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JulijaS [17]
3 years ago
14

Write as a mathematical expression he product of 5 and x

Mathematics
1 answer:
Vaselesa [24]3 years ago
8 0
The answer to the question

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45 POINTS WILL PICK BRAINLIEST
jonny [76]
The answer to your question is c :)

6 0
3 years ago
Read 2 more answers
Find P(rolling 4) with one number cube.
olasank [31]
A die has 6 sides and 4 on one of the sides. so P(4) = 1/6
5 0
4 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
Please help me i don’t understand
Vesna [10]

Answer:

Below

Step-by-step explanation:

All you need to do to find angle x is to subtract 48 from 90 :)

So, 90 - 48 = 42

Angle x = 42 degrees

Hope this helps!

4 0
3 years ago
Read 2 more answers
Consider f (x) = StartRoot x squared minus 1 EndRoot and g (x) = StartRoot x squared + 1 EndRoot. What value(s) of x would make
dalvyx [7]

Answer:

Any value of x

<em></em>

Step-by-step explanation:

Given

f(x) = \sqrt{x^2 - 1}

g(x) = \sqrt{x^2 + 1}

Required

What value of x is  f(g(x)) = g(f(x))

Solving for f(g(x))

f(x) = \sqrt{x^2 - 1}

f(g(x)) = \sqrt{(\sqrt{x^2 + 1})^2 - 1}

Solve the inner square

f(g(x)) = \sqrt{(x^2 + 1 - 1}

f(g(x)) = \sqrt{x^2 } }

f(g(x)) = x

Solving g(f(x))

g(x) = \sqrt{x^2 + 1}

g(f(x)) = \sqrt{(\sqrt{x^2 - 1})^2 + 1}

g(f(x)) = \sqrt{x^2 - 1 + 1}

g(f(x)) = \sqrt{x^2 }

g(f(x)) = x

Equate f(g(x)) and g(f(x))

f(g(x)) = g(f(x))

x = x

<em>This implies that </em>f(g(x)) = g(f(x))<em> at any value of x</em>

8 0
4 years ago
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