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omeli [17]
3 years ago
11

Help me these are due today pls pls pls help

Mathematics
2 answers:
ddd [48]3 years ago
6 0

answer: someone alredy answered so here! (brainliest please!?)

Step-by-step explanation:

Ivan3 years ago
3 0

Answer:

She bought 7 pounds of almonds

Step-by-step explanation:

Divide 24.50 by 3.50, because if 1 pound is 3.50 and she spent 24.50 then you divide the two to get the answer.

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Ilia_Sergeevich [38]
I'd say 100 even, don't know if this is a real question or not.
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3 years ago
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Find the degree of the monomial 3x4y3​
Alex73 [517]

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4 0
2 years ago
The sum of two numbers is "-1" One number is 12 more than the other one. Find the numbers.
ruslelena [56]

Answer:

x = 5.5

y = - 6.5

Step-by-step explanation:

Let one of the numbers = x

Let the other number = y

x + y = - 1

x = y + 12    Put the second equation into the first one

y + 12 + y = - 1       Subtract 12 from both sides

y + y = - 1 - 12        Combine both left and right sides

2y = - 13                Divide by 2

2y/2 = - 13/1

y = - 6.5

x + y = - 1

x - 6.5 = - 1            Add 6.5 to both sides

x = 5.5

7 0
3 years ago
Hi guys . pls help me with this​
daser333 [38]

Step-by-step explanation:

The upstream speed is S / t₁, and the downstream speed is S / t₂.

If we say f is the speed of the fish in calm water, and r is the speed of the river, then:

f − r = S / t₁

f + r = S / t₂

If we say T is the time it takes to cross the river, then the speed perpendicular to the river is ℓ/T, the speed parallel to the river is r, and the overall speed is f.

Using Pythagorean theorem:

f² = (ℓ/T)² + r²

f² − r² = (ℓ/T)²

(f − r) (f + r) = (ℓ/T)²

(S / t₁) (S / t₂) = (ℓ/T)²

S² / (t₁ t₂) = (ℓ/T)²

(t₁ t₂) / S² = (T/ℓ)²

√(t₁ t₂) / S = T/ℓ

T = ℓ√(t₁ t₂) / S

6 0
3 years ago
Garry wants to have $750,000 in 30 years. How much does he have to invest per month in an annuity with an annual interest rate o
Nikolay [14]
I think the answer is Garry has to invest 2,035 dollars a month for 30 years to reach his goal of 750,000 dollars
4 0
3 years ago
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