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3 years ago

Could someone please help me with this

Mathematics

1 answer:

natka813 [3]3 years ago

3 0

Answer:

24 m/s

Step-by-step explanation:

Go to 20s on the graph(located on the x-axis) then go up. The point is right under 25m/s, therefore it is 24 m/s.

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Which expressions are equivalent to 2-(-6+3)+4c

Artyom0805 [142]

Answer:

c. none of the above

Step-by-step explanation:

(-6+3)= -3

2--3=2+3=5

5+4c cant add them since they arent like terms

final answer 5+4c and that option isnt here

3 0

3 years ago

The perimeter of (the distance around) ABCD is 66, and DC is twice as long as CB. How long is AB?

ArbitrLikvidat [17]

Answer:

$AB=22\ units$

Step-by-step explanation:

we know that

In this problem ABCD is a rectangle

DC=AB

BC=AD

Let

BC ---> the length of rectangle

DC ---> the width of rectangle

we know that

The perimeter of rectangle is equal to

$P=2(BC+DC)$

we have

$P=66\ units$

$66=2(BC+DC)$

simplify

$33=BC+DC$ ----> equation A

$DC=2BC$ ----> equation B

substitute equation B in equation A

$33=BC+2BC$

Solve for BC

Combine like terms

$33=3BC$

Divide by 3 both sides

$BC=11\ units$

<em>Find the value of DC</em>

$DC=2BC$ ----> $DC=2(11)=22\ units$

Remember that

DC=AB

therefore

$AB=22\ units$

3 0

3 years ago

Need more help 5/6 divided by 2

podryga [215]

Answer:

5/12

Step-by-step explanation:

5/6 divided by 2

= 5/6 divided by 2/1

= 5/6 x 1/2

= 5 x 1 over 6 x 2

= 5 over 6 x 2

= 5 / 12

8 0

3 years ago

I need this answer!!!

Kamila [148]

Answer:

2 feet per second

Step-by-step explanation:

6/3 is 2

3 0

3 years ago

Read 2 more answers

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

7 0

3 years ago

Could someone please help me with this​

Could someone please help me with this