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AnnyKZ [126]
3 years ago
11

Sandra​, who is paid​ time-and-a-half for hours worked in excess of 40​ hours, had gross weekly wages of $ 608 or 56 hours worke

d. What is her regular hourly​ rate?
Mathematics
1 answer:
vodomira [7]3 years ago
4 0
Time-and-a-half doesn't make enough sense to answer the question. If you fix it, or comment below the correction I will gladly help.
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HALP PLEASE
Sergeeva-Olga [200]

Answer:

A: -2

Step-by-step explanation:

Hopefully this helps!

5 0
3 years ago
M Enter the number that belongs in<br> the green box
gladu [14]

Answer:

m<C = 180 - (90+32)

= 180 - 122

= 58°

4 0
2 years ago
The equation line l is given by the equation y= 7/4x+3. line m goes through the point ( –1,4) and is perpendicular to line l whi
mr_godi [17]

We determine line m as follows:

*First, by theorem we have the following:

m_1=-\frac{1}{m_2}

Here m1 & m2 are the slopes of two perpendicular lines. For all lines that are perpendicular that is true, so we calculate the slope of line m using the slope of the function given [Which has a slope of 7/4]:

m_1=-\frac{1}{\frac{7}{4}}\Rightarrow m_1=-\frac{4}{7}

So, the slope of line m is -4/7. Now, using this slope and the point (-1, 4) we replace in the following expression:

y-y_1=m_1(x-x_1)

Here x1, y1 & m1 are the x-component of the point, the y-component of the point, and the slope of the line respectively, so we replace and solve for y:

y-4=-\frac{4}{7}(x-(-1))\Rightarrow y-4=-\frac{4}{7}x-\frac{4}{7}\Rightarrow y=-\frac{4}{7}x+\frac{24}{7}

And that last function of y is the line m.

8 0
1 year ago
A rectangle is plotted on the coordinate grid. It has vertices at (-3, 5), (4, 5), (4, -1), and (-3, -1). Which choices below ar
zhannawk [14.2K]

A rectangle is plotted on the coordinate grid. It has vertices at (-3, 5), (4, 5), (4, -1), and (-3, -1). Which choices below are the dimensions of the rectangle?

Answer

(-3,5)

3 0
2 years ago
Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane
liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

Answer:

Rate = 0.935042^\circ /cm

Step-by-step explanation:

Given

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

T(x,y) =x\sin2y

r = 1m

v = 2m/s

Express the given point P as a unit tangent vector:

P = (\frac{1}{2}, \frac{\sqrt 3}{2})

u = \frac{\sqrt 3}{2}i - \frac{1}{2}j

Next, find the gradient of P and T using: \triangle T = \nabla T * u

Where

\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})}  = (sin \sqrt 3)i + (cos \sqrt 3)j

So: the gradient becomes:

\triangle T = \nabla T * u

\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] *  [\frac{\sqrt 3}{2}i - \frac{1}{2}j]

By vector multiplication, we have:

\triangle T = (sin \sqrt 3)*  \frac{\sqrt 3}{2} - (cos \sqrt 3)  \frac{1}{2}

\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)

\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5

\triangle T = 0.935042

Hence, the rate is:

Rate = \triangle T = 0.935042^\circ /cm

3 0
2 years ago
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