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BigorU [14]
3 years ago
12

Sue's ice cream cone has 6 more scoops than Tessa's ice cream cone. Sue's ice cream cone has 9 scoops. How many scoops are on Te

ssa's ice cream cone? scoops
Mathematics
2 answers:
marshall27 [118]3 years ago
5 0
Tessa has 3 scoops on her cone.

9-6=3
aivan3 [116]3 years ago
4 0

Answer:

3

Step-by-step explanation:

9 - 6 = 3

Hope that helped :)

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Rosa and Albert receive the same amount of allowance each week . The table shows what part of their allowance they each spent on
Anettt [7]
1/2 x > 0.4x

where Robert's spending is greater than Rosa's spending.

Allowance of Rosa = Allowance of Robert

let x = allowance

Rosa : video games = 0.4(x) ; pizza = 2/5 (x)
Robert : video games = 1/2 (x) ; pizza = 0.25 (x)

let us assume that their allowance is 100 each week. so, x = 100

Rosa : video games = 0.40(100) = 40
           pizza 2/5 (100) = 200/5 = 40
total spending: 40 + 40 = 80

Robert : video games = 1/2 (100) = 50
             pizza 0.25 (100) = 25
total spending: 50 + 25 = 75

Spending on video games
Rosa = 40
Robert = 50

Robert spent more of his allowance on video games than Rosa.

1/2 x > 0.4x
7 0
3 years ago
O is the centre of the circle, EF is a tangent, angle BCE=28 degrees, angle ACD=31 degrees.
Yuki888 [10]

Answer:

a. 28˚

b. 76˚

c. 104˚

d. 56˚

Step-by-step explanation

a. because angle between tangent and chord is equal to the angle(s) in alternate segment.

b. because angles in triangles add up to 180˚, 180-28=152 and because isosceles triangle, 152/2=76˚

c. because angles in triangles add up to 180˚ and opposite angles in a cyclic quadrilateral add up to 180˚, 31+76=107, 180-107=73, 73-28=45, angles in triangle so 180-(31+45)=104˚

d. 28*2=56˚ because angles at circumference are half angles at centre

4 0
3 years ago
Prove algebraically that the sum of the squares of two consecutive intgers is always an odd number
musickatia [10]

Answer:

See below.

Step-by-step explanation:

Let's let our first integer be n.

Then, our second, consecutive integer must be (n+1).

We want to prove that the sum of the square of two consecutive integers is always odd.

So, let's square our two expressions and add them up:

(n)^2+(n+1)^2

Square. Use the perfect square trinomial pattern. So:

=n^2+(n^2+2n+1)

Combine like terms:

=2n^2+2n+1

Notice that the first term 2n² has a 2 in front. Anything multiplied by 2 is even. So, regardless of what integer n is, 2n² is <em>always</em> even.

Our second term 2n also has a 2 in front. So, whatever n is, 2n is also <em>always</em> even.

So far, 2n² and 2n are both even. Therefore, the sum of 2n² and 2n is <em>also </em>even.

Lastly, we have the constant term 1. It doesn't matter what n is, we know that (2n² + 2n) is definitely and is always even. So, if we add 1 to this, we will get an odd number.

Q.E.D.

3 0
3 years ago
A line passes through the points (6, 9) and (8, 10). What is its equation in point-slope<br> form?
olchik [2.2K]

Answer:

y = 1/2x+6

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
2y - 2(5y) + 3× - 4 (78) = 11
Lera25 [3.4K]

y= -947/8

Just isolate the variables, even though this is a long, painstaking, agonizing process

:)

6 0
3 years ago
Read 2 more answers
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