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4vir4ik [10]
2 years ago
15

Step 1: –10 + 8x < 6x – 4

Mathematics
1 answer:
Galina-37 [17]2 years ago
6 0
X< 3 because whenever you multiply or divide by a negative number, the symbol switches.
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Round 475,805 to the nearest ten thousand.
Gnom [1K]

475,805 rounded to the nearest ten thousand would be 476,000.

8 0
3 years ago
In a certain neighborhood, there's a sagging power line between two utility poles. The utility poles are 50 feet tall and 120 fe
e-lub [12.9K]
<h3>Answer:  33.75 feet</h3>

In fraction form, this value is equal to 135/4

33.75 ft is equivalent to 33 ft, 9 inches.

===============================================

Explanation:

Refer to the diagram below.

The key point to start with is point H, which is the vertex of the parabola.

Recall that vertex form is

y = a(x-h)^2 + k

What we'll do is plug in the vertex (h,k) = (60,30) which is the location of point H. We'll also plug in (0,45) which is the y intercept, aka the location of point C.

So,

y = a(x-h)^2 + k

y = a(x-60)^2 + 30 .... plug in vertex

45 = a(0-60)^2 + 30 .... plug in y intercept coordinates

45 = a(-60)^2 + 30

45 = a(3600) + 30

45 = 3600a + 30

45-30 = 3600a

3600a = 15

a = 15/3600

a = 1/240

This then means:

y = a(x-h)^2 + k

y = (1/240)(x-60)^2 + 30

This is the equation of our parabola. Plug in x = 30 to determine the height of point K

y = (1/240)(x-60)^2 + 30

y = (1/240)(30-60)^2 + 30

y = (1/240)(-30)^2 + 30

y = (1/240)(900) + 30

y = 15/4 + 30

y = 15/4 + 120/4

y = 135/4

y = 33.75

Therefore, the height of the power line, when it is 30 feet away from one of the poles, is 33.75 feet. This is the y coordinate of point K.

Side note: 33.75 ft = 33 ft + 0.75 ft = 33 ft + 12*0.75 in = 33 ft + 9 inches

8 0
2 years ago
What is orthogonal in linear algebra
o-na [289]
This is a square matrix whose entries are real and rows and columns are orthogonal unit vectors.
8 0
3 years ago
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Find the value of the trigonometric ratio. Simplify the ratio if possible.
sweet [91]

Step-by-step explanation:

Given that,

BC = 8

Ac = 15

We can find AB using the pythagoas theorem.

AB=\sqrt{AC^2+BC^2}\\\\=\sqrt{15^2+8^2}\\\\AB=17

We know that,

\cos\theta=\dfrac{B}{H}, B is base and H is Hypotenuse

\cosB=\dfrac{8}{17}

Hence, this is the required solution.

3 0
3 years ago
A student has a range of values from 0.000001 to 1,000,000, but limited paper to graph these values on. How might the student be
erica [24]

Answer: The student could use large intervals (what he counts by)

Step-by-step explanation:

4 0
2 years ago
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