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raketka [301]
3 years ago
7

Line A passes through the points (220, 125) and (300, 75). Line B passes through the points (0, 0) and (30, 25). At what point d

o the lines intersect?
Mathematics
1 answer:
yanalaym [24]3 years ago
5 0

Answer:

No, Isiah is not correct. The GCF of the coefficients is 1, and there are no common variables among all three terms of the polynomial. 5b4 is a factor of -25a2b5 and -35b4, but not a3. Additionally, a2 is a factor of a3 and -25a2b5, but not -35b4

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WILL GIVE BRAINLIST SOON AS I CAN SOMEBODY PLEASE HELP ME ​
fgiga [73]

Answer:

A= 75.06 Sq Cm

Step-by-step explanation:

Please refer to the picture attached with this.

In ΔABD

AB = 8.5 , AD = 6 and say BD = x

Applying Pythagoras theorem in this triangle

Pythagoras theorem says

a^2+b^2=c^{2}

where c is the side opposite to the right angle, and b and a are the other two sides of the right angled triangle.

x^2+6^2=8.5^{2}

x^2=72.25-36

x^2=36.25

x=6.02

Hence side a = x +19

a = 6.02+19

a = 25.02

Now we apply the formula for area of a triangle which is given as

A=\frac{1}{2}\times b \times h

Where b is the base , here we have base as a = 25.02

h is the height , here h = 6

Putting these values in Formula we get

A=\frac{1}{2}\times 25.02 \times 6

A= 25.02 \times 3

A=75.06

Hence Area of the triangle is 75.06 sq cm

5 0
3 years ago
Mutiply the expression 2 (7n-11)
attashe74 [19]
14n-22 is the answer. Use the distributive property. 
4 0
3 years ago
GEOMETRY HELP PLEASE
Darya [45]
A line adds up to equal 180 degrees. And adjacent angles also add to equal 180 since they share a side on a line. So add them together and make them equal 180
9x - 7 + 8.5x + 4 = 180
17.5x - 3 = 180
17.5x = 183
x = 10.5
4 0
3 years ago
A gas is said to be compressed adiabatically if there is no gain or loss of heat. When such a gas is diatomic (has two atoms per
Tems11 [23]

Answer:

The pressure is changing at \frac{dP}{dt}=3.68

Step-by-step explanation:

Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

We know that the volume is decreasing at the rate of \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} and we want to find at what rate is the pressure changing.

The equation that model this situation is

PV^{1.4}=k

Differentiate both sides with respect to time t.

\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)

Apply this rule to our expression we get

V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0

Solve for \frac{dP}{dt}

V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}

when P = 23 kg/cm2, V = 35 cm3, and \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} this becomes

\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68

The pressure is changing at \frac{dP}{dt}=3.68.

7 0
3 years ago
GE produces 5 different sizes of TVs using 2 machines. Different sizes of TVs take different amount of time to get processed at
OLga [1]

Answer:

Option # C: 5

Step-by-step explanation:

Decision variables, as the name imply are entities that can have different values in the given scenario and which in turn produce different outcomes. For our question, the variables are the sizes of TVs available, as the production time and profit depend on the sizes. Hence, TV sizes (i.e. 5) are our decision variables.

Furthermore, it does not matter on which machine the TVs are being manufactured as they will take identical time for same TV size. Thus we do not consider the number of machines as our decision variables.

8 0
3 years ago
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