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3 years ago

How do you write 26/10 as a percentage

1 answer:

5 0

26/10 as a percentage would be 260%.
The way we explain this is you multiply 26 * 10 and 10 * 10. You get 260/100.
260/ 100 as a percentage is 260%.

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Convert the equation to y=mxtb 2x-3y=-6 please show the steps .

stellarik [79]

You subtract the x from both sides of the equation so it will then be -3y = -2x - 6. You then divide each number by -3, -3y divided by -3, -2x divided by -3, and the - 6 divided by negative 3. It would equal y = 2/3x + 2

7 0

3 years ago

Is p= -1 a solution to the equation 6(p-4)-30p ?

kirill [66]

Answer:

Yes

Step-by-step explanation:

6(-1-4)-30(-1) can be simplified into 6(-5)+30 which becomes -30+30, and then 0.

3 0

3 years ago

5² x 5³ = Choices: 5 55 5^5 5^6

Vesna [10]

Answer:

5^5

Step-by-step explanation:

when multiplying two expression like this you should add up their powers

in this case it's 2 + 3

4 0

3 years ago

Read 2 more answers

"Based on the following construction, which of the following are true? choose all that apply."

umka21 [38]

From the diagram,

Angle

Bisecting an angle means dividing the angle into two equal part.

1/2 m

$\begin{gathered} \text{ } \\ \\ \end{gathered}$

4 0

1 year ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

4 0

3 years ago