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Drupady [299]
3 years ago
13

Find the initial point of the vector that is equivalent to vector u = (1,2) and whose terminal point is A(2,0)

Mathematics
1 answer:
ahrayia [7]3 years ago
7 0
Subtracting the functions  

2,0-1,-2-----(1,-2)
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Evelyn went shopping with half of her monthly allowance. She spent $39.95 on a dress and 3/7 of the remainder on some accessorie
Gnoma [55]

Answer:

Evelyn monthly allowance is $187.

Step-by-step explanation:

Let the monthly allowance be 'x'.

Given:

Money taken with for shopping = \frac{x}{2}

Money spent of dress = $39.95

Also Given:

3/7 of the remainder on some accessories.

Money spent on accessories is equal to \frac{3}{7} times Difference of monthly allowance and Money spent of dress.

framing in equation form we get;

Money spent on accessories = \frac{3}{7}\times(\frac{x}{2}-39.95)

Also Given:

Money left after shopping = $30.60

Now we know that;

Money taken with for shopping is equal to sum of Money spent of dress and Money spent on accessories and Money left after shopping.

framing in equation form we get;

\frac{x}{2}=39.95+\frac{3}{7}(\frac{x}{2}-39.95)+30.60\\\\\frac{x}{2}=39.95+\frac{3x}{14}- 17.12 +30.60

Combining like terms we get;

\frac{x}{2}-\frac{3x}{14}=39.95-17.12+30.60\\\\\frac{x}{2}-\frac{3x}{14}= 53.43

Now we will make denominator common using L.C.M

\frac{x\times 7}{2\times7}-\frac{3x\times1}{14\times1}= 53.43\\\\\frac{7x}{14}-\frac{3x}{14}= 53.43\\\\\frac{7x-3x}{14}=53.43\\\\\frac{4x}{14}=53.43\\\\4x=53.43\times14\\\\x= \frac{53.43\times14}{4} = \$187

Hence Evelyn monthly allowance is $187.

3 0
4 years ago
Peter, Bridget and Caroline share some sweets in the ratio 2:5:2. Peter gets 14 sweets. How many did Bridget get?
babunello [35]

Answer:

35

Step-by-step explanation:

The 14 sweets Peter gets is 2 parts of the ratio, then

14 ÷ 2 = 7 ← number of sweets in 1 part of the ratio , then

5 parts = 5 × 7 = 35 ← number of sweets Bridget gets

7 0
3 years ago
Which of the following graphs shows the solution set for the inequality below? 3|x + 1| < 9
Bas_tet [7]

Step-by-step explanation:

The absolute value function is a well known piecewise function (a function defined by multiple subfunctions) that is described mathematically as

                                 f(x) \ = \ |x| \ = \ \left\{\left\begin{array}{ccc}x, \ \text{if} \ x \ \geq \ 0 \\ \\ -x, \ \text{if} \ x \ < \ 0\end{array}\right\}.

This definition of the absolute function can be explained geometrically to be similar to the straight line   \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  , however, when the value of x is negative, the range of the function remains positive. In other words, the segment of the line  \textbf{\textit{y}} \ = \ \textbf{\textit{x}}  where \textbf{\textit{x}} \ < \ 0 (shown as the orange dotted line), the segment of the line is reflected across the <em>x</em>-axis.

First, we simplify the expression.

                                             3\left|x \ + \ 1 \right| \ < \ 9 \\ \\ \\\-\hspace{0.2cm} \left|x \ + \ 1 \right| \ < \ 3.

We, now, can simply visualise the straight line,  y \ = \ x \ + \ 1 , as a line having its y-intercept at the point  (0, \ 1) and its <em>x</em>-intercept at the point (-1, \ 0). Then, imagine that the segment of the line where x \ < \ 0 to be reflected along the <em>x</em>-axis, and you get the graph of the absolute function y \ = \ \left|x \ + \ 1 \right|.

Consider the inequality

                                                    \left|x \ + \ 1 \right| \ < \ 3,

this statement can actually be conceptualise as the question

            ``\text{For what \textbf{values of \textit{x}} will the absolute function \textbf{be less than 3}}".

Algebraically, we can solve this inequality by breaking the function into two different subfunctions (according to the definition above).

  • Case 1 (when x \ \geq \ 0)

                                                x \ + \ 1 \ < \ 3 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 3 \ - \ 1 \\ \\ \\ \-\hspace{0.9cm} x \ < \ 2

  • Case 2 (when x \ < \ 0)

                                            -(x \ + \ 1) \ < \ 3 \\ \\ \\ \-\hspace{0.15cm} -x \ - \ 1 \ < \ 3 \\ \\ \\ \-\hspace{1cm} -x \ < \ 3 \ + \ 1 \\ \\ \\ \-\hspace{1cm} -x \ < \ 4 \\ \\ \\ \-\hspace{1.5cm} x \ > \ -4

           *remember to flip the inequality sign when multiplying or dividing by

            negative numbers on both sides of the statement.

Therefore, the values of <em>x</em> that satisfy this inequality lie within the interval

                                                     -4 \ < \ x \ < \ 2.

Similarly, on the real number line, the interval is shown below.

The use of open circles (as in the graph) indicates that the interval highlighted on the number line does not include its boundary value (-4 and 2) since the inequality is expressed as "less than", but not "less than or equal to". Contrastingly, close circles (circles that are coloured) show the inclusivity of the boundary values of the inequality.

3 0
3 years ago
A regular pentagon with side length s has perimeter 5s. Dakota built a dog pen in her backyard for her dog's new puppies. The pe
sashaice [31]
The answer is 243 hope this helps
4 0
3 years ago
Question 2(Multiple Choice Worth 1 points)
DanielleElmas [232]

Answer: 19

Step-by-step explanation:

g(4) = 4² + 3

g(4) = 16 + 3

g(4) = 19

5 0
2 years ago
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