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jolli1 [7]
3 years ago
10

Can someone plss help me

Mathematics
1 answer:
puteri [66]3 years ago
7 0
0/3 is the same thing as 0, so write those nxt to each other. Then 1/3, 2/3. Lastly 3/3 is the same as 1, so write those nxt to each other.
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2 (10- 22x0) = (2x + 2)
Lorico [155]

Answer:

x=9

Step-by-step explanation:

2(10-22×0)=(2x+2)

Applying BODMAS

2(10-0)=(2x+2)

2(10)=2x+2

2x+2=20

2x=20-2

2x=18

x=9

3 0
3 years ago
What equation has a solution of 2/3 for x
Ahat [919]

Answer:

D

Step-by-step explanation:

Hope this helped! Please tell me if I'm wrong!

6 0
2 years ago
Consider the initial value problem:
Inessa [10]

Answer:

\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2

Step-by-step explanation:

The given initial value problem is;

y''-5y'+6y=-5\sin(2t)---(1)\\y(0)=-5,y'(0)=2

Let

u(t)=y(t)----(2)\\v(t)=y'(t)----(3)

Differentiating both sides of equation (1) with respect to t, we obtain:

u'(t)=y'(t)---(4)\\ \\\implies u'(t)=v(t)---(5)

Differentiating both sides of equation (2) with respect to t gives:

v'(t)=y''(t)----(6)

From equation (1),

y''=5y'-6y-5\sin(2t)\\y''=5v(t)-6u(t)-5\sin(2t)\\\implies v'(t)=5v(t)-6u(t)-5\sin(2t)----(7)

Putting t=0 into equation (2) yields

u(0)=y(0)\\\implies u(0)=-5

Also putting t=0 into equation (3)

u'(0)=y'(0)\\u'(0)=2

The system of first order equations is:

\left \{ {{u'=v} \atop {v'=5v-6u-5\sin(2t)}} \right. \\u(0)=-5;u'(0)=2

3 0
3 years ago
Please answer this . It’s of class 8 Thank you
Katen [24]

Answer:

= (3/5)-³

= (5/3)³

= (5/3) × (5/3) × (5/3)

= (125/27)

Hope it helps.

3 0
3 years ago
Read 2 more answers
Find the value of x. Round to the nearest tenth.
elena55 [62]

Using tan(x) = 4/3

X = arctan(4/3)

X = 53.1 degrees

4 0
3 years ago
Read 2 more answers
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