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Zina [86]
3 years ago
8

What is the value of 3x-3 when x equals 3

Mathematics
2 answers:
choli [55]3 years ago
8 0

Answer:

6

Step-by-step explanation:

when you substitute 3 in for x you get 3(3) - 3

solve the equation

3(3) = 9

9 - 3 = 6

Anvisha [2.4K]3 years ago
5 0

Answer:

Your answer is 6.

Step-by-step explanation:

Substitute the value of the variable into the equation and simplify.

You might be interested in
Speed describes a change in ____ relative to a change in ______. a speed can never be ______
kumpel [21]
1) acceleration

2) speed

3) velocity

8 0
3 years ago
Make and test a conjecture about the given quantity.
larisa [96]

Answer:

18. The conjecture is: The product of two negative even integers is an even integer.

Testing: Let integers be -4 and -6. [Two negative even integers]

Then (-4) × (-6) = 24 [Even integer]

19. The conjecture is: Sum of two negative odd integers is a negative even integer.

Testing: Let the two negative odd integers be -3 and -7.

Then -3 + (-7) = -10 is a negative even integer

20. The conjecture is: Sum of two even integers and two odd integers is an even integer.

Testing: Let integers be 4, 12, 13, 17 (Two even integers and two odd integers)

Then 4 + 12 + 13 + 17 = 46 (Even integer)

21. The statement is:  p------->q [If you ran 1 mile, then you ran 5280 feet]

Decide whether statement is true: So if p is True then q is also True.

Hence p------->q is True. So the given statement is True.

Step-by-step explanation:

18. The product of two negative even integers.

Let the integers be -2m and -2n, m and n are positive integers

Product in (-2m) × (-2n) = 4mn = 2(2mn)

= 2p, p = 2mn

The conjecture is: The product of two negative even integers is an even integer.

Testing: Let integers be -4 and -6. [Two negative even integers]

Then (-4) × (-6) = 24 [Even integer]

19. The sum of two negative odd integers.

Let the integers be -(2m + 1) and -(2n + 1), m and n are positive integers

Then -(2m + 1) + [-(2n + 1)] = -2m - 1 - 2n - 1

= -2m - 2n - 2

= -2(+m + 2 + 1)

= -2p, p = m + n + 1

The conjecture is: Sum of two negative odd integers is a negative even integer.

Testing: Let the two negative odd integers be -3 and -7.

Then -3 + (-7) = -10 is a negative even integer

20. The sum of two even integers and two odd integers.

Let the integers be 2m, 2n, 2p + 1, 2q + 1 (m, n, p, q are positive integers)

Then, 2m + 2n + 2p + 1 + 2q + 1 = 2(m + n + p + q + 1)

= 2r, r = m + n + p + q + 1

The conjecture is: Sum of two even integers and two odd integers is an even integer.

Testing: Let integers be 4, 12, 13, 17 (Two even integers and two odd integers)

Then 4 + 12 + 13 + 17 = 46 (Even integer)

21. Let p: You ran 1 mile

Let q: You 5280 feet

The statement is:  p------->q [If you ran 1 mile, then you ran 5280 feet]

Since 1 mile = 5280 feet

So if p is True then q is also True.

Hence p------->q is True. So the given statement is True.

6 0
3 years ago
Please help with by finding the surface area of numbers 122 to 124 which are the top one ands please read 125 a and 126 to find
Annette [7]
Surface area of cylinder is 2 (pi)rh+2(pi)r^2
this is the area of the side [(2 (pi)rh] and the two circles together [2 (pi)r^2]

122. r=24
h=15
2 (3.14)24×15+2 (3.14)24^2
6.28×24×15+6.28×24^2
2260.8+3617.28
5878.08mm

123. r=5
h=8
2 (3.14)5×8+2 (3.14)5^2
6.28×5×8+6.28×5^2
251.2+157
408.2cm

124. the surface area of a cylinder that has a hole is going to be
surface area of big cylinder-2small area circles (the holes) + the area of the holes sides

hopefully that isn't too confusing

big radius: 8
height: 14

little radius: 2.5
height: 14

2 (3.14)8×14+2 (3.14)8^2
6.28×8×14+6.28×8^2
703.36+401.92

1105.28in (area of big cylinder)

2 (3.14)2.5^2 (two small holes)
6.28×2.5^2
39.25in (area of holes)

2 (3.14)×2.5×14 (area of side)
6.28×2.5×14
219.8in

1105.28-39.25+219.8
1066.03+219.8

1285.83in total surface area

125. h=20
r=6 since diameter is a foot or 12 in, half of that is 6
2 (3.14)6×20+2 (3.14)6^2
6.28×6×20+6.28×6^2
753.6+226.08
979.68in

126. first radius= 10
first height = 6

second radius= 6
second height= 10

2 (3.14)10×6+2 (3.14)10^2
6.28×10×6+6.28×10^2
376.8+628
1004.8in (surface area of first cylinder)

2 (3.14)6×10+2 (3.14)6^2
6.28×6×10+6.28×6^2
376.8+226.08
602.88in (surface area of second cylinder)

1004.8-602.88= 401.92

the surface area of the cylinder with radius 10 and height 6 is GREATER that the surface area of the cylinder with radius 6 and height 10..
by 401.92in
8 0
3 years ago
Angelo brought apples and bananas at the fruit stand. He bought 20 pieces of fruit and spent $11.50 apples cost $.50 and bananas
Jlenok [28]

Angelo bought 14 apples and 6 bananas

<em><u>Solution:</u></em>

Let "a" be the number of apples bought

Let "b" be the number of bananas bought

Cost of 1 apple = $ 0.50

Cost of 1 banana = $ 0.75

<em><u>He bought 20 pieces of fruit. Therefore,</u></em>

number of apples bought + number of bananas bought = 20

a + b = 20 -------- eqn 1

<em><u>He spent $ 11.50. Therefore, we frame a equation as:</u></em>

number of apples x Cost of 1 apple + number of bananas x Cost of 1 banana = 11.50

a \times 0.50 + b \times 0.75 = 11.50

0.50a + 0.75b = 11.50 ------- eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

From eqn 1,

a = 20 - b ------ eqn 3

<em><u>Substitute eqn 3 in eqn 2</u></em>

0.50(20 - b) + 0.75b = 11.50

10 - 0.5b + 0.75b = 11.50

0.25b = 11.50 - 10

0.25b = 1.5

Divide both sides of equation by 0.25

<h3>b = 6</h3>

<em><u>Substitute b = 6 in eqn 3</u></em>

a = 20 - 6

<h3>a = 14</h3>

Thus he bought 14 apples and 6 bananas

3 0
4 years ago
Question: F=d+e+t, solve for t.
Komok [63]
           F = d + e + t   Flip the sides to put t on the left
d + e + t = F              Subtract d from both sides
      e + t = F - d        Subtract e from both sides
            t = F - d - e
7 0
3 years ago
Read 2 more answers
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