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3 years ago

Compare the values of the 2s and 5s in 55,220

1 answer:

4 0

The value of the first 5 is 5 ten thousands and the second 5 i 5 thousands while the value of the first 2 is 2 hundreds and the second 2 is 2 tens.

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On Saturday, many tickets fir a hockey game were sold. Children and adult were in the ratio of 4:7. There are 88 tickets in tota

ollegr [7]

Answer:

<em>56 adult tickets were sold</em>

Step-by-step explanation:

Ratio of ticket sold to Children and adult is in the ratio 4 : 7

Total ratio = 4 +7 = 11

Ratio of adult = 7

Total ticket sold = 88

Number of adult tickets sold = Ratio of adult/Total ratio * Total ticket sold

Number of adult tickets sold = 7/11 * 88

Number of adult tickets sold = 7 * 8

Number of adult tickets sold = 56 tickets

<em>Hence 56 adult tickets were sold</em>

3 0

3 years ago

Find the domain of the function in interval notation: f(x) =1/ x + 22

cricket20 [7]

That's the answer

x = - 1/22

6 0

3 years ago

How do you find the angle measure of a right triangle using all 3 sides?

r-ruslan [8.4K]

Use sin, cos, tan formula

7 0

3 years ago

How do you write 2sin5cos5 as a single trigonomic ratio?

erastova [34]

Answer:

ertyui

Step-by-step explanation:

6 0

3 years ago

One of the roots of the quadratic equation dx^2+cx+p=0 is twice the other, find the relationship between d, c and p

scZoUnD [109]

Answer:

$c^2 = 9dp$

Step-by-step explanation:

Given

$dx^2 + cx + p = 0$

Let the roots be $\alpha$ and $\beta$

So:

$\alpha = 2\beta$

Required

Determine the relationship between d, c and p

$dx^2 + cx + p = 0$

Divide through by d

$\frac{dx^2}{d} + \frac{cx}{d} + \frac{p}{d} = 0$

$x^2 + \frac{c}{d}x + \frac{p}{d} = 0$

A quadratic equation has the form:

$x^2 - (\alpha + \beta)x + \alpha \beta = 0$

So:

$x^2 - (2\beta+ \beta)x + \beta*\beta = 0$

$x^2 - (3\beta)x + \beta^2 = 0$

So, we have:

$\frac{c}{d} = -3\beta$ -- (1)

and

$\frac{p}{d} = \beta^2$ -- (2)

Make $\beta$ the subject in (1)

$\frac{c}{d} = -3\beta$

$\beta = -\frac{c}{3d}$

Substitute $\beta = -\frac{c}{3d}$ in (2)

$\frac{p}{d} = (-\frac{c}{3d})^2$

$\frac{p}{d} = \frac{c^2}{9d^2}$

Multiply both sides by d

$d * \frac{p}{d} = \frac{c^2}{9d^2}*d$

$p = \frac{c^2}{9d}$

Cross Multiply

$9dp = c^2$

$c^2 = 9dp$

Hence, the relationship between d, c and p is: $c^2 = 9dp$

8 0

3 years ago