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3 years ago

Carlos divided 56 1/4 ounces of granola into bags weighing 3 3/4 ounces. How many bags can he make?

Mathematics

2 answers:

Vikki [24]3 years ago

7 0

56.25 / 3.75 = 15 bags

Setler [38]3 years ago

3 0

56.25/3.75=15 bags. Hope this helps!

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Need help with this asap!!

Bingel [31]

Answer:

≈ 53.7° (Option 3)

Step-by-step explanation:

In the figure given above , the lengths of other two sides (excluding hypotenuse) are known to us. So , it would be convenient to use 'tan'.

Length of side opposite to angle 'x' = 15 cm

Length of side adjacent to angle 'x' = 11 cm

Using 'tan' formula - $\tan x = \frac{Side \: opposite \: to \: x}{Side \: adjacent \: to \: x}$ , lets solve this.

$=> \tan x = \frac{15}{11}$

Using inverse tan function , lets find x.

$=> x = {\tan}^{-1}(\frac{15}{11} )$ ≈ 53.74°

7 0

3 years ago

In the picture below, line PQ is parallel to line RS, and the lines are cut by a transversal, line TU. The transversal is not pe

natta225 [31]

Answer:<W=<E the answer is W and E

Step-by-step explanation:

i got it right on study island

4 0

3 years ago

How do i set this up using a trig equation

PolarNik [594]

In this use SOHCAHTOA and according to this we will use CAH on your calculator do cos(72)times 4= 1.23 should be the answer

3 0

3 years ago

What is the probability of you picking a weekend day from the days of the week?

svet-max [94.6K]

Answer:

3/7

Step-by-step explanation:

its 3/7 because there are 3 weekend days and there are 7 days in total in one week, so out of the week there is 3 weekend days.

i hope this helps :)

6 0

3 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

3 years ago