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3 years ago

I need help with this someone!!

Mathematics

2 answers:

hichkok12 [17]3 years ago

5 0

Answer:

Your proportional constant would be 3 in this case.

Step-by-step explanation:

To find it yourself, try and find what number the y-axis equals when x=1. Hope this helps!

Alex3 years ago

4 0

Answer:

Do u have answer choices?

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PLEASEEEE HELP!!!☹️☹️ i will mark brainliest

Nikitich [7]

Z=110 because if you add up all the angles inside the triangle it would =180 and take the angle by z and subtract it from 180 you get your answer

6 0

2 years ago

Please HELP!!!!! I need to know the missing value. PLease!!!!!!!!!!!!!!

GuDViN [60]

Answer:

w = 21.4 ft

Step-by-step explanation:

Area of Trapezoid = Area = (a+b)/2 * h

Where a and b are the bases of trapezoid, and h is the height.

In the figure we are given:

a= 32.3 ft

b=w=?

h= 22.9 ft

Area = 614.865 ft²

Putting values in formula

Area = (a+b)/2 * h

614.865 = (32.3 + w) / 2 * 22.9

1229.73 = 32.3 + w * 22.9

1229.73 / 22.9 = 32.3 +w

53.7 -32.3 = w

=> w = 21.4 ft

7 0

3 years ago

. Evaluate each of the following without using a calculator. Show all steps clearly. - 5^2

Verizon [17]

Answer:

Step-by-step explanation:

-5^2 expanded is -5 x -5

- two negatives equal a positive

-visual representation of 5 x 5

O O O O O

7 0

1 year ago

If the distance covered by an object in time t is given by s(t)=t^2+5t , where s(t) is in meters and t is in seconds, what is th

alisha [4.7K]

Answer:

E. 44 meters

Step-by-step explanation:

The function that models the distance covered by the object is

$s(t)=t^2+5t$

where s(t) is in meters and t is in seconds.

The distance covered by the object after 1 second is

$s(1)=1^2+5(1)=6m$

The distance covered by the object after 1 second is

$s(1)=5^2+5(5)=50m$

The distance covered between 1 second and 5 seconds is

50-6=44m

6 0

3 years ago

There are 130 people in a sport centre.

Mashcka [7]

Answer:

Step-by-step explanation:

Using set notation to answer the question. let:

n(U) be the total people in a sport centre =130

n(G) be number of people that use the gym. = 76

n(P) be number of people that use the swimming pool. = 60

n(T) be number of people that use the track = 32

n(G∩P) be number of people that uses the gym and the pool = 23

n(P∩T) be number of people that uses the pool and the track = 8

n(G∩T) be number of people that uses the gym and the track = 320

n(G∩P∩T) be number of people that uses all three facilities = 6

Using the relationship below;

n(U)= n(GUPUT)+n(GUPUT)'

where n(GUPUT)' is the number of people that doesn't use any facility.

Before we can get that, we need to know n(GUPUT) using the formula;

n(GUPUT) = n(G) + n(P)+n(T)-n(G∩P)-n(P∩T)-n(G∩T)+n(G∩P∩T)

n(GUPUT) = 76+60+32-23-8-20+6

n(GUPUT) = 123

n(GUPUT)' = n(U)- n(GUPUT)

n(GUPUT)' = 130 - 123

n(GUPUT)' = 7

This means that 7 person doesn't use any facility.

Probability that a person selected at random doesn't use any facility = n(GUPUT)' /n(U) = 7/130

3 0

3 years ago