All categories

3 years ago

Solve and check please help

Mathematics

2 answers:

Virty [35]3 years ago

4 0

Answer:

7) 24 8) 75 9) .9

Step-by-step explanation:

You multiply the denominatior with the whole number.

Leno4ka [110]3 years ago

4 0

Answer:

x = 24 b = 75 s = 0.9

Step-by-step explanation:

7. Multiply both sides by 2 and you get x = 24

8. Multiply both sides by 5 and you get b = 75

9. Multiply both sides by 0.3 and you get s = 0.9

You might be interested in

swat32

Answer:

55a=

step-by-step explanation:

5 0

2 years ago

4) If a = 28 and b = 4, then a/b = ______

exis [7]

Answer:

Step-by-step explanation:

You divided both 28 and 4 and you will get 7 as your answer.

7 0

3 years ago

A nugget of gold is placed in a graduated cylinder that contains 85 mL of water.the water level rises to 225 mL after the nugget

Rufina [12.5K]

To get the volume of an object, one of the oldest techniques is to fill a container with water of which we know the volume (in this case 85mL), and drop the object in. The water will rise and show a new volume (225mL). If we subtract the old volume (85) from the new volume (225), we will get the volume of the object (225 - 85 = 140).

8 0

3 years ago

place the decimal point in the answer to make correct. explain your reasoning. 3.9853 × 8.033856= ??????

tatiyna

It is 32.0173263 because you add spaces to the decimal point and of course you don't put the zeros in

7 0

3 years ago

Consider f and c below. f(x, y) = x2 i + y2 j c is the arc of the parabola y = 2x2 from (−1, 2) to (1, 2) (a) find a function f

Korvikt [17]

If there is some scalar function $f(x,y)$ such that

$\nabla f(x,y)=\mathbf f(x,y)=x^2\,\mathbf i+y^2\,\mathbf j$

then we want to find $f$ such that

$\dfrac{\partial f}{\partial x}=x^2\implies f(x,y)=\dfrac{x^3}3+g(y)$
$\dfrac{\partial f}{\partial y}=y^2=\dfrac{\mathrm dg}{\mathrm dy}\implies g(y)=\dfrac{y^3}3+C$
$\implies f(x,y)=\dfrac{x^3}3+\dfrac{y^3}3+C$

So the vector field $\mathbf f(x,y)$ is conservative, which means the fundamental theorem applies; the line integral of $\mathbf f$ along any path $\mathcal C$ parameterized by some vector-valued function $\mathbf r(t)$ over $a\le t\le b$ is given by

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=\int_{t=a}^{t=b}\mathbf f(\mathbf r(t))\cdot\dfrac{\mathrm d\mathbf r}{\mathrm dt}\,\mathrm dt=f(\mathbf r(b))-f(\mathbf r(a))$

In this case,

$\displaystyle\int_{\mathcal C}\mathbf f\cdot\mathrm d\mathbf r=f(1,2)-f(-1,2)=\dfrac23$

5 0

3 years ago