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Stolb23 [73]
3 years ago
14

The area of the triangle with sides 60 cm,100 cm and 140 cm is of the form 1500 square root x cm².

Mathematics
1 answer:
Ede4ka [16]3 years ago
5 0

Answer:

150

Step-by-step explanation:

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Reflect the shape A in the line x=1<br> What are the coordinates of the vertices of the image?
mestny [16]

Answer:

(0,4) (0,7) (-2,4)

Step-by-step explanation:

The line x=1 is a vertical line that passes through the point (1,0). So we then place the new triangle at equal distance from the line.

I hope this helps!

please mark brainliest!!

6 0
2 years ago
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A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far
ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3} (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2}) (1b)

Where:

x, y - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

r_{1}, r_{2} - Length of each vector, in kilometers.

\theta_{1}, \theta_{2} - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that r_{1} = 42\,km, r_{2} = 28\,km, \theta_{1} = 32^{\circ} and \theta_{2} = 154^{\circ}, then the resulting coordinates of the final position of the surveyor is:

(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})

(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]

(x,y) = (10.452, 34.531)\,[km]

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}

\theta \approx 16.840^{\circ}

And the distance from the camp is calculated by the Pythagorean Theorem:

r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}

r = 36.078\,km

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

5 0
3 years ago
The circumference (C) of a circle is 16cm. Which formula can you use to find the diameter (d) if you know that C= πd?
Dima020 [189]

The formula is d=C/π.

The diameter is 2 times the radius

The formula for the circumference using the radius is 2πr.

in order to do this backwards, we would have to do 16÷2÷π, but we're not looking for the radius.

Therefore, we take out the ÷2 part, which would be 16÷π

16 is the circumference

16÷π=d

d=C÷π

8 0
3 years ago
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What is 42 divided into 84 + 8 - 676 ???
puteri [66]

Answer: -5


Step-by-step explanation:

84 +  8 -676

3 0
3 years ago
Please help whoever can with #13 and thank you
bixtya [17]
These are so great! They are a perfect combination of Physics and pre-calculus! Your max height of that projectile is going to occur at the max value of the parabola, or at its vertex.  So we need to find the vertex.  The coordinates of the vertex will give us the x value, which is the time in seconds it takes to reach y which is the max height.  Do this by completing the square.  Begin by setting the equation equal to 0 and then moving the 80 over to the other side.  Then factor out the -16.  This is all that: -16( t^{2}-4t)=-80.  Take half the linear term which is 4 and square it and add it in to both sides.  Half of 4 is 2, 2 squared is 4, so add 4 into the set of parenthesis and to the -80.  -16( t^{2}-4t+4)=-80-64. The -64 on the right comes from the fact that when you added 4 into the parenthesis, you had the -16 out in front which is a multiplier.  -16 * 4 - -64.  So what you really added in was -64.  Now the perfect square binomial we created in that process was -16(x-2) ^{2} =-144.  When we move the 144 back over by addition we find that the vertex of the polynomial is (2, 144).  And that tells us that it takes 2 seconds for the projectile to reach its max height of 144 feet.  To find the time interval in which the object's height decreases occurs from its max height of 144 to where the graph of the parabola goes through the x-axis to the right of the max.  To find where the graph goes through the x-axis, or the zeroes of the graph, you factor the polynomial.  When you do that using the quadratic formula you get that x = -1 and 5.  So at its max height it is at 2 seconds, and by 5 seconds it hits the ground.  So the time interval of its height decreasing is from 2 seconds to 5 seconds, or a total of 3 seconds.  I think you need the 2 and 5, from the wording of your problem.
8 0
3 years ago
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