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Sauron [17]
2 years ago
13

P-please help (30 points)

Mathematics
1 answer:
liubo4ka [24]2 years ago
8 0

Answer:

No .1

Ans ; 2

No.2

Ans ; 3

Step-by-step explanation:

Multiply or add and if the value is whole no. then its W else L

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marco and suzi each mulyiplied 0.721 X 100 Marco got 7.21 for his product suiz got 72.1 for her product which student multiplied
Marianna [84]
Size got it right because it’s multiplied by 100
5 0
2 years ago
Help me please i have another page to due as well it’s all tiring
lutik1710 [3]

Pythagorean Theorem: a^2 + b^2 = c^2

-A and B are legs of the triangle

-C is the hypotenuse

#1.

6^2 + 8^2 = c^2

36 + 64 = c^2

100 = c^2

c = 10

#2.

5^2 + 12^2 = c^2

25 + 144 = c^2

169 = c^2

c = 13

#4.

10^2 + 4^2 = c^2

100 + 16 = c^2

116 = c^2

c = 4\sqrt{29}

#5.

15^2 + 9^2 = c^2

225 + 81 = c^2

306 = c^2

c = 3\sqrt{34}

#6.

120^2 + b^2 = 150^2

14400 + b^2 = 22500

b^2 = 8100

b = 90

#7.

144^2 + b^2 = 194^2

20736 + b^2 = 37636

b^2 = 16900

b = 130

Hope this helps!! :)

4 0
3 years ago
Elliott has 5 1/2 bags of birdseed.He uses 3 2/5 bags to feed the birds at the park.He puts the rest of the birdseed in the bird
andre [41]

Answer:

Elliott put 2.1 bags of birdseed in the feeders

Step-by-step explanation:

(5 1/2) - (3 2/5) = 2.1

I hope I helped! :)

Would appreciate Brainliest! ;)

8 0
2 years ago
Awner this asap pls my cousins homework more coming up
mixas84 [53]
The answer is option 1
3 0
3 years ago
Find the solution of the given initial value problem. ty' + 2y = sin t, y π 2 = 9, t > 0 y(t) =
Helen [10]

For the ODE

ty'+2y=\sin t

multiply both sides by <em>t</em> so that the left side can be condensed into the derivative of a product:

t^2y'+2ty=t\sin t

\implies(t^2y)'=t\sin t

Integrate both sides with respect to <em>t</em> :

t^2y=\displaystyle\int t\sin t\,\mathrm dt=\sin t-t\cos t+C

Divide both sides by t^2 to solve for <em>y</em> :

y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac C{t^2}

Now use the initial condition to solve for <em>C</em> :

y\left(\dfrac\pi2\right)=9\implies9=\dfrac{\sin\frac\pi2}{\frac{\pi^2}4}-\dfrac{\cos\frac\pi2}{\frac\pi2}+\dfrac C{\frac{\pi^2}4}

\implies9=\dfrac4{\pi^2}(1+C)

\implies C=\dfrac{9\pi^2}4-1

So the particular solution to the IVP is

y(t)=\dfrac{\sin t}{t^2}-\dfrac{\cos t}t+\dfrac{\frac{9\pi^2}4-1}{t^2}

or

y(t)=\dfrac{4\sin t-4t\cos t+9\pi^2-4}{4t^2}

6 0
2 years ago
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