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3 years ago

True or false???????????

Mathematics

2 answers:

Sedbober [7]3 years ago

6 0

The answer is true because they are both positive numbers

AleksandrR [38]3 years ago

5 0

True.
no matter what value you put in x, it will produce positive real numbers.

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What is the sum of two numbers is 40 and their difference is 10

Butoxors [25]

Answer:

The two numbers are 25 and 15.

Step-by-step explanation:

Let's find out the two numbers, this way:

x = first number

y = second number

This is the equations system:

x + y = 40

x - y = 10

Solving for x in the first equation:

x + y = 40

x = 40 - y

Solving for y in the second equation:

40 - y - y = 10

40 - 2y = 10

-2y = 10 - 40

-2y = - 30

y = -30/-2 = 15

Solving for x:

x + y = 40

x + 15 = 40

x = 40 - 15

x = 25

The two numbers are 25 and 15.

8 0

4 years ago

Tools -

gregori [183]

Answer:

Option A . 1708 bags

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Find the volume of the silo

The volume of the silo is equal to the volume of a cylinder plus the volume of a cone

we have

---> the radius is half the diameter

substitute

Remember that

The silo is approximately 85% full

The volume of grain in the silo is

step 2

Find the number of bags of grain

Divide the volume of grain in the silo by the volume of one sack

7 0

3 years ago

Calculate the limit of the function with L'Hospital rule

mr_godi [17]

Answer:

L=24

Step-by-step explanation:

L'Hopital's Rule for Evaluating Limits:

Rule is that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ takes $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form, then,

$\lim_{x \to a} \frac{f(x)}{g(x)}= \lim_{x \to a} \frac{f'(x)}{g'(x)}$

where $f'(x)=\frac{df(x)}{dx}$ and $g'(x)=\frac{dg(x)}{dx}$

Now coming to the problem,

$L= \lim_{x \to \frac{\pi}{6} } \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3} )}$

Here $f(x)=cot^{3}x-3cotx$ and $g(x)=cos(x+\frac{\pi}{3} )$

Substituting $x=\frac{\pi}{6}$ in f(x) and g(x),

$f(\frac{\pi}{6})=cot^{3}\frac{\pi}{6}-3cot\frac{\pi}{6}\\=3\sqrt{3}-3\sqrt{3}\\ =0$

$g(\frac{\pi}{6})=cos(\frac{\pi}{6}+\frac{\pi}{3})\\=cos\frac{\pi}{2}\\=0$

Since L takes the form $\frac{0}{0}$ , using l'hopital's rule

$L= \lim_{x \to \frac{\pi}{6}} \frac{cot^{3}x-3cotx}{cos(x+\frac{\pi}{3})}= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}x(-cosec^{2}x)-3(-cosec^{2}x)}{-sin(x+\frac{\pi}{3})}$

now substituting $x=\frac{\pi}{6}$ ,

$L= \lim_{x \to \frac{\pi}{6}} \frac{3cot^{2}\frac{\pi}{6}(-cosec^{2}\frac{\pi}{6})-3(-cosec^{2}\frac{\pi}{6})}{-sin(\frac{\pi}{6}+\frac{\pi}{3})}\\=\frac{3*3^{2}(-2^{2})+3(2^{2})}{-1}\\=24$

6 0

3 years ago

Not sure about the rest need help

never [62]

I tried to type the answer it said I was wrong

4 0

3 years ago

If a reflection over the line y=x occurs, then the rule for this reflection is

Anni [7]

If you reflect over the line y = -x, the x-coordinate and y-coordinate change places and are negated (the signs are changed).

7 0

3 years ago