All categories

1 year ago

I need some help with these please :')

Mathematics

1 answer:

Nataly [62]1 year ago

5 0

Step-by-step explanation:

problem 1.

1. 6y = 30

2. y = 5

3. 3x + 2(5) = 16

4. 3x+10=16

5. 3x=6

6. x=2

solution: (2,5)

problem 2.

1. x=7

2. 4(2)-2y=18

3. 28-2y=18

4. -2y= -10

5. y=5

solution (7,5)

problem 3

10x=10

x=1

7(1) +y =-2

y=-9

(1,9)

problem 4

0= -6

no solution

problem 5

0=0

Infinite many

You might be interested in

. For each of these intervals, list all its elements or explain why it is empty. a) [a, a] b) [a, a) c) (a, a] d) (a, a) e) (a,

Eva8 [605]

Answer:

Elements are of the form

(i) $[a,a]=\{[x,y] : a\leq x\leq a, a\leq y\leq a; a\in \mathbb R\}$

(ii) $[a,b)=\{[x,y) :a\leq x$

(iii) $(a,a]=\{(x,y] :a$

(iv) $(a,a)=\{(x,y): a$

(v) (a,b) where a>b= $\{(x,y) : a>x>b,a>y>b;a>b,a,b \in \mathbb R\}$

(vi) [a,b] where a>b= $\{[x,y] : a\geq x\geq b,a\geq y\geq b;a>b,a,b \in \mathbb R\}$

Step-by-step explanation:

Given intervals are,

(i) [a,a] (ii) [a,a) (iii) (a,a] (iv) (a,a) (v) (a,b) where a>b (vi) [a,b] where a>b.

To show all its elements,

(i) [a,a]

Imply the set including aa from left as well as right side.

Its elements are of the form.

$\{[a,a] : a\in \mathbb R\}=\{[0,0],[1, 1],[-1,-1],[2,2],[-2,-2],[3,3],[-3,-3],........\}$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a] represents singleton sets, and singleton sets are empty so is [a,a].

(ii) [a,a)

This means given interval containing a by left and exclude a by right.

Its elements are of the form.

$[ 1, 1),[-1,-1),[2,2),[-2,-2),[3,3),[-3,-3),........$

Since there is a singleton element a of real numbers withis the set, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set [a,a) represents singleton sets, and singleton sets are empty so is [a.a).

(iii) (a,a]

It means the interval not taking a by left and include a by right.

Its elements are of the form.

$( 1, 1],(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Since there is a singleton element a of real numbers, this set is empty.

Because there is no increment so if $a\in \mathbb R$ then the set (a,a] represents singleton sets, and singleton sets are empty so is (a,a].

(iv) (a,a)

Means given set excluding a by left as well as right.

Since there is a singleton element a of real numbers, this set is empty.

Its elements are of the form.

$( 1, 1),(-1,-1],(2,2],(-2,-2],(3,3],(-3,-3],........$

Because there is no increment so if $a\in \mathbb R$ then the set (a,a) represents singleton sets, and singleton sets are empty, so is (a,a).

(v) (a,b) where a>b.

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which not take x and y by any side of the interval.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$(a,b)=\{(5,0),(5,1),(5,2).....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

(vi) [a,b] where $a\leq b$ .

Which indicate the interval containing a, b such that increment of x is always greater than increment of y which include both x and y.

That is the graph is bounded by value of a and it contains elements like it we fixed a=5 then,

$[a,b]=\{[5,0],[5,1],[5,2].....\}$ e.t.c

So this set is connected and we know singletons are connected in $\mathbb R$ . Hence given set is empty.

8 0

2 years ago

All questions to be solved using linear combination.

monitta

1)
I:x-y=-7
II:x+y=7

add both equations together to eliminate y:
x-y+(x+y)=-7+7
2x=0
x=0

insert x=0 into II:
0+y=7
y=7

the solution is (0,7)

2)
I: 3x+y=4
II: 2x+y=5

add I+(-1*II) together to eliminate y:
3x+y+(-2x-y)=4+(-5)
x=-1

insert x=-1 into I:
3*-1+y=4
y=7

the solution is (-1,7)

3)

I: 2e-3f=-9
II: e+3f=18

add both equations together to eliminate f:
2e-3f+(e+3f)=-9+18
3e=9
e=3

insert e=3 into I:
2*3-3f=-9
-3f=-9-6
-3f=-15
3f=15
f=5

the solution is (3,5)

4)
I: 3d-e=7
II: d+e=5

add both equations together to eliminate e:
3d-e+(d+e)=7+5
4d=12
d=3

insert d=3 into II:
3+e=5
e=2

the solution is (3,2)

5)
I: 8x+y=14
II: 3x+y=4

add I+(-1*II) together to eliminate y
8x+y+(-3x-y)=14-4
5x=10
x=2

insert x=2 into II:
3*2+y=4
y=4-6
y=-2

the solution is (2,-2)

8 0

2 years ago

Find the domain and range for the function graphed below

Nana76 [90]

Answer:

Domain: 3, 2, 1, 0, -1 , -2 , -3

Range: 5 , 4 , 3 , 2 , 1 , 0 , -1 , -2

Step-by-step explanation:

The domain values are all of the x-values, and the range value are all of the y-values. Domain ad range can also be know as input and output.

3 0

2 years ago

The legth of a rectangle is 15 inches less then twice its width, w. If the perimeter of the rectangle is 96 inches what is the l

KATRIN_1 [288]

Answer:

length of the rectangle=27"

Step-by-step explanation:

length be = l; width be = b

2b-15= l

perimeter= 2(l+b)

=2(2b-15+b)=96

=2(3b-15) =96

3b-15 .=48

3b=48+15=63

b=21. Therefore l = (2 x 21) - 15=42-15 =27"

8 0

2 years ago

Hola alguien para hablar estoy aburrida y me gustaría hablar con alguien

kkurt [141]

Answer:

i know english only

Step-by-step explanation:

sorry but you stupit you not good at englis

7 0

1 year ago

Read 2 more answers