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Free_Kalibri [48]
3 years ago
6

A group of students must collect at least $120 to organize a science fair. They have already collected $30. Which graph best rep

resents all remaining amounts of money, in dollars, that the students should still collect to organize the science fair?
Number line graph with closed circle on 120 and shading to the right

Number line graph with closed circle on 90 and shading to the right.

Number line graph with closed circle on 150 and shading to the right

Number line graph with closed circle on 30 and shading to the right.

Mathematics
2 answers:
Bezzdna [24]3 years ago
8 0
Number line graph with closed circle on 30 and shading to the right
because you already have 30 dollars and they need at least 120 but if they get more then 120 they can still hold the science fair.
Nookie1986 [14]3 years ago
5 0

Answer:

The correct option is 2.

Step-by-step explanation:

It is given that a group of students must collect at least $120 to organize a science fair. They have already collected $30.  

Let the remaining amounts of money, in dollars, that the students should still collect to organize the science fair be x.

x+30\geq 120

Subtract 30 from both the sides.

x+30-30\geq 120-30

x\geq 90

It means the remaining amounts of money is greater than or equal to 90.

Therefore the number line graph with closed circle on 90 and shading to the right and option 2 is correct.

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If a new soccer ball is labeled $45, what would the new price be if the sign above it says "63% off"?
Paul [167]

Answer:

New price of the ball after discount = $16.65

Step-by-step explanation:

Selling price of the soccer ball = $45

Discount labeled on the ball = 63% off

Therefore, selling price after the discount = Original selling price - %discount

= $45 - (63% of $45)

= $45 - $28.35

= $16.65

New price of the ball after discount = $16.65

3 0
3 years ago
I need help on this and I need to know how to do the steps for it
cestrela7 [59]
I got you, my answer is x<4

5 0
3 years ago
Suppose you know the perimeter of n octagons arranged as shown. What would you do to find the perimeter if 1 more octagon was ad
kogti [31]

As shown in the figures given :

For Figure 1 : perimeter = 8 units   [As can be seen in the figure]

For figure 2(with 2 octagons) : perimeter = 8 × 2 - 1 = 15 units    [since 1 side is common ]

For figure 2(with 3 octagons) : perimeter = 8 × 3 - 2 = 22 units  [since 2 sides is common ]

If one more octagon is added

then perimeter = 8 × 4 - 3 = 29 units    [since 3 sides will be common ]


3 0
3 years ago
Can someone help me with the ones that aren’t done
garri49 [273]

5:

C = 5/9(F - 32)

C = 5/9(F) - 17.777

-5/9(F) + C = -17.777

-5/9(F) = -C - 17.777

F = 5/9(C) + 32.0306

7 0
3 years ago
Read 2 more answers
Suppose m = 2 + 6i, and | m + n | = 3√10, where n is a complex number.
Ksju [112]

Answer: a) √50

b) n = 1 + 7i

Step-by-step explanation:

first, the modulus of a complex number z = a + bi is

IzI = √(a^2 + b^2)  

The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:

m = 2 + 6i

n = a  + bi

Im + nI = 3√10

Im + n I = √(a^2 + b^2 + 2^2 + 6^2)= 3√10

            = √(a^2 + b^2 + 40) = 3√10

             a^2 + b^2 + 40 = 3^2*10 = 9*10 = 90

             a^2 + b^2 = 90 - 40 = 50

            √(a^2 + b^2 ) = InI = √50

The modulus of n must be equal to the square root of 50.

now we can find any values a and b such a^2 + b^2 = 50.

for example, a = 1 and b = 7

1^2 + 7^2 = 1 + 49  = 50

Then a possible value for n is:

n = 1 + 7i

6 0
3 years ago
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