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2 years ago

What is the missing exponent 25=5

2 answers:

7 0

First let's look at the equation:

25=5

So, now we know that we are going to need to find what would = 25

So 25/5=5
As well as 5*5=25

So now we know that the missing exponent will be 5, since it fits in there so perfectly!! :)

Your answer is 5

Ber [7]2 years ago

5 0

Answer:

the missing exponent is 2

Step-by-step explanation:

Well 5^2 is equals to 5×5 which we know is qual

to 25.

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What is the slope intercept form of y= -2

Mashcka [7]

The slope-intercept form: y = mx + b

(m - slope, b - y-intercept)

We have y = -2. It's the slope intercept form where m = 0 and b = -2.

y = -2 → y = 0x - 2

8 0

3 years ago

A cubical water tank is 1.2 m long. What is the capacity of the tank in litres ?

torisob [31]

Answer:it turned out I was not gay after all

Step-by-step explanation:I was really into girls

3 0

2 years ago

The density of a certain material is such that it weighs 6 pounds per pint of volume.

ella [17]

Answer:

Density = 5751.66 gram per litre

Step-by-step explanation:

It is given that the density of a certain material 6 pounds per pint.

The conversion from pint to liter is as follows :

1 pint = 0.473 litre

The conversion from pounds to kg is as follows :

1 pound = 453.59 kg

We need to express the density from pounds per pint to grams per liter.

1 pound per pint = 958.611 gram per litre

So,

6 pound per pint = 958.611 × 6 gram per litre

6 pound per pint = 5751.66 gram per litre

3 0

2 years ago

How do you solve interquartile range?

77julia77 [94]

By Hand
Step 1:
Put the numbers in order.
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27.

Step 2:
Find the median.
1, 2, 5, 6, 7, 9, 12, 15, 18, 19, 27.

Step 3:
Place parentheses around the numbers above and below the median.
Not necessary statistically, but it makes Q1 and Q3 easier to spot.
(1, 2, 5, 6, 7), 9, (12, 15, 18, 19, 27).

Step 4:
Find Q1 and Q3
Think of Q1 as a median in the lower half of the data and think of Q3 as a median for the upper half of data.
(1, 2, 5, 6, 7), 9, ( 12, 15, 18, 19, 27). Q1 = 5 and Q3 = 18.

Step 5:
Subtract Q1 from Q3 to find the interquartile range.
18 – 5 = 13.

5 0

2 years ago

Read 2 more answers

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$

6 0

3 years ago