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3 years ago

I need help plz show work!!!

Mathematics

1 answer:

Degger [83]3 years ago

4 0

500 cents in all Is equivalent to $5
6-cent: 20
8-cent: 30
10-cent: 10
20-cent: 2(states in word problem)

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The radius of a circle is 16 feet. What is the circle's area?<br> Use 3.14 for .<br> square feet

Alla [95]

Answer:

803.84 feet squared

Step-by-step explanation:

A=pi*r^2

16^2=256

256*pi=803.84

7 0

3 years ago

Energy released as a gas changes into a liquid is called what? (#9)

Zinaida [17]

Answer:

Energy will be released to change from liquid to solid (fusion), gas to liquid (condensation), or gas to solid. Latent heat of evaporation is the energy used to change liquid to vapor. IMPORTANT: The temperature does not change during this process, so heat added goes directly into changing the state of the substance

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Help with 5 and 6 please !! (13 points!!)

melamori03 [73]

Hi i can help you but i do not understand your question

8 0

4 years ago

Read 2 more answers

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

3 years ago

In J, K is the midpoint. The coordinates of ) are (2, 2), and the coordinates of Kare (10, 11). What are the coordinates of L?

DiKsa [7]

Answer:

L(18, 20)

Step-by-step explanation:

In JL, K is the midpoint. The coordinates of J are (2, 2), and the

coordinates of K are (10, 11). What are the coordinates of L?

Solution:

If O(x, y) is the midpoint between two points A( $x_1,y_1$ ) and B( $x_2,y_2$ ). The equation to determine the location of O is given by:

$x=\frac{x_1+x_2}{2} \\\\y=\frac{y_1+y_2}{2}$

Since JL is a line segment and K is the midpoint. Given the location of J as (2, 2) and K as (10, 11). Let ( $x_2,y_2$ ) be the coordinate of L. Therefore:

$10=\frac{2+x_2}{2} \\\\20=2+x_2\\\\x_2=18$

$11=\frac{2+y_2}{2} \\\\22=2+y_2\\\\y_2=20$

Therefore L = (18, 20)

4 0

3 years ago