + (x + 2)^2 = 100

<h3>How to determine the legs of the triangle?</h3>

The complete question is in the attached image

The given parameters are:

Area = 24

Legs = x and x + 2

The area of the triangle is calculated as:

Area = 0.5 * Base * Height

This gives

0.5 * x * (x + 2) = 24

So, we have:

0.5(x)(x + 2) = 24

Divide through by 0.5

(x)(x + 2) = 48

Expand

x^2 + 2x = 48

Subtract 48 from both side

x^2 + 2x - 48 = 0

Hence, the equations to calculate the legs are 0.5(x)(x + 2) = 24, x^2 + 2x - 48 = 0 and x^2 + (x + 2)^2 = 100