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AfilCa [17]
2 years ago
11

Write two expressions to show a number increased by 11 ​

Mathematics
1 answer:
erma4kov [3.2K]2 years ago
4 0

Answer:

0+11=11

14+11=25

Step-by-step explanation:

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A brownie recipe calls for 1 cup of sugar and a cup of flour to make one batch of brownies. To make multiple batches, the equati
marysya [2.9K]

Answer:

I cant see ths graph but look for one with a positive relationship between flour and sugar meaning that if flour is on one axis sugar should be on the other they should both go up by one every time

Step-by-step explanation:

The graph should say for example at 3 cups of flour that their is 3 cups of sugar. It should look upwards in growth and use whole positive numbers.

3 0
2 years ago
Is -253 in water deeper than -280 in water
lora16 [44]
-253 IS deeper than -280 in water. As your going into the negatives, The smaller numbers are bigger than the big numbers. It is the opposite when it comes to positive numbers.
3 0
3 years ago
If 5 burgers and 4 orders of fries cost $30.76 and 8 burgers and 6 orders of fries cost $48.28, what is the cost of a burger and
padilas [110]

Step-by-step explanation:

5B+4F= $30.76

×3

15B+12F= $92.28

----------------------------

8B+6F= $48.28

×2

16B+12F= $96.56

----------------------------

Cost of one burger:

$97.56-$92.28= $4.28 (final ans)

Cost of one fries:

$30.76-5($4.28)= $9.36 (for 5 fries)

$9.36÷5= $1.87 (final ans)

3 0
2 years ago
Which is the same? please help​
defon

Answer: The answer is D

Step-by-step explanation: BRAINLIEST PLS

3 0
3 years ago
Does 23^-1 (mod 1000) exist? If yes solve it.
sweet [91]

Yes, 23 has an inverse mod 1000 because gcd(23, 1000) = 1 (i.e. they are coprime).

Let <em>x</em> be the inverse. Then <em>x</em> is such that

23<em>x</em> ≡ 1 (mod 1000)

Use the Euclidean algorithm to solve for <em>x</em> :

1000 = 43×23 + 11

23 = 2×11 + 1

→   1 ≡ 23 - 2×11   (mod 1000)

→   1 ≡ 23 - 2×(1000 - 43×23)   (mod 1000)

→   1 ≡ 23 - 2×1000 + 86×23   (mod 1000)

→   1 ≡ 87×23 - 2×1000 ≡ 87×23   (mod 1000)

→   23⁻¹ ≡ 87   (mod 1000)

3 0
3 years ago
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