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3 years ago

PLEASE HELP!10 POINTS

Mathematics

1 answer:

zlopas [31]3 years ago

6 0

Hello,

just see my picture

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Challenge: How many flowers, spaced every 4 in., are needed to surround a circular garden with a

meriva

Answer:

314 flowers

Step-by-step explanation:

In order to find how long the perimeter of the circular garden is, we need to find the circumference of the garden.

The circumference of a circle is radius*2*pi. So, in this case, the circumference would be 200*2*3.14. This equals 1256 inches.

Next, we know that flowers are planted every 4 inches, so all we need to do is to find how many 4-inch segments are in 1256 inches.

1256/4=314

Thus, 314 flowers are needed.

6 0

4 years ago

Tom is a cashier at HEB, he makes $13.25 per hour. If he works 22 hours this week what will be his gross income?

mamaluj [8]

Answer:

$291.50

Step-by-step explanation:

For 22 hours, he earns 22 times his hourly pay.

22×$13.25 = $291.50

6 0

4 years ago

What is the equation for the line perpendicular to the line represented by the equation y = 13x – 2 that passes through the poin

laiz [17]

The equation of the perpendicular line is:

y = (-1/13)*x - 87/13

<h3>How to get the line equation?</h3>

For a general linear equation:

y = a*x + b

Another linear equation that is perpendicular to the above one is given by:

y = (-1/a)*x + c

Where a, b, and c are real numbers.

In this case, we want to find a perpendicular line to:

y = 13*x - 2

Then it will be something like:

y = (-1/13)*x + c

To find the value of c, we use the fact that it must pass through the point (4, -7), then:

-7 = (-1/13)*4 + c

Now we can solve that for c:

-7 + 4/13 = c

-87/13

Then the linear equation is:

y = (-1/13)*x - 87/13

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

4 0

2 years ago

What is 233 to 1 significant figure

Mademuasel [1]

233 to 1 significant figure :

233 ≈ 200 (1 s.f.)

6 0

3 years ago

Read 2 more answers

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

4 years ago