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Wewaii [24]
3 years ago
12

I put the question below ty

Mathematics
1 answer:
Liula [17]3 years ago
8 0
The answer to your question is B
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Solve 5(11x - 2) = 155
julia-pushkina [17]

Answer:

5(11x-2) = 155

55x - 10 = 155

55x = 165

x = 165/55

x = 3

Step-by-step explanation:

hope it helps

5 0
3 years ago
a big cruise ship dropped anchor off the Caribbean island of antigua. the heavy anchor dropped into the water at a rate of 2.5 m
AysviL [449]

Falling at the rate of 2.5 m/s for 45 seconds, the anchor fell

                                     (2.5 m/s) x (45 s)  =  112.5 meters .

If it wound up only 40 meters underwater, then it must have fallen

                                      (112.5 m  -  40 m)  =  72.5 meters

from its storage position on the big cruise ship, 
before it ever reached the water.

Since it was falling at the rate of  2.5 m/s, it took

                                     (72.5 m) / (2.5 m/s)  =  29 seconds

to get down to the water surface after it started dropping.

6 0
3 years ago
Can someone answer this question please please help me I really need it if it’s correct I will mark you brainliest .
Allisa [31]

Answer:

We divided the figure into 1 triangle and 2 rectangles.

Adding up the area of them, we get the total area:

A = 14 x (10 + 8 + 12 - 16)/2 + 8 x 5 + 6 x (12 + 8)

   = 258 (yd2)

Hope this helps!

:)

5 0
3 years ago
Write a number using the following clues:
Korvikt [17]

Answer:

Theres a lot of numbers you can choose from 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96,

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Special right triangles.
Oliga [24]

QUESTION 9

The given right angle triangle has its two legs equal to x\:cm and the hypotenuse is 2\:units.


Using the Pythagoras theorem, we have

x^2+x^2=2^2


This implies that;

2x^2=2^2


2x^2=4


x^2=2


x=\sqrt{2}


x=1.414


x\approx1.4\:units


QUESTION 10


The given right angle triangle has its two legs equal to x\:units and the hypotenuse is 7\:units.


Using the Pythagoras theorem, we have

x^2+x^2=7^2


This implies that;

2x^2=7^2


2x^2=49


x^2=24.5


x=\sqrt{24.5}



x=4.95\:units


QUESTION 11

Sam divided the square backyard into two sections along the 40ft diagonal.

Let one of the sides of the garden be x\:units, then the other side of the garden is also  x\:units since it was a square backyard.


The diagonal is the hypotenuse which is 40ft and the two legs are  x\:units each.


Applying the Pythagoras Theorem, we have;

x^2+x^2=40^2


2x^2=1600


x^2=800


\Rightarrow x=\sqrt{800}


\Rightarrow x=28.28


The length of one side of the garden is approximately 28cm.


QUESTION 12

The hypotenuse of Nicole's right triangular support is 92cm long.


It was given that the two lengths of the triangular support are of equal length.


Let the two lengths of the triangular support be l\:cm each.


We then apply the Pythagoras theorem to obtain;

l^2+l^2=92^2


\Rightarrow 2l^2=8464


\Rightarrow l^2=4232


\Rightarrow l=\sqrt{4232}


\Rightarrow l=65.05cm


Therefore Nicole needs l+l=65.05cm+65.05cm\approx130cm of wood to complete the support.

The correct answer is A.


QUESTION 13

A 45-45-90 triangle is an isosceles right triangle.


Let the length of the two legs be m\:inches each.

It was given that the hypotenuse of this triangle is 12\:inches.


Applying the Pythagoras Theorem, we have;

m^2+m^2=12^2


\Rightarrow 2m^2=144


\Rightarrow m^2=72


\Rightarrow m=\sqrt{72}


\Rightarrow m=8.49


The length of one leg of the hypotenuse is approximately 8.5 inches to 1 decimal place.


QUESTIONS 14.

It was given that the distance from home plate to first base is 90 feet.


Since the baseball field form a square, the length of the baselines are all 90ft.

We want to calculate the distance from home plate to 2nd base,which is the diagonal of the square baseball field.

Let this distance be d\:ft, then using the Pythagoras theorem;

d^2=90^2+90^2


d^2=8100+8100


d^2=16200


\:d=\sqrt{16200}


\:d=127.279


Therefore the distance from home plate to second base is 127 ft to the nearest foot.


QUESTION 15.

The distance from third base to  first base is also 127 ft to the nearest foot.


The reason is that the  diagonals of a square are equal in length.


3 0
3 years ago
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