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Yuri [45]
3 years ago
6

How many solutions does this system of equations have?

Mathematics
2 answers:
steposvetlana [31]3 years ago
7 0
It has no solutions bc the lines ( when graphed ) are parallel. so the answer is b.
julsineya [31]3 years ago
6 0
0.Just minus the 2nd equation with the first equation.You will grt 0=6 which doesn't make sense
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What is the answer to my question
Vedmedyk [2.9K]
672 cm³
Step-by-step explanation:
10x7x9=630
2x3x7=42
630+42=672
5 0
3 years ago
Hi can someone help me with this
Ad libitum [116K]

Answer:

25 units

Step-by-step explanation:

Applying Pythagoras' Theorem,

(BD)^2= (CD)^2 + (BC)^2

(BC)^2= 65^2 - 60^2

(BC)^2= 625

BC= √625= 25

5 0
3 years ago
38. What is the surface area of a conical grain storage tank that has a height of 62 meters and a diameter of 24 meters? Round t
svetoff [14.1K]
1. The surface area of a cone S is S= \pi  r^{2}+  \pi rl, where r is the radius of the base and l is the slant height.

2. The slant height l is found as follows:

l^{2}= OC^{2}+ OB^{2}
l^{2}= 12^{2}+ 62^{2}=144+3844=3988

l= \sqrt{3988}=63.15

3. 

A=\pi r^{2}+ \pi rl=3.14(12)^{2}+3.14*12*63.15

= 452.16+2379.492 =2831.652 ( m^{2} )

 = 452.16+2379.492 =2831.652 ( m^{2} )

4. The answer is C

8 0
3 years ago
Read 2 more answers
Use distributive property and area models to solve these equations . Show your work. 4 x 9.4 6.4 x 1.2 3.5 x 4.9 8.4 x 5.2
Verizon [17]
The Awnser is 9.4 6.4
6 0
3 years ago
A new automated production process averages 1.4 breakdowns per day. Because of the cost associated with a breakdown, management
dalvyx [7]

Answer:

The required probability is 0.167

Step-by-step explanation:

Consider the provided information.

Let x be the number of breakdown per day.

A new automated production process averages 1.4 breakdowns per day.

λ=1.4

Probability of having three or more breakdowns during a day is:

P(x\geq 3)=1-[f(0)+f(1)+f(2)]

The Poisson probability function is: f(x)=\frac{\lambda^xe^{-\lambda}}{x!}

Therefore the required probability is:

P(x\geq 3)=1-[\frac{\left(1.4^{0}e^{-1.4}\right)}{0!}+\frac{\left(1.4^{1}e^{-1.4}\right)}{1!}+\frac{\left(1.4^{2}e^{-1.4}\right)}{2!}]

P(x\geq 3)\approx1-0.833

P(x\geq 3)=0.167

Hence, the required probability is 0.167

4 0
3 years ago
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