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4 years ago

What is the GCF of 15 and 30

2 answers:

6 0

The greatest common factor (GCF) of 15 and 30 is: 15

Factors of 15: 1, 3, 5, 15
Factors of 30: 1, 2, 3, 5, 6, 10, 15

The common factors between the two numbers are 1, 3, 5, and 15, but considering 15 is the highest number out of them all, it is the greatest common factor.

BaLLatris [955]4 years ago

5 0

The greatest common factor is 15

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One cubic meter of water weighs approximately 2200 pounds. There are 100 cm in one meter. If you have a box-shaped fish tank tha

Diano4ka-milaya [45]

Answer:

158.4 pounds

Step-by-step explanation:

first find the volume of the tank

$V = 30 \times 30 \times 80$

$V = 72000 cm^3$

to convert this to m^3 we know that 1m = 100cm

$V = 72000 cm^3 \times \left(\dfrac{1\,\text{m}}{100\,\text{cm}}\right)^3$

$V = 72000 cm^3 \times \left(\dfrac{1\,\text{m}^3}{100^3\,\text{cm}^3}\right)$

the cm^3 cancel out.

$V = \dfrac{72000}{100^3} m^3$

$V = 0.072\,m^3$

by simply converting the side-lengths to meters beforehand can give you this answer directly:

$V = 0.3 \times 0.3 \times 0.8 = 0.072 m^3$

it is given that 1 meter cube of water weighs about 2200 pounds.

$1\,m^3 = 2200 \,\text{lb}$

multiply both sides with 0.072 to find our answer:

$1\,m^3 \times 0.072 = 2200 \times 0.072 \,\text{lb}$

$0.072\,m^3 = 158.4 \,\text{lb}$

hence the weight of the water in the tank would be 158.4 pounds!

3 0

3 years ago

A plane flying horizontally at an altitude of 1 mi and a speed of 470 mi/h passes directly over a radar station. Find the rate a

kiruha [24]

Answer:

407 mi/h

Step-by-step explanation:

Given:

Speed of plane (s) = 470 mi/h

Height of plane above radar station (h) = 1 mi

Let the distance of plane from station be 'D' at any time 't' and let 'x' be the horizontal distance traveled in time 't' by the plane.

Consider a right angled triangle representing the above scenario.

We can see that, the height 'h' remains fixed as the plane is flying horizontally.

Speed of the plane is nothing but the rate of change of horizontal distance of plane. So, $s=\frac{dx}{dt}=470\ mi/h$

Now, applying Pythagoras theorem to the triangle, we have:

$D^2=h^2+x^2\\\\D^2=1+x^2$

Differentiating with respect to time 't', we get:

$2D\frac{dD}{dt}=0+2x\frac{dx}{dt}\\\\\frac{dD}{dt}=\frac{x}{D}(s)$

Now, when the plane is 2 miles away from radar station, then D = 2 mi

Also, the horizontal distance traveled can be calculated using the value of 'D' in equation (1). This gives,

$2^2=1+x^2\\\\x^2=4-1\\\\x=\sqrt3=1.732\ mi$

Now, plug in all the given values and solve for $\frac{dD}{dt}$ . This gives,

$\frac{dD}{dt}=\frac{1.732\times 470}{2}=407.02\approx 407\ mi/h$

Therefore, the distance from the plane to the station is increasing at a rate of 407 mi/h.

6 0

3 years ago

For the function f(x)=x2−24÷3+12 f(1) f(0) f(−1)

LuckyWell [14K]

Put the values of x to the equation of the function.

$f(x)=\dfrac{x^2-24}{3}+12$

$f(1)=\dfrac{1^2-24}{3}+12=\dfrac{1-24}{3}+12=\dfrac{-23}{3}+12=-7\dfrac{2}{3}+12=4\dfrac{1}{3}\\\\f(0)=\dfrac{0^2-24}{3}+12=\dfrac{-24}{3}+12=-8+12=4\\\\f(-1)=\dfrac{(-1)^2-24}{3}+12=\dfrac{1-24}{3}+12=\dfrac{-23}{3}+12=-7\dfrac{2}{3}+12=4\dfrac{1}{3}$

3 0

3 years ago

Find the value of the variables. Diagram above. Tysm please help very urgent!

serious [3.7K]

Answer:

If you look at the different triangles in the star they give you two angles....

all of the points in a triangle would be equal and you just subtract 24 from 105

x is 24 and y is 81

Step-by-step explanation:

8 0

3 years ago

What is the difference of

inysia [295]

Answer:65,

Step-by-step explanation:

8 0

3 years ago