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2 years ago

What is -2 × 1 1/4=

Mathematics

1 answer:

RoseWind [281]2 years ago

5 0

Answer: -11/2

Step-by-step explanation:

Just set it up so it looks like this:

-2/1 x 11/4 and multiply across the top, which would be -22 and then the across the bottom, which would equal 4. So it’s -22/4. BUT you can simplify it down to -11/2 by dividing both sides of the fraction by 2. So then answer is -11/2

Hope that makes sense! :)

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770,070 in expanded form

Shalnov [3]

700,000 + 70,000 + 70

is your answer

hope this helps :D

4 0

3 years ago

Read 2 more answers

A cone-shaped paper drinking cup is to be made to hold 27 cm of water. Find the height h and radius r of the cup that will use t

andreyandreev [35.5K]

Answer:

r = 2.632 cm
h = 3.722 cm

Step-by-step explanation:

The formula for the volume of a cone of radius r and height h is ...

V = (1/3)πr²h

Then r² can be found in terms of h and V as ...

r² = 3V/(πh)

The lateral surface area of the cone is ...

A = (1/2)(2πr)√(r² +h²) = πr√(r² +h²)

The square of the area is ...

T = A² = π²r²(r² +h²)

Substituting for r² using the expression above, we have ...

T = π²(3V/(πh))((3V/(πh) +h²) = 9V²/h² +3πVh

We want to find the minimum, which we can do by setting the derivative to zero.

dT/dh = -18V²/h³ +3πV

This will be zero when ...

3πV = 18V²/h³

h³ = 6V/π . . . . . multiply by h³/(3πV)

For V = 27 cm³, the value of h that minimizes paper area is ...

h = 3∛(6/π) ≈ 3.7221029

The corresponding value of r is ...

r = √(3V/(πh)) = 9/√(π·h) ≈ 2.6319242

The optimal radius is 2.632 cm; the optimal height is 3.722 cm.

_____

The second derivative test applied to T finds that T is always concave upward, so the value we found is a minimum.

Interestingly, the ratio of h to r is √2.

8 0

3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :