3 years ago

Can someone plss help me

1 answer:

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:

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