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3 years ago

Choose the inequality that could be used to solve the following problem.

Mathematics

2 answers:

kirill115 [55]3 years ago

7 0

Answer:

Correct choice is C

Step-by-step explanation:

Let x be unknown number.

Three times a number x is $3\cdot x=3x.$

Three times a number is no less than negative six means that the number $3x$ can be equal to -6, can be greater than -6, but cannot be less than -6.

Thus, the inequality that could be used to solve the problem is

$3x\ge -6.$

Maslowich3 years ago

4 0

Answer:

3x ≥ -6

Step-by-step explanation:

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In a candy bar, 165 calories are from sugar. There are 250 calories in the candy bar. What percent of the calories are from suga

icang [17]

You can simply find out the percentage of the calories from the sugar with this formula :

165/250 x 100%

You will find out that the percentage would be 66 %

hope this helps

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3 years ago

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Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

5 0

3 years ago

How many solutions can be found for the linear equation? 4(x + 5) - 5 = 8x + 18/ 2

mylen [45]

Answer: The correct option is A.

Explanation:

The given equation is,

$4(x+5)-5=\frac{8x+18}{2}$

Multiply both sides by 2.

$2[4(x+5)-5]=\frac{8x+18}{2}$

$8(x+5)+2(-5)=8x+18$

$8x+40-10=8x+18$

$8x+30=8x+18$

Subtract both sides by 8x.

$30=18$

This statement is false for any value of x, therefore the system of equation have no solution and option A is correct.

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3 years ago

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What is the longest side of a right angled triangle called?

Anarel [89]

Answer:

The hypotenuse

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3 years ago

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Identify the outliers of the data set. Then determine if the outlier increases or decreases the value of the mean.

Juli2301 [7.4K]

<h3>Answer: D) 76; increases</h3>

========================================================

Explanation:

The main group or cluster of values spans from 32 to 42 (inclusive).

Then off on its own is the value 76, which we consider an outlier. This value is fairly far from the group. As a rule, large outliers pull on the arithmetic mean to make it larger than it should be. Think of it like the outlier pulling on the mean as if it was done through a magnet or gravitational pull.

Similarly, small outliers pull the mean to the left to make it smaller than it should be. We don't have any small outliers in this case.

--------------

Let's consider the set

A = {32, 34, 35, 36, 37, 38, 38, 41, 42, 42, 46}

where I've sorted the values and I replaced 76 with 46.

Computing the mean of set A gets us

(32+34+35+36+37+38+38+41+42+42+46)/11 = 38.27 approximately

--------------

Now let's form this set

B = {32, 34, 35, 36, 37, 38, 38, 41, 42, 42, 76}

which is the original set your teacher gave you. It's nearly identical to set A, except that the 46 is now 76 again.

Compute the mean of set B

(32+34+35+36+37+38+38+41+42+42+76)/11 = 41

---------------

Set A has a mean of roughly 38.72 and set B has a mean of 41. We see that the mean has increased.

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3 years ago