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3 years ago

What is the value of x?

Mathematics

1 answer:

AveGali [126]3 years ago

5 0

x=27+360n,153+360n, for any integer n

I hope this help. Have a blessed day!

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Please help I need it

ohaa [14]

Answer:

6.300

Step-by-step explanation:

it is not 0.500 or above so you would round down to the nearest thousandth

5 0

2 years ago

A line passes through the point 6, -8 and has a slope of -3/2

rewona [7]

Answer: y=-3/2x+-8

Step-by-step explanation:

6 0

3 years ago

At the beginning of an experiment, a scientist has 300 grams of radioactive goo. After 150 minutes, her sample has decayed to 37

monitta

Answer:

Half-life of the goo is 49.5 minutes

$G(t)= 300e^{-0.014t}$

191.7 grams of goo will remain after 32 minutes

Step-by-step explanation:

Let $M_0\,,\,M_f$ denotes initial and final mass.

$M_0=300\,\,grams\,,\,M_f=37.5\,\,grams$

According to exponential decay,

$\ln \left ( \frac{M_f}{M_0} \right )=-kt$

Here, t denotes time and k denotes decay constant.

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\\\ln \left ( \frac{37.5}{300} \right )=-k(150)\\-2.079=-k(150)\\k=\frac{2.079}{150}=0.014$

So, half-life of the goo in minutes is calculated as follows:

$\ln \left ( \frac{50}{100} \right )=-kt\\\ln \left ( \frac{50}{100} \right )=-(0.014)t\\t=\frac{-0.693}{-0.014}=49.5\,\,minutes$

Half-life of the goo is 49.5 minutes

$\ln \left ( \frac{M_f}{M_0} \right )=-kt\Rightarrow M_f=M_0e^{-kt}$

So,

$G(t)= M_f=M_0e^{-kt}$

Put $M_0=300\,\,grams\,,\,k=0.014$

$G(t)= 300e^{-0.014t}$

Put t = 32 minutes

$G(32)= 300e^{-0.014(32)}=300e^{-0.448}=191.7\,\,grams$

7 0

3 years ago

A helicopter was -28.4 feet below sea level in a canyon and ascended 35.2 feet. What is the location of the helicopter after it

Brilliant_brown [7]

Answer:

6.8 ft

Step-by-step explanation:

A helicopter was -28.4 feet below sea level in a canyon and ascended 35.2 feet.

Ascending means add

-28.4 + 35.2

6.8 ft

5 0

4 years ago

Read 2 more answers

Triangle GFH has vertices G(2, –3), F(4, –1), and H(1, 1). The triangle is rotated 270° clockwise using the origin as the center

Rufina [12.5K]

Answer:

See attachment.

Step-by-step explanation:

The given triangle,GFH has vertices G(2, –3), F(4, –1), and H(1, 1).

When this triangle is rotated 270° clockwise using the origin as center of rotation; then the coordinates of the image triangle, G'F'H' can be obtained using the mapping;

$(x,y)\to (-y,x)$

This implies that the new triangle will have vertices at;

G'(3, 2), F'(1, 4), and H'(-1, 1).

The correct choice is shown in the attachment.

7 0

3 years ago

Read 2 more answers