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docker41 [41]
3 years ago
8

PLEASE HELP ME WITH THIS PROBLEM!!! I NEED TO TURN IT IN TODAY

Mathematics
1 answer:
Murrr4er [49]3 years ago
8 0

Answer:

U=19

A=8

B=5

Step-by-step explanation:

Cardinality refers to the number of elements in a set.

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Help solve with subsitution x+y=1, 2x+3y=-4
Anettt [7]
{   x + y = 1    * ( - 2 )
  2 x + 3y = -4

- 2 x - 2 y = - 2
  2 x + 3 y = - 4
_______________
         y = - 6

Subsititute y = - 6 
<span>
2 x + 3 y = - 4

2 x + 3 * (-6) = -4

2 x - 18 = - 4

2 x = - 4 + 18

2 x = 14

x = 14 / 2

x = 7

The solution  to the system of equations are:

x = 7 and y = - 6

hope this helps!
</span>


8 0
3 years ago
-6x+6y=6<br> - 6x+3y=-12 also steps on how to solve it
Masja [62]

Answer:

x=5, y=6. (5, 6).

Step-by-step explanation:

-6x+6y=6

-6x+3y=-12

---------------

simplify -6x+6y=6 into -x+y=1 and y=1-(-x)=1+x

-6x+3(1+x)=-12

-6x+3+3x=-12

-3x=-12-3

-3x=-15

3x=15

x=15/3

x=5

-6(5)+6y=6

-30+6y=6

6y=6-(-30)

6y=6+30

6y=36

y=36/6

y=6

3 0
3 years ago
What are the domain, range, and asymptote of h(x) = (1.4)^x + 5?
Elena L [17]

Answer:

A.Domain:{x\x is a real number }

Range:{y\y>5}

Asymptote :y=5

Step-by-step explanation:

We are given that a function

h(x)=(1.4)^x+5

We have to find the domain, range and asymptote of h(x).

It is exponential function therefore, it is defined for all values of x.

Hence, domain  of h(x)={x| x is a real number}

Substitute x=0 then we get

h(0)=(1.4)^0+5=1+5=6

(a^0=1)

Hence, range of h(x)={y|y>5}

For exponential function,

Horizontal asymptote:a^x\rightarrow 0 when x\rightarrow-\infty

Apply limit x\rightarrow-\infty

\lim_{x\rightarrow-\infty}h(x)=\lim_{x\rightarrow-\infty}(1.4)^x+5=0+5=5

e^{-\infty}=0

\lim_{x\rightarrow \infty}(1.4)^x+5=\infty

Hence, the horizontal asymptote y=5.

6 0
3 years ago
Read 2 more answers
Complete the table. Round your entries to the nearest thousandth.
Flauer [41]

Answer:

there is no table...... ??

4 0
3 years ago
Read 2 more answers
Find an equation for the plane that contains the point p(−1,3,5)p(−1,3,5) and has the normal vector n=2i+4j−3kn=2i+4j−3k.
coldgirl [10]
Given a plane:
normal vector <2,4,-3>
passes through point (-1,3,5)

The equation of the plane is
&Pi; :  2(x-(-1))+4(y-3)-3(z-5)=0
expand and simplify
&Pi; :  2x+4y-3z+5=0
6 0
3 years ago
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