All categories

2 years ago

Helppp me with these questions plz guys

Mathematics

1 answer:

SSSSS [86.1K]2 years ago

3 0

Number 41 is letter c I think tell me if it helps you

You might be interested in

Complete this statement:

klasskru [66]

Answer:

$45x^3a+27xa^2= 9xa(5x^2+3a)$

Step-by-step explanation:

Given:

$45x^3a+27xa^2= 9xa$

We need to complete this Statement.

By Solving the above equation we get;

We will take some common factor out so we will get;

$45x^3a+27xa^2= 9xa(5x^2+3a)$

Hence the Complete statement is $45x^3a+27xa^2= 9xa(5x^2+3a)$

5 0

2 years ago

A and B are partners sharing profit and loss in the ratio of 3:2. A new partner C is admitted and 1/4 share off profit is given.

Gre4nikov [31]

Answer:

Correct option is: 9:6:5

Step-by-step explanation:

I'm not sure but hope it helps.

6 0

2 years ago

A factory can make 450 gizmos every hour. If the factory operates 24 hours per day, how many gizmos can they make in one year?

kiruha [24]

I think the answer to your question is
3,942,000

7 0

3 years ago

Solve the equation shown for x. 2/3 x -4 1/2 = -8

liq [111]

The answer would be -5 1/4

4 0

2 years ago

Read 2 more answers

Jack looks at a clock tower from a distance and determines that the angle of elevation of the top of the tower is 40°. John, who

zmey [24]

Answer:

18.7939 m

Step-by-step explanation:

-Let x be the distance between John and clock tower.

-Let y be the vertical distance from the eyes of the two men standing to the top of the clock tower.

#Taking the right triangle ACD:

$\Tan \ theta=\frac{Perpendicular \ Height}{Base}\\\\Tan \ 60\textdegree=\frac{y+1.5}{x}\\\\y=x \ Tan \ 60\textdegree -1.5$

#Taking the right triangle ABD:

$\Tan \ theta=\frac{Perpendicular \ Height}{Base}\\\\Tan \ 40\textdegree=\frac{y+1.5}{x+20}\\\\y=(x+20)\ Tan \ 40\textdegree -1.5$

#We equate the two yo solve for x and y;

$(x+20)\ Tan \ 40\textdegree -1.5=x\ Tan \ 60\textdegree -1.5\\\\(x+20)\ Tan \ 40\textdegree=x\ Tan \ 60\textdegree\\\\0.8391x+16.7820=1.7321x\\\\x=18.7939$

Hence, John's distance from the tower's base is 18.7939 m

8 0

2 years ago