Find the value of y when x= -12.<br> y=x/2 +9<br>y =

Which postulate or theorem can be used to

sveticcg [70]

Answer:

AA Similarity Postulate

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is proportional and its corresponding angles are congruent

step 1

Verify the proportion of the corresponding sides

$\frac{PS}{PQ}=\frac{PT}{PR}$

substitute

$\frac{45}{20}=\frac{36}{16}$

$2.25=2.25$ ----> is true

Corresponding sides are proportional

Triangle PQR is similar to Triangle PST

That means

Corresponding angles must be congruent

side QR is parallel side ST

and

$m\angle PQR=m\angle PST$ ----> by corresponding angles

$m\angle PRQ=m\angle PTS$ --> by corresponding angles

so

PQR is similar to PST by AA Similarity Postulate

4 0

4 years ago

A particular isotope has a half-life of 74 days. If you start with 1 kilogram of this isotope, how much will remain after 150

shtirl [24]

$\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}$

$\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}$

6 0

3 years ago

Robert and Nelson are in a

Arte-miy333 [17]

15 km
12+3=15 in total if the school is the meeting point of these two distances.

3 0

3 years ago

The angle of depression from a helicopter to a landing pad is 37 degrees. if the helicopter is 1250 feet from the ground. what i

djyliett [7]

The horizontal distance from the helicopter to the landing pad is 1658.81 feet

<em><u>Solution:</u></em>

The figure is attached below

Triangle ABC is a rightangled triangle

A helicopter is flying at point A and landing pad is at point c

Angle of depression of the helicopter is 37 degrees so angle of elevation of this helicopter from landing pad will be same as 37 degrees

The helicopter is 1250 feet from the ground

Therefore, AB = 1250 feet

To find: horizontal distance from the helicopter to the landing pad

BC is the horizontal distance from the helicopter to the landing pad

BC = ?

By the definition of tan,

$tan \theta = \frac{opposite}{adjacent}$

$tan 37 = \frac{AB}{BC}\\\\tan 37 = \frac{1250}{BC}\\\\ 0.75355 = \frac{1250}{BC}\\\\BC = \frac{1250}{0.75355}\\\\BC = 1658.81$

Thus the horizontal distance from the helicopter to the landing pad is 1658.81 feet

5 0

3 years ago

The mass of a cube,

Nonamiya [84]

The mass in kg is 106.27 if the mass of a cube, M (kg), is proportional to the cube of the length of its edge, L(m).

<h3>What is a proportional relationship?</h3>

It is defined as the relationship between two variables when the first variable increases, the second variable also increases according to the constant factor.

We have:

The mass of a cube, M (kg), is proportional to the cube of the length of its edge, L(m).

M ∝ L³

After removing proportional sign

M = cL³

Plug M = 50 kg and L = 70 cm = 0.7 m

50 = c(0.7)³

c = 145.77 kg/m³

If L = 0.9 m, then M

M = (145.77 kg/m³)(0.9 m)³

M = 106.266 ≈ 106.27 kg

Thus, the mass in kg is 106.27 if the mass of a cube, M (kg), is proportional to the cube of the length of its edge, L(m).

Learn more about the proportional here:

brainly.com/question/14263719

#SPJ1

3 0

2 years ago

Find the value of y when x= -12. y=x/2 +9y =​

Find the value of y when x= -12. y=x/2 +9y =