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2 years ago

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steve gaimes is paid overtime for all time worked past 40 hours in a week. his regular time pay rate is $12 an hour and his over

time rate is $18 an hour last week steve worked 47.3 hours how many regular time hours did steve work last week

1 answer:

Elena-2011 [213]2 years ago

6 0

Answer:

40 hours

Step-by-step explanation:

47.3-40= 7.3

He worked 7.3 hours overtime meaning he worked 40 regular hours.

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Trapezoid A B C D is shown. Sides A B and D C are parallel. Sides A D and B C are congruent. Angle B is (3 x) degrees and angle

photoshop1234 [79]

Answer: The value of x in trapezoid ABCD is 15

Step-by-step explanation: The trapezoid as described in the question has two bases which are AB and DC and these are parallel. Also it has sides AD and BC described as congruent (that is, equal in length or measurement). These descriptions makes trapezoid ABCD an isosceles trapezoid.

One of the properties of an isosceles trapezoid is that the angles on either side of the two bases are equal. Since line AD is equal to line BC, then angle D is equal to angle C. It also implies that angle A is equal to angle B.

With that bit of information we can conclude that the angles in the trapezoid are identified as 3x, 3x, 9x and 9x.

Also the sum of angles in a quadrilateral equals 360. We can now express this as follows;

3x + 3x + 9x + 9x = 360

24x = 360

Divide both sides of the equation by 24

x = 15

Therefore, in trapezoid ABCD

x = 15

4 0

2 years ago

Read 2 more answers

Ebola has an Ro of 2. How<br> many third-wave cases can<br> doctors expect?

stellarik [79]

Answer:

4 I guess

Step-by-step explanation:

Because

3-1=2

2^2=2*2=4

4 0

2 years ago

Lera brought a watermelon that weighed b kg to a party. Nick brought a box of candies. How much is the weight of the box of cand

Dennis_Churaev [7]

Answer:

1.5b (kg)

Step-by-step explanation:

Let's begin by listing out the variables we were given:

weight of the watermelon = b (kg),

weight of watermelon = (2/5) * weight of candies

weight of candies = 1 ÷ (2/5) = 1 ÷ 0.4

weight of candies = 2.5b (kg)

How much is the weight of the box of candies greater than the weight of the watermelon is given by:

weight of the box of candies - weight of watermelon= 2.5b - b = <u>1.5b</u> (kg)

<u>Therefore, the weight of the box of candies is greater than the weight of the watermelon by 1.5b (kg) </u>

7 0

2 years ago

Given that a function, g, has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8, select the st

notka56 [123]

Options

(A)g(5) = 12
(B)g(1) = -2
(C)g(2) = 4
(D)g(3) = 18

Answer:

(D)g(3) = 18

Step-by-step explanation:

Given that the function, g, has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8

Then the following properties must hold

The value(s) of x must be between -1 and 4
The values of g(x) must be between 0 and 18.
g(-1)=2
g(2)=9

We consider the options and state why they are true or otherwise.

<u>Option A: g(5)=12</u>

The value of x=5. This contradicts property 1 stated above. Therefore, it is not true.

<u>Option B: g(1) = -2 </u>

The value of g(x)=-2. This contradicts property 2 stated above. Therefore, it is not true.

<u>Option C: g(2) = 4 </u>

The value of g(2)=4. However by property 4 stated above, g(2)=9. Therefore, it is not true.

<u>Option D: g(3) = 18</u>

This statement can be true as its domain is in between -1 and 4 and its range is in between 0 and 18.

Therefore, Option D could be true.

3 0

2 years ago

Learning Task 3. Find the equation of the line. Do it in your notebook.

Wewaii [24]

Answer:

1) The equation of the line in slope-intercept form is $y = 5\cdot x +9$ . The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ . The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) The equation of the line in slope-intercept form is $y = 3\cdot x +4$ . The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ . The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ . The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

Step-by-step explanation:

1) We begin with the slope-intercept form and substitute all known values and calculate the y-intercept: ( $m = 5$ , $x = -1$ , $y = 4$ )

$4 = (5)\cdot (-1)+b$

$4 = -5 +b$

$b = 9$

The equation of the line in slope-intercept form is $y = 5\cdot x +9$ .

Then, we obtain the standard form by algebraic handling:

$-5\cdot x + y = 9$

The equation of the line in standard form is $-5\cdot x + y = 9$ .

2) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = 3$ , $y_{1} = 4$ , $x_{2} = -2$ , $y_{2} = 2$ )

$3\cdot m + b = 4$ (Eq. 1)

$-2\cdot m + b = 2$ (Eq. 2)

From (Eq. 1), we find that:

$b = 4-3\cdot m$

And by substituting on (Eq. 2), we conclude that slope of the equation of the line is:

$-2\cdot m +4-3\cdot m = 2$

$-5\cdot m = -2$

$m = \frac{2}{5}$

And from (Eq. 1) we find that the y-Intercept is:

$b=4-3\cdot \left(\frac{2}{5} \right)$

$b = 4-\frac{6}{5}$

$b = \frac{14}{5}$

The equation of the line in slope-intercept form is $y = \frac{2}{5}\cdot x +\frac{14}{5}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{2}{5}\cdot x +y = \frac{14}{5}$

$-2\cdot x +5\cdot y = 14$

The equation of the line in standard form is $-2\cdot x +5\cdot y = 14$ .

3) By using the slope-intercept form, we obtain the equation of the line by direct substitution: ( $m = 3$ , $b = 4$ )

$y = 3\cdot x +4$

The equation of the line in slope-intercept form is $y = 3\cdot x +4$ .

Then, we obtain the standard form by algebraic handling:

$-3\cdot x +y = 4$

The equation of the line in standard form is $-3\cdot x +y = 4$ .

4) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -3$ , $y_{1} = 0$ , $x_{2} = 0$ , $y_{2} = 6$ )

$-3\cdot m + b = 0$ (Eq. 3)

$b = 6$ (Eq. 4)

By applying (Eq. 4) on (Eq. 3), we find that the slope of the equation of the line is:

$-3\cdot m+6 = 0$

$3\cdot m = 6$

$m = 2$

The equation of the line in slope-intercept form is $y = 2\cdot x + 6$ .

Then, we obtain the standard form by algebraic handling:

$-2\cdot x +y = 6$

The equation of the line in standard form is $-2\cdot x +y = 6$ .

5) We begin with a system of linear equations based on the slope-intercept form: ( $x_{1} = -1$ , $y_{1} = -2$ , $x_{2} = 5$ , $y_{2} = 3$ )

$-m+b = -2$ (Eq. 5)

$5\cdot m +b = 3$ (Eq. 6)

From (Eq. 5), we find that:

$b = -2+m$

And by substituting on (Eq. 6), we conclude that slope of the equation of the line is:

$5\cdot m -2+m = 3$

$6\cdot m = 5$

$m = \frac{5}{6}$

And from (Eq. 5) we find that the y-Intercept is:

$b = -2+\frac{5}{6}$

$b = -\frac{7}{6}$

The equation of the line in slope-intercept form is $y = \frac{5}{6}\cdot x -\frac{7}{6}$ .

Then, we obtain the standard form by algebraic handling:

$-\frac{5}{6}\cdot x +y =-\frac{7}{6}$

$-5\cdot x + 6\cdot y = -7$

The equation of the line in standard from is $-5\cdot x + 6\cdot y = -7$ .

6 0

1 year ago