Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 22 Exercise 1 Answer
Given :
To write: The formula for area A of a parallelogram.
Evaluate to get the answer.
The formula of the area A of the parallelogram is:
A=bh ………(1)
The formula of the area A of the parallelogram is A=bh.
Given :
Using Eq(1) we can solve for b:
A=bh (parallelogram formula)
30 in2 =b⋅(5in)plug−in given values
b.(5in)&=30in2 (isolate b, divide by 5)
b.(5in)/5in =30in2\5in (solve for b)
b=6in.
Read and Learn More Big Ideas Math Algebra 1 Student Journal 1st Edition Solutions
The required solution is b=6in.
Given :
To find The value of b without substituting values in the formula.
Evaluate to get the answer.
Using Eq(1) we can solve for b:
A =bh (parallelogram formula)
bh=A (isolate b, divide by h)
bh/h=A/h (solve for b)
b=A/h.
For part b, we plug in the given values first before we can solve for b, on the other hand, for part c, the result is just a formula for b using Eq(1).
The main difference between part b and part c is that we were able to get the value of b in part b however, for part c, we only obtained the formula for solving b.
The main difference between part b and part c is that we were able to get the value of b in part b, however, for part c we only obtained the formula for solving b.
Big Ideas Math Algebra 1 Chapter 1 Exercise 1.5 Solution
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 23 Exercise 2 Answer
Given :
To find The area A of a trapezoid.
Evaluate to get the answer.
The area of the trapezoid is given by : A=(b1+b2/2)h
Solve for h:
A=(b1+b2/2)h formula
h=A(b1+b2/2)
h=63/(8)+(10)/2
h=63/18/2
h=63/9
h=7cm.
The required solution for area A of a trapezoid is 7cm.
Given :
To find Circumference C of circle.
Evaluate to get the answer.
The circumference of the circle is given by: C=2πr
Solve for r:
C =2πr (formula)
r=C/2π (isolate r)
r=24π/2π (solve for r)
r=12 ft
The required solution for Circumference C of circle is 12ft.
Given :
To find Volume V of a rectangular prism.
Evaluate to get the answer.
The volume of a rectangular prism is given by: V=Bh
Solve for h:
V=Bh (formula)
h=V/B (isolate h)
h=75/15 (solve for h)
h=5yd2
The required solution for volume of a rectangular prism is 5yd2.
Given :
To find Volume V of the cone.
Evaluate to get the answer.
The volume of a cone is given by: V=1/3πr2 h
Solve for h:
V=1/3πr2h (formula)
h=3V/πr2 (isolate h)
B=πr2
h=3.24/12π (solve for h)
h=6cm.
The required solution for volume of a cone is 6cm.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 23 Exercise 3 Answer
You can convert units easily and accurately with one simple rule:
Measurement by a carefully chosen form of the number 1.
Numbers of units treated exactly the same as coefficients with variables.
we can multiply anything by 1 and not change its value.
If you had a sample of a substance with a mass of 0.0034 grams, and you wanted to express that mass in mg, you could use the following dimensional analysis.
⇒ 1m gm=10-3 gm
∴ ⇒ 0.0034gm×1000mg/ 1g
=3.4g.
Any formula is made of dependent variables and independent variables. Suppose, you are given a measurement of one variable, you can rearrange the formula using operations such as addition subtraction, multiplication, division, and other properties of equality and solve for the unknown measurement.
Solving Linear Equations Exercise 1.5 Big Ideas Math
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 1 Answer
Given: The expression is y−2x=15
Evaluate to get the answer.
On solving the given equation :
Solve for y:
y−2x =15 given
y−2x+2x=15+2x (add 2x to both sides, isolate y)
y−0=15+2x
y=15+2x
The required solution is y=15+2x.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 2 Answer
Given: Expression is 4x+y=2.
Evaluate to get the answer.
On solving the given equation :
Solve for y:
4x+y=2 given
4x+y−4x=2−4x (subtract 4x to both sides, isolate y)
y+4x−4x=2−4x
y−0=2−4x
y=2−4x
The required solution is y=2−4x.
Algebra 1 Student Journal Chapter 1 Exercise 1.5 Answers
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 3 Answer
Given: The expression is 5x−2 =8+5y.
