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vampirchik [111]
3 years ago
8

PLS HELP ME ASAP GUYS FOR 1, 2, and 3 PLS (SHOW WORK!!) + LOTS OF POINTS!

Mathematics
1 answer:
pantera1 [17]3 years ago
8 0
1. B forms a proportion, this is what I did, i simply 10/18 to 5/9 and i try to see if 5/9 is equivalent to 25/45 and it is. 5/9*5/5= 25/45

2. You multiply 27*4=108 then 108/12=9 so a=12

3. you multiply 16*21=336 then 336/12=28 so b=28

Hope this helps!

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12 cm<br> 8 cm<br> 10.5 cm<br> Find the volume
topjm [15]

The Volume is 1008

<u>Multiply</u>

12*8*10.5=1008

Hope This Helps!

8 0
3 years ago
What is the sum of 3/10s and 5/100s <br> (Fractions)
vampirchik [111]

Answer:

80/100

Step-by-step explanation:

this is simplified ones

8/10

4/5

4 0
3 years ago
What is the equation in slope-intercept form of a line that is perpendicular to y=2x+2 and passes through the point (4, 3)?
andriy [413]

Answer:

y =  -  \frac{1}{2} x + 5

Step-by-step explanation:

<u>slope-intercept </u><u>form</u>

y= mx +c, where m is the slope and c is the y-intercept

Given line: y= 2x +2

slope= 2

The product of the slopes of perpendicular lines is -1. Let the slope of the unknown line be m.

m(2)= -1

m= -1 ÷2

m= -½

Substitute the value of m into the equation:

y= -½x +c

To find the value of c, substitute a pair of coordinates.

When x= 4, y= 3,

3= -½(4) +c

3= -2 +c

c= 3 +2

c= 5

Thus, the equation of the line is y= -½x +5.

7 0
3 years ago
Find all the solutions of the given system of equations. 3x-4y=2 9x-12y=6
Mkey [24]
I will try to solve your system of equations.
3
x
−
4
y
=
2
;
9
x
−
12
y
=
6
Step: Solve
3
x
−
4
y
=
2
for x:
3
x
−
4
y
+
4
y
=
2
+
4
y
(Add 4y to both sides)
3
x
=
4
y
+
2
3
x
3
=
4
y
+
2
3
(Divide both sides by 3)
x
=
4
3
y
+
2
3
Step: Substitute
4
3
y
+
2
3
for
x
in
9
x
−
12
y
=
6
:
9
x
−
12
y
=
6
9
(
4
3
y
+
2
3
)
−
12
y
=
6
6
=
6
(Simplify both sides of the equation)
6
+
−
6
=
6
+
−
6
(Add -6 to both sides)
0
=
0
Answer:
Infinitely many solutions.
8 0
4 years ago
Set Question involving real and natural numbers
Allushta [10]

Only two real numbers satisfy x² = 23, so A is the set {-√23, √23}. B is the set of all non-negative real numbers. Then you can write the intersection in various ways, like

(i) A ∩ B = {√23} = {x ∈ R | x = √23} = {x ∈ R | x² = 23 and x > 0}

√23 is positive and so is already contained in B, so the union with A adds -√23 to the set B. Then

(ii) A U B = {-√23} U B = {x ∈ R | (x² = 23 and x < 0) or x ≥ 0}

A - B is the complement of B in A; that is, all elements of A not belonging to B. This means we remove √23 from A, so that

(iii) A - B = {-√23} = {x ∈ R | x² = 23 and x < 0}

I'm not entirely sure what you mean by "for µ = R" - possibly µ is used to mean "universal set"? If so, then

(iv.a) Aᶜ = {x ∈ R | x² ≠ 23} and Bᶜ = {x ∈ R | x < 0}.

N is a subset of B, so

(iv.b) N - B = N = {1, 2, 3, ...}

3 0
3 years ago
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