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Kazeer [188]
4 years ago
10

Hello ^ Please help! Thanks

Mathematics
2 answers:
Ivan4 years ago
8 0
The angle measure of a straight line is 180°, and so if the you know that one angle off of it is (5x)°, then you know that the other section is (180 - 5x)°.

Because it's an isosceles triangle, there are two identical angles, which happen to be the angles with the measure of (5x - 30)° and (180 - 5x)°.
Since they have the same measure, 5x - 30 = 180 - 5x. Solve:

5x-30=180-5x\\\\10x=210\\\\x=\frac{210}{10}=21

Thus, x = 21°.
AURORKA [14]4 years ago
6 0

The answer is  x = 21°

I hope I helped :)

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I need help on this algebra
Gnesinka [82]
Since the population increases by 7 percent each year, we can multiply 850 by 1.07 seven times. In other words, we can do 850 • 1.07^7. This is equal to about 1365.
3 0
3 years ago
Draw a line segment of length 8 cm divide it in the ratio 5:3​<br><br><br><br>plzzzz helpppp
liraira [26]

Answer:

Draw the line segment AB = 7 cm.

Draw ray BX making an acute ∠ABX.

Along BX. mark off five points B1, B2, B3, B4 and B5. Join B2 to A.

Through B5 draw B5P || B2A, intersecting BA produced at P.

The point P so obtained is the required point which divides AB externally in the ratio 3:5.

7 0
3 years ago
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Consider the planes 4x+3y+4z=1 and 4x+4z=0. (A) Find the unique point P on the y−axis which is on both planes. ( 0 equation edit
Nana76 [90]

Answer:

Step-by-step explanation:

A) From the order of the exercise we already know that the intersection points lies on the Y-axis, so its coordinates are P(0;y;0). In order to find it, we only need to substitute the equation 4x+4z=0 into the equation 4x+3y+4z=1. Then,

1=4x+3y+4z = 3y + (4x+4z)= 3y+0.

From the expression above it is easy to obtain that y=1/3, and the intersection point is P(0;1/3;0).

B) To obtain the parallel vector to both planes we use the cross product of the normal vector of the planes.

\left[\begin{array}{ccc}i&j&k\\4&3&4\\4&0&4\end{array}\right] = 12i-0j+12k

As we want a unit vector, we must calculate the modulus of u:

|u|=\sqrt{12^2+0^2+12^2} = \sqrt{2\cdot 12^2}=12\sqrt{2}.

Thus, the wanted vector is \frac{u}{12\sqrt{2}}. Therefore,

u = \frac{1}{\sqrt{2}}i-\frac{1}{\sqrt{2}}k = \frac{\sqrt{2}}{2}i-\frac{\sqrt{2}}{2}k.

C) In order to obtain the vector equation of the intersection line of both planes, we just need to put together the above results.

r = \frac{1}{3}j +\lambda \left( \frac{\sqrt{2}}{2}i- \frac{\sqrt{2}}{2}k \right)

where \lambda is a real number.

8 0
4 years ago
What do all points that lie on the y-axis have in common in terms of their coordinates?
vitfil [10]
For all points that lie on the y-axis, the x-coordinate is zero.
5 0
2 years ago
What is the circumference of the field if the area is 14400
Nuetrik [128]

Answer:

About 425

Step-by-step explanation:

Tell me if you want the explanation.

5 0
3 years ago
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