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3 years ago

I tried to work this out, help

Mathematics

2 answers:

PilotLPTM [1.2K]3 years ago

5 0

Two angles are complementary if the give 90° when summed.

Since angle 1 is right, angles 2 and 3 form a right angle as well, and thus they are complementary.

Oksanka [162]3 years ago

4 0

Answer:

∠2

Step-by-step explanation:

Complementary angles are angles that add up to 90°

So ∠3 + ∠2 = 90°

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jolli1 [7]

Answer:

373

Step-by-step explanation:

5 0

3 years ago

Graph the equation y= x^2 - 2x - 8 on the accompanying set of axes. You must plot 5 points including the roots and the vertex.

Vanyuwa [196]

Answer:

I'm not so sure about this one so sorry if this is wrong

Step-by-step explanation:

7 0

3 years ago

Combine the radicals. 3√5-8√5+2√5

weeeeeb [17]

Answer: $-3\sqrt{5}$

-3 times the square root of 5

=============================================

Explanation:

Let $x = \sqrt{5}$

Replace all the root 5 terms with x and we go from

$3\sqrt{5}-8\sqrt{5}+2\sqrt{5}$

$3x-8x+2x$

From here, combine like terms to get

$3x-8x+2x = -5x+2x = -3x$

and the last thing to do is replace the x with sqrt(5)

$-3x = -3\sqrt{5}$

Meaning that,

$3x-8x+2x = -3x$

$3\sqrt{5}-8\sqrt{5}+2\sqrt{5} = -3\sqrt{5}$

7 0

2 years ago

SOMEONE PLEASE HELP ME ASAP PLEASE!!!!!

hammer [34]

I think you got confused with the 50 and 90.
Read my NOTE there it will surely help you answering other questions like this

5 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

2 years ago