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3 years ago

Find the absolute value of 7+4|10x-4|>111

Mathematics

1 answer:

Lerok [7]3 years ago

8 0

Answer:

x > 3 or x < -2.2

Step-by-step explanation:

7+ 4|10x-4| > 111

4|10x-4| > 104

|10x-4| > 26

10x-4 > 26 or 10x-4 < -26

10x > 30 or 10x < -22

x > 3 or x < -2.2

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Please help!:

Gnesinka [82]

The answer to this is A. 650. You find the surface area of the two cubes, then the rectangular prism, then you add them together.<span />

6 0

3 years ago

Find: (6m^5 + 3 - m^3 - 4m) - (-m^5 + 2m^3 - 4m + 6)

vitfil [10]

Answer:

7m^5 - 3m^3 - 3

Step-by-step explanation:

5 0

3 years ago

Your job is to paint lines in a long, straight parking lot. You determine the dividing lines need to be parallel. The length of

kozerog [31]

Answer:

a. 9 ft

b. 90 ° right angled

c. Right angle

d. 90°

e, Right angle

f. Angles on a straight line

g. 18 spots

Step-by-step explanation:

Here we have maximization question;

a. The separation distance of the dividing lines in a parking lot need to be far apart enough as to accommodate a vehicle with room to open the doors, therefore, it should be between 8.5 to 10 ft wide which gives a mean parking space width of approximately 9 ft

b. The angle of lines of the parking lot to the curb that will accommodate the most cars is 90°, because it reduces the width occupied by a car

c. The angle is right angled

d. Since the adjacent angle + calculated angle = angles on a straight line = 180 °

Therefore, adjacent angle = 90°

e. The angle is right angled

f. Angles on a straight line

g. The number of spots will be 162/9 = 18 spots.

8 0

4 years ago

Find the area of triangle ABC with vertices A(2,1), B (12,2), C (12,8). Hence, or otherwise find the perpendicular distance from

Lapatulllka [165]

<h2>The area of a triangle is =54 square units</h2><h2>The perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$ </h2>

Step-by-step explanation:

Given a triangle ABC with vertices A(2,1),B(12,2) and C(12,8)

$x_1=2,y_1=1,x_2=12,y_2=2,x_3=12 and y_3=8$

The area of a triangle is= $\frac{1}{2} [x_1(y_2-y_3) +x_2 (y_3- y_1)+x_3(y_1-y_2)]$

= $|\frac{1}{2} [2(2-8+12(8-1)+12(1-2)]|$

= $|-54|$ = 54 square units

The length of AC = $\sqrt{(x_1-x_3)^{2} +(y_1-y_3)^2}$

= $\sqrt{(2-12)^{2} +(1-8)^2}$

= $\sqrt{149}$ units

Let the perpendicular distance from B to AC be = x

According To Problem

$\frac{1}{2} \times x \times \sqrt{149} = 54$

⇔ $x =\frac{108}{\sqrt{149} }$ units

Therefore the perpendicular distance from B to AC is = $\frac{108}{\sqrt{149} } units$

5 0

3 years ago

X^2-x-7=0 complete the square

lesya692 [45]

Hello there!
Use the formula $( \frac{b}{2} )^2$ in order to created a new term. Solve for x by using this term to complete the square.
$x= \frac{1}{2}± \frac{ \sqrt{29} }{2}$

Hope this helps! :)
~Zain

6 0

3 years ago