Jane is 7 years older than John.If the sum of their ages is 43.How old is Jane and how old is John​

Answers

Answer 1
Answer:

Answer:

If

Jane= x

John = y

y+7= x

x-y=7 ---- (1)

x+y=43 --- (2)

(1) + (2),

x-y+x+y=7+43

2x=50

x= 50÷2

x= 25

substituting x= 25 in (2),

25+y=43

y= 43-25

y= 18

Jane is 25 years old and John is 18 years old.

Step-by-step explanation:


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Please answer I’ll rate brainlyest

"What is the measure of angle ABC?" is this a statement

Answers

Answer:42.5

Step-by-step explanation:

no that is a question, statements are different

HELPPPPPPP PLS MATH!! MKARKING BRAINNIEST!!!!!!!!!!

Answers

Answer:

A

Step-by-step explanation:

Because i said so

If a substance has a half life of 850 years, and there are 35 milligrams present now, how long will it take for the substance to be reduced to 10 miligrams calculus

Answers

This is a problem involving differential equation. The rate of decaying substance is directly proportional to the amount of substance.
dP/dt = kA
where k = proportionality constant, A = amount of subtance.

Integrating that equation yields,
A = ce^(kt)
where c = constant of integration.
The constant of integration is represented mostly as an initial amount of substance.

We have given two initial condition:
(1) t = 0, A = 35 milligrams (present)
(2) t = 850 years, A = 35/2 milligrams (half-life)

We are going to solve for c:
35 = ce^0
c = 35

Next, solving for k:
35/2 = 35e^(k*850)
1/2 = e^(k*850)
Multiplying natural logarithm on both sides to bring down k:
ln (1/2) = k*850
k = ln (1/2) / 850
k = -0.00081547

Finally, the time when the amount of substance has 10 milligrams remaining is:
10 = 35e^(-0.00081547*t)
10/35 = e^(-0.00081547*t)
Multiplying natural logarithm on both sides to bring down t:
ln (2/7) = -0.00081547*t
t = ln (2/7) / -0.00081547
t = 1536.25 years (ANSWER)

Keshawn is asked to compare and contrast the domain and range for the two functions.f(x) = 5x
g(x) = 5^x
Which statements could he include in his explanation? Check all that apply.
The domain of both functions is all real numbers.
The domain of f(x) is x > 5.
The domain of g(x) is x > 5.
The range of both functions is y > 5.
The range of f(x) is y > 0.
The range of g(x) is y > 0.

Answers

Answer:

Hence, option 1 is correct.

Hence, last option is correct.

Step-by-step explanation:

We have given two function

f(x)=5x  and g(x)=5^x

Domain is the values x can take and

And range is the values y can take

Hence, x can take any real value in both the functions in f(x) and g(x)

Hence, option 1 is correct

Now, for range

range of f(x) can be any real number

But range of g(x) can be positive number

Hence, last option is correct.

Rest all options are incorrect.

He could include two statements:
-  The domain of both functions is all real numbers.
-  The range of g ( x ) is y > 0 

The measure of W is 134°, the measure of Y is 46°, and the measure of Z is 134°.What is the measure of X?
a 40°
b 46°
c 134°
d 90°

Answers

The angles in any quadrilateral must add up to 360 degrees.
A trapezoid is a quadrilateral, because it has four sides.

W+X+Y+Z=360 \n  \n (134)+X+(46)+(134)=360 \n  \n 314+X=360 \n  \n X=46

Angle X is (b) 46 degrees.
B. 46 because the measure of all of the angles in a quadrilateral is 360. If you add 134+46+134 it equals 314, and 360-314 is 46.
 

An object is traveling around a circle with a radius of 10 cm. If in 20 seconds a central angle of 1/3 radian is swept out, what is the linear speed of the object?

Answers

The linear speed of the object is (1)/(6)

  • The Radius, r = 10 cm
  • The Fraction of Radius swept out = (1)/(3)
  • The distance = 10 / (1)/(3)
  • The distance covered = (10)/(3)

The linear speed :

  • Radius swept out ÷ time taken
  • (10)/(3) * (1)/(20) = (1)/(6)

The linear speed of the object is (1)/(6)

Learn more : brainly.com/question/15154430?referrer=searchResults

Firstly, we will find the distance covered by the object in the 20 seconds:

\text{angle}(in~radians)=\frac{\text{arch}}{\text{radius}}\Longrightarrow (1)/(3)=(d)/(10~cm)\iff d=(10)/(3)~cm

We must know that v=(d)/(\Delta t). So:

v=(d)/(\Delta t)\Longrightarrow v=((10)/(3)~cm)/(20~s)\iff \boxed{v=(1)/(6)~cm/s}