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3 years ago

Write the perfect squares that are greater than 150 and less than 250

Mathematics

2 answers:

Shkiper50 [21]3 years ago

8 0

Answer:

169, 196, 225

Step-by-step explanation:

13*13=169

14*14=196

15*15=225

Can I get brainliest!

Fiesta28 [93]3 years ago

3 0

169 (13x13)
196 (14x14)
225 (15x15)

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Which of the following is an asset

Andrej [43]

Answer: B. House mortgage

Step-by-step explanation: Anything tangible or intangible that can be sold or profitable is considered an asset. Hope it helps! xo

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2 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 $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

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

7 0

2 years ago

The Hudson Bay tides vary between 3 feet and 9 feet. The tide is at its lowest point when time (t) is 0 and completes a full cyc

BaLLatris [955]

Answer:

$y(t)= 6-3cos(\dfrac{2\pi}{14}t )$

Step-by-step explanation:

The function that could model this periodic phenomenon will be of the form

$y(t) = y_0+Acos(wt)$

The tide varies between 3ft and 9ft, which means its amplitude $A$ is

$A =\dfrac{(9-3)ft}{2} \\\\\boxed{A = 3ft}$

and its midline $y_0$ is

$y_o=3+3 \\\\\boxed{y_o= 6ft}$ .

Furthermore, since at $t=0$ the tide is at its lowest ( 3 feet ), we know that the trigonometric function we must use is $-cos(\omega t)$ .

The period of the full cycle is 14 hours, which means

$\omega t =2\pi$

$\omega (t+14)= 4\pi$

giving us

$\boxed{\omega = \dfrac{2\pi}{14}.}$

With all of the values of the variables in place, the function modeling the situation now becomes

$\boxed{y(t)= 6-3cos(\dfrac{2\pi}{14}t ).}$

8 0

3 years ago

What is the volume of a cone when the radius is 6 and the height is 13

lidiya [134]

Answer:

<h2>489.84 units³</h2>

Step-by-step explanation:

Given,

Radius ( r ) = 6

Height ( h ) = 13

Volume of cone = ?

Now, Let's find the volume of cone:

$= \pi \: {r}^{2} \frac{h}{3}$

plug the values

$= 3.14 \times {6}^{2} \times \frac{13}{3}$

Evaluate the power

$= 3.14 \times 36 \times \frac{13}{3}$

Calculate

$489.84$ units³

Hope this helps..

Best regards!!

6 0

2 years ago

Coach Pettaway is buying basketball equipment for his team. He gets a

soldier1979 [14.2K]

First, let’s all acknowledge that whoever comes up with problems like this WANTS kids to hate math...smh

I’m sure there is a prettier way to solve this, but here’s what I did:

8(2.25) + 3(22.50) =
18 + 67.50 = 85.50 per “set” of balls/jerseys
400/85.50 = 4.678 = number of “sets” he can buy. Round down to 4 so we have room for tax.

85.5 x 4 “sets”= $342
Tax on 342 is 0.06 x 342 = 20.52

$342 + 20.52 = $362.52 spent

Basketballs = 4 sets x 8 balls per set= 32
Jerseys = 4 sets x 3 jerseys per set= 12

32 basketballs, 12 jerseys, $362.52 spent

3 0

2 years ago