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slega [8]
2 years ago
8

Please help, fast‍♀️‍♀️

Mathematics
1 answer:
kramer2 years ago
4 0

Inequalities are used to express unequal expressions.

The inequalities from the word problems are:

  • \mathbf{m - 3.5 \le -2}.
  • \mathbf{0 \ge 2x + 1}.
  • \mathbf{-\frac 12 \ge 2k - 4}

The statements from the inequalities are:

  • -4 is not a solution to \mathbf{x + 8 < -3}
  • -6 is not a solution to \mathbf{10 \le 3 - m}
  • -1 is not a solution to \mathbf{-3x \le -12.5}

  • Graph b represents \mathbf{x > -7}

<h3>The word problems</h3>

<u>1. A number minus 3.5 is less than or equal to -2</u>

The statement can be broken down into the following expressions

\mathbf{A\ number\ minus\ 3.5 \to m - 3.5}

\mathbf{less\ than\ or\ equal\ to\ -2 \to \le -2}

So, when the expressions are brought together, we have:

\mathbf{m - 3.5 \le -2}

<u>2. Zero is greater than or equal to twice a number x plus 1</u>

The statement can be broken down into the following expressions

\mathbf{Zero\ is\ greater\ than\ or\ equal\ to \to 0 \ge }

\mathbf{twice\ a\ number\ x\ plus\ 1\  \to 2x + 1}

So, when the expressions are brought together, we have:

\mathbf{0 \ge 2x + 1}

<u />

<u>3. -1/2 is at least twice a number k minus 4</u>

The statement can be broken down into the following expressions

\mathbf{-\frac 12\ is\ at\ least \to -\frac 12 \ge }

\mathbf{twice\ a\ number\ k\ minus\ 4\  \to 2k - 4}

So, when the expressions are brought together, we have:

\mathbf{-\frac12 \ge 2k - 4}

None of the options is correct

<h3>The solutions</h3>

<u>4. Tell whether -4 is a solution to x + 8 < -3</u>

We have:

\mathbf{x + 8

Subtract 8 from both sides

\mathbf{x + 8 - 8

\mathbf{x

The above inequality means that:

<em>x is less than -11</em>

-4 is not a solution, because -4 is greater than -11

<u>5. Tell whether -6 is a solution to 10 <= 3 - m</u>

We have:

\mathbf{10 \le 3 - m}

Subtract 3 from both sides

\mathbf{10 -3\le 3 - 3 - m}

\mathbf{7 \le  - m}

Multiply both sides by -1 (the inequality sign changes)

\mathbf{-7 \ge m}

Make m the subject

\mathbf{m \le -7}

The above inequality means that:

<em>m is less than -7</em>

-6 is not a solution, because -6 is greater than -7

<u>6. Tell whether -1 is a solution to -3x <= -12.5</u>

We have:

\mathbf{-3x \le -12.5}

Divide both sides by -3 (the inequality sign changes)

\mathbf{x \ge 4\frac16}

The above inequality means that:

<em>x is greater than or equal to </em>\mathbf{4\frac16}<em />

-1 is not a solution, because -1 is less than \mathbf{4\frac16}<em />

<h3>The graph</h3>

The inequality is given as: \mathbf{x > -7}

The less than sign (>) means that:

  • The graph would use an open circle
  • The arrow must point to the right

Only graph b satisfies this condition

Hence, the graph of \mathbf{x > -7} is graph b

Read more about inequalities at:

brainly.com/question/15137133

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PLEASE HELP IM DESPERATE precalculus. also 50 points!! please god help
kompoz [17]

Answer:

For B thru F these options will vary but here how you do it

B. Step 1 Draw the 4 Quadrants.

Then Draw the Triangle in the lower right quadrant which we call quadrant 4. Label the X axis as Adjacent and positive. Label the Y axis as Opposite and negative. Label the Slanted side as the hypotune and AS POSITVE SINCE HYPOTENUSE IS ALWAYS POSITIVE.

FOR C. IN QUADRANT 2, PLOT A POINT AT 0,12 AND AT (-5,0). CONNECT THE DOTS AND IT FORMS A TRIANGLE. Label the X axis as adjacent and negative and the y axis as positve and opposite and label the slanted side hypotunese and positive.

FOR D Draw a straight line along the x axis then draw a slanted line passing through (5,-1). In between them put the theta symbol in there.

The labeling is the same for C.

For E. Since tan must be positve and secant must be positve, our triangle must be in the 1st Quadrant. Draw any right triangle as long it is in the first quadrant

The x axis is adjacent and positve. The y axis is opposite and positve. The hypotenuse is the slanted side and it is positve.

For F. Since sin is negative and cos is positve the triangle is in the 4th quadrant. Draw any triangle in the 4th quadrant and the labeling is the same for Problem B.

2. We can find the sec of cos by flipping cosine.

\cos( \frac{x}{y} )  =  \sec( \frac{y}{x} )

\cos( \frac{1}{2} )  =  \sec(2 )

Sec is 2.

To find the cotangent, first let find the sin then tan.

We can use the identity

\cos( {theta}^{2} )  +  \sin( {theta}^{2} )  = 1

Let plug in the number

\cos( \frac{ {1}^{2} }{{2}^{2} } )  +  \sin(x {}^{2} )  = 1

\cos( \frac{1}{4} )  +  \sin(x {}^{2} )  = 1

\sin(x {}^{2} )  =   1 -  \frac{1}{4}

\sin(x {}^{2} )  =  \frac{3}{4}

\sin(x)  =  \frac{ \sqrt{3} }{ \sqrt{4} }

\sin(x)  =  \frac{ \sqrt{3} }{2}

Since sin is negative, sin x=

-  \frac{ \sqrt{3} }{2}

Now let apply the formula

\frac{ \sin(x) }{ \cos(x) }  =  \tan(x)

\frac{ \frac{ -  \sqrt{3} }{2} }{ \frac{1}{2} }  =  \tan(x)

-  \sqrt{3}

Now let find cotangent we can the reciprocal of

tan.

\tan=  -  \sqrt{3}

\cot =   - \frac{1}{ \sqrt{3} }

Rationalize denominator

\frac{ - 1}{ \sqrt{3} }  \times   \frac{ \sqrt{3} }{ \sqrt{3} }  =  \frac{ \sqrt -{ 3} }{3}

cotangent equal

-  \frac{ \sqrt{ 3} }{3}

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PLS HELP ME ON TTHIS QUESTION I WILL MARK YOU AS BRAINLIEST IF YOU KNOW THE ANSWER PLS GIVE ME A STEP BY STEP EXPLANATION WHILE
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Answer:

  D (compute the mean)

Step-by-step explanation:

A box plot is a visual representation of the 5-number summary of a set of data. It is plotted against a number line.

The 5-number summary of the data set is most easily determined from an ordered set of the data, so ordering the data is usually the first step. Since the 5-number summary is plotted against a number line, the next steps are finding the 5-number summary, and creating a suitable number line.

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The business of adding the data and dividing by the number of data items is a description of computing the mean of the data. That value is not part of the 5-number summary, and is not needed in the creation of the box plot.

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in such a case the easiest approach might be using the point-slope form of a line equation :

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so, in our case that would be

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artcher [175]

Answer:

None

Step-by-step explanation:

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