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2 years ago

What is the length of segment bd

Mathematics

2 answers:

gizmo_the_mogwai [7]2 years ago

8 0

Answer:

10 if not mistaking, sorry if im wrong

Step-by-step explanation:

nata0808 [166]2 years ago

3 0

Answer:

BD = 16

Step-by-step explanation:

AB and AC are both radii

AB = AC = 6 + 4 = 10

BE = 8 because it's the leg of a right triangle with 6² + 8² = 10²

BD = 2 * BE = 2 * 8 = 16

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The lengths of the sides of a rectangle are shown. Write

Simora [160]

Answer:

4t+6

Step-by-step explanation:

To find the perimeter of the rectangle you add the length of the sides.

Perimeter= t+ t+ t+ 3+ t+ 3= <u>4t+6</u>

Perimeter=2(t)+ 2(t+3)= 2t+ 2t+ 6= <u>4t+6</u>

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2 years ago

Which of the following systems has infinitely many solutions?

stich3 [128]

Answer:

Step-by-step explanation:

4 0

2 years ago

Ples help me find slant assemtotes

FrozenT [24]

A polynomial asymptote is a function $p(x)$ such that

$\displaystyle\lim_{x\to\pm\infty}(f(x)-p(x))=0$

$(y+1)^2=4xy\implies y(x)=2x-1\pm2\sqrt{x^2-x}$

Since this equation defines a hyperbola, we expect the asymptotes to be lines of the form $p(x)=ax+b$ .

Ignore the negative root (we don't need it). If $y=2x-1+2\sqrt{x^2-x}$ , then we want to find constants $a,b$ such that

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

We have

$\sqrt{x^2-x}=\sqrt{x^2}\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=|x|\sqrt{1-\dfrac1x}$
$\sqrt{x^2-x}=x\sqrt{1-\dfrac1x}$

since $x\to\infty$ forces us to have $x>0$ . And as $x\to\infty$ , the $\dfrac1x$ term is "negligible", so really $\sqrt{x^2-x}\approx x$ . We can then treat the limand like

$2x-1+2x-ax-b=(4-a)x-(b+1)$

which tells us that we would choose $a=4$ . You might be tempted to think $b=-1$ , but that won't be right, and that has to do with how we wrote off the "negligible" term. To find the actual value of $b$ , we have to solve for it in the following limit.

$\displaystyle\lim_{x\to\infty}(2x-1+2\sqrt{x^2-x}-4x-b)=0$

$\displaystyle\lim_{x\to\infty}(\sqrt{x^2-x}-x)=\frac{b+1}2$

We write

$(\sqrt{x^2-x}-x)\cdot\dfrac{\sqrt{x^2-x}+x}{\sqrt{x^2-x}+x}=\dfrac{(x^2-x)-x^2}{\sqrt{x^2-x}+x}=-\dfrac x{x\sqrt{1-\frac1x}+x}=-\dfrac1{\sqrt{1-\frac1x}+1}$

Now as $x\to\infty$ , we see this expression approaching $-\dfrac12$ , so that

$-\dfrac12=\dfrac{b+1}2\implies b=-2$

So one asymptote of the hyperbola is the line $y=4x-2$ .

The other asymptote is obtained similarly by examining the limit as $x\to-\infty$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-ax-b)=0$

$\displaystyle\lim_{x\to-\infty}(2x-2x\sqrt{1-\frac1x}-ax-(b+1))=0$

Reduce the "negligible" term to get

$\displaystyle\lim_{x\to-\infty}(-ax-(b+1))=0$

Now we take $a=0$ , and again we're careful to not pick $b=-1$ .

$\displaystyle\lim_{x\to-\infty}(2x-1+2\sqrt{x^2-x}-b)=0$

$\displaystyle\lim_{x\to-\infty}(x+\sqrt{x^2-x})=\frac{b+1}2$

$(x+\sqrt{x^2-x})\cdot\dfrac{x-\sqrt{x^2-x}}{x-\sqrt{x^2-x}}=\dfrac{x^2-(x^2-x)}{x-\sqrt{x^2-x}}=\dfrac
 x{x-(-x)\sqrt{1-\frac1x}}=\dfrac1{1+\sqrt{1-\frac1x}}$

This time the limit is $\dfrac12$ , so

$\dfrac12=\dfrac{b+1}2\implies b=0$

which means the other asymptote is the line $y=0$ .

4 0

2 years ago

Hi, please help if your are able please

Simora [160]

The answer is -28

hope this helps :)
(can you please mark me as brainliest?)

8 0

2 years ago

Read 2 more answers

K is the midpoint of Line JL, JL=4x-2 and JK=7. find X

ipn [44]

X=4. If k is the midpoint then JK is equal to KL. Since JK is 7 then KL is also 7 meaning JK is 14. JK is represented by the expression 4x-2 so you just set that expression equal to 14 and solve for x. Hope this helped!

7 0

2 years ago