Evaluate to get the answer.
On solving the given equation :
Solve for y:
5x−2 =8+5y given ,
5x−2−8=8+5y−8 (subtract both sides by 8, isolate 5y)
5x−10=5y (divide both sides by 5)
5y/5= 5x−10/5 (simplify)
y=5x/5-10/5
y=x-2
The required solution is y=x−2.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 4 Answer
Given: The expression is y+x=11.
Evaluate to get the answer.
On solving the given equation :
Solve for y:
y+x=11 given,
y+x−x=11−x (subtract x on both sides, isolate y)
y-0=11−x
y=11-x
The required solution is y=11−x.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 5 Answer
Given: The expression is 3x−y=−4
Evaluate to get the answer.
On solving the given equation :
Solve for y:
3x−y=−4 given
3x−y−3x=−4−3x (subtract 3x to both sides, isolate y).
(−1)⋅−y=(−1)⋅(−4−3x) (multiply both sides by−1)
y=4+3x
The required solution is y=4+3x.
Big Ideas Math Linear Equations Exercise 1.5 Help
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 6 Answer
Given: The expression is 3x+1=7−4y.
Evaluate to get the answer.
On solving the given equation:
Solve for y:
3x+1 =7−4y given
3x+1−7=7−4y−7 (subtract 7 to both sides, isolate y)
3x−6/−4=−4y/−4 (divide both sides by−4)
−3x−6/4=y
y=−3x−6/4
The required solution is y=−3x−6/4.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 7 Answer
Given: The equation y=10x−4x.
To find The value of x.
Evaluate to get the final answer.
We need to isolate the variable x from other variable y and we get the result.
The value of x can be evaluated as:
y=10x−4x
y=(10−4)x
y=6x
The value of x in the equation y=10x−4x is:
y=6x.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 8 Answer
Given: The equation q=3x+9xz.
To find The value of x.
Evaluate to get the final answer.
We need to isolate the variable x from other variables q,z and we get the result.
The value of x can be evaluated as:
q=3x+9xz
q=x(3+9z)
x=q/3+9z
The value of in the equation q =3x+9xz is:
x=q/3+9z.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 9 Answer
Given: The equation r=4+7x−sx.
To find The value of x.
Evaluate to get the final answer.
We need to isolate the variable x from other variables, s and we get the result.
The value of x can be evaluated as:
r=4+7x−sx
r−4=7x−sx
r−4=(7−s)x
x=7−s/r−4
The value of x in the equation r=4+7x−sx is:
x=7−s/r−4.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 10 Answer
Given: The equation y+4x=10x−6.
To find The value of x.
Evaluate to get the final answer.
We need to isolate the variable from other variables, x and we get the result.
The value of x can be evaluated as:
y+4x=10x−6
10x−4x=y+6
6x=y+6
x=y+6/6
The value of x in the equation y+4x=10x−6 is:
x=y+6/6.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 11 Answer
Given: The equation 4g+r=2r−2x.
To find The value of x.
Evaluate to get the final answer.
We need to isolate the variable x from other variables g,r and we get the result.
The value of x can be evaluated as:
4g+r=2r−2x
2x=2r−r−4g
2x=r−4g
x=r−4g/2
The value of x in the equation 4g+r=2r−2x is:
x=r−4g/2.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 12 Answer
Given: The equation 3z+8=12+3x−z.
To find The value of x.
Evaluate to get the final answer.
We need to isolate the variable x from other variables z and we get the result.
The value of x can be evaluated as:
3z+8=12+3x−z
3x=3z+z+8−12
3x=4z−6
x=4z−6/3
The value of x in the equation 3z+8=12+3x−z is:
x=4z−6/3.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 13 Answer
Given: The equation A=1/2 bh.
To find The value of b.
Evaluate to get the final answer.
We need to isolate the variable b from other variables h, A and we get the result.
The value of b can be evaluated as:
A=1/2 bh
2A=bh
b=2A/h
The value of b found after evaluation is 2A/h.
Chapter 1 Exercise 1.5 Step-By-Step Solutions Big Ideas Math
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 25 Exercