All categories

3 years ago

What is the solution to q+(-9)=12

2 answers:

7 0

q−9=12

Step 1: Add 9 to both sides.

q−9+9=12+9

q=21

Answer:

q=21

Volgvan3 years ago

4 0

Q + (-9) = 12
-9 + q = 12
-9 + 9 + q = 12 + 9
9 + 9 = 0
0 + q = 12 + 9
q = 12 + 9
12 + 9 = 21
q = 21
Hope this helps.

You might be interested in

HELP ME FIND SLOPE!! PLZ

Murrr4er [49]

Answer:

Step-by-step

the slope is 1, 1

explanation:

rise over run it is a positive slope btw

5 0

2 years ago

PLEASE HELP ME! i think its A

natali 33 [55]

Answer:

The answer is 3rd option.

Step-by-step explanation:

In a graph, domain is always represented by the x-axis. So by looking at the graph, the minimum value of x is -2. We can know that x ≥ -2.

5 0

3 years ago

Read 2 more answers

I need help please I will mark you brainlists

V125BC [204]

Answer:

2. ΔABC is isosceles

Step-by-step explanation:

Given : m∠A = 50

m∠BCD = 180 - (m∠BDC + m∠DBC)

= 180 - (100 + 30)

= 50

then

m∠A = m∠BCD

therefore ΔABC is isosceles

6 0

3 years ago

The earth rotates once per day about an axis passing through the north and south poles, an axis that is perpendicular to the pla

forsale [732]

Answer:

Speed at Equator = 463.97 meters per second

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

Step-by-step explanation:

The formula is:

$v=\frac{2 \pi R}{T}$

Where

v is speed

R is radius

T is time

and another formula for centripetal acceleration:

$a_c=\frac{4 \pi^{2} R}{T^2}$

Now,

at equator, the radius is radius of earth (given), time in seconds is

T = 24 * 60 * 60 = 86,400

THus,

$v_E=\frac{2 \pi (6.38*10^{6}}{86,400}=463.97$

Speed at Equator = 463.97 meters per second

Centripetal Acceleration:

$a_{cE}=\frac{v_E^2}{R_E}=\frac{463.97}{6.38*10^{6}}=3.37*10^{-2}$

Centripetal Acceleration at Equator = $3.37*10^{-2}$ meters per second squared

At 30.0° north of the equator:

$R_N=R_E Cos (30)= (6.38*10^6)Cos(30)=5.53*10^6$

Now,

Speed = $v_{30N}=\frac{2 \pi (5.53*10^6)}{86,400}=401.79$

Speed at 30 degrees north of equator = 401.79 meters per second

Centripetal Acceleration:

$a_{30N}=\frac{v_E^2}{R_E}=\frac{401.79}{5.53*10^6}=2.92*10^{-2}$

Centripetal Acceleration at 30 degrees north of equator = $2.92*10^{-2}$ meters per second squared

4 0

3 years ago

Keon and Lesha are comparing their daily calorie intake. Each day, Lesha chooses healthy 100-calories snacks such as apples and

Maksim231197 [3]

Scale for the horizontal axis is in increments of 1. Scale for the vertical axis in in increments of 100. Black – Lesha, Blue - Keon Part B:What does the rate of change for each line represent? The rate of change represents the amount of calories per snack.Part C:What is Lesha’s rate of change? Answers will vary.Example: 100 calories/snack. Part D:How does Lesha’s rate of change compare to Keon’s rate of change? Why are the two rates different from one another? In this case it doubles - the rates are different because with healthy snacks the caloric intake per snack is less than unhealthy snacks.

7 0

3 years ago

Read 2 more answers