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victus00 [196]
3 years ago
7

Ali had his car repaired at Ace Auto. He was charged AED 50 per hour of labor plus AED 150 for parts. His total bill for the rep

air was AED 375. How many hours was Ali charged for?
Mathematics
1 answer:
AnnZ [28]3 years ago
6 0

Step-by-step explanation:

x = hours Ali was charged for

150 + 50 x = 375

- 150 -150

50x = 225

÷ 50. ÷50

x = 4.5 hours

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Need help with math problem if do get 5 star
777dan777 [17]

Answer:

d) 63 square units

Step-by-step explanation:

x = 3 units (base - b)

y = 6 units (width - w)

h = 7 units (length and height - l and h)

  • Area of a rectangle: l*w
  • Area of a triangle: 1/2*b*h

For the triangles:

Area of a triangle: 1/2*b*h

A = 1/2 * 3 * 7

A = 1/2 * 21

A = 10.5

*Since there are two triangles, it would be 10.5 * 2 = 21

For the rectangle:

Area of a rectangle: l*w

A = 7 * 6

A = 42

Now, we add them together:

A = 42 + 21

A = 63 square units

Therefore, the area of the parallelogram is 63 square units

Hope this helps!

6 0
2 years ago
The (Figure 1) shows two thin beams joined at right angles. The vertical beam is 19.0 kg and 1.00 m long and the horizontal beam
il63 [147K]

A) The center of gravity of the two joined beams is;

<u><em>(x, y) = (25/44), (19/88)</em></u>

B) The gravitational torque on the joined beams about an axis through the corner is;

<u><em>τ = 245 N.m</em></u>

We are given;

Mass of vertical beam; m₂ = 19 kg

Mass of horizontal beam, m₁ = 25 kg

Length of horizontal beam; L₁ = 2 m

Length of vertical beam; L₂ = 1 m

A) Formula for center of gravity of the two joined beams is;

(x, y) = [(m₁x₁ + m₂x₂)/(m₁ + m₂)],  [(m₁y₁ + m₂y₂)/(m₁ + m₂)]

Where;

(x₁, y₁) is the <em>center of gravity</em> on the <em>horizontal beam</em>

(x₂, y₂) is the <em>center of gravity</em> on the <em>vertical beam</em>

(x₁, y₁) = (L₁/2, 0)

Plugging in the relevant values gives;

(x₁, y₁) = (2/2, 0)

(x₁, y₁) = (1, 0)

(x₂, y₂) = (0, L₂/2)

Plugging in the relevant values gives;

(x₂, y₂) = (0, 1/2)

(x₂, y₂) = (0, 0.5)

Thus, center of gravity of the 2 joined beams is;

(x, y) = [((25 × 1) + (19 × 0))/(25 + 19)],  [((25 × 0)  + (19 × 0.5)))/(25 + 19)]

(x, y) = (25/44), (19/88)

B) Formula for the gravitational torque on the joined beams about an axis through the corner is given as;

τ = (m₁g)x₁ + (m₂g)x₂

Plugging in the relevant values;

τ = (25 × 9.8)1 + (19 × 9.8)0

τ = 245 N.m

Read more at; brainly.com/question/14825257

8 0
2 years ago
Given cos theta = - 2/5 and sin theta &lt; 0, find the six trigonometric values. I need help with this
Lena [83]

Answer:

\sin\,\theta =-\frac{\sqrt{21} }{5}

\tan\,\theta =\frac{\sqrt{21} }{2}

\sec\,\theta = \frac{-5}{2}

cosec\,\theta =\frac{-5}{\sqrt{21} }

\cot\,\theta =\frac{2}{\sqrt{21} }

Step-by-step explanation:

\cos\theta =\frac{-2}{5}

As both sin\,\theta,

\theta lies in the third quadrant.

In the third quadrant,

\sin\theta

\sin\,\theta =-\sqrt{1-\cos^2\,\theta} \\=-\sqrt{1-(\frac{-2}{5})^2 } \\\\=-\sqrt{1-\frac{4}{25} }\\\\=-\sqrt{\frac{25-4}{25} }\\\\=-\frac{\sqrt{21} }{5}

\tan\,\theta = \frac{\sin\,\theta}{\cos\,\theta }\\\\=\frac{\frac{-\sqrt{21} }{5} }{\frac{-2}{5} }\\\\=\frac{\sqrt{21} }{2}

\sec\,\theta =\frac{1}{\cos\,\theta }\\\\=\frac{1}{\frac{-2}{5} }\\\\=\frac{-5}{2}

\ cosec \,\theta = \frac{1}{sin\,\theta }\\\\=\frac{1}{\frac{-\sqrt{21} }{5} }\\\\=\frac{-5}{\sqrt{21} }

\cot\,\theta =\frac{1}{\tan\,\theta}\\\\=\frac{1}{\frac{\sqrt{21} }{2} }\\\\=\frac{2}{\sqrt{21} }

3 0
3 years ago
AB contains the points A(7, -11) and B(8,17). find the slope of a line perpendicular to AB
Molodets [167]
If a line were to be perpendicular from said points, the slope of it would be -1/28. Hope this helps!
6 0
3 years ago
Read 2 more answers
What is the absolute value of |6|
motikmotik

Answer:

6

Step-by-step explanation:

7 0
3 years ago
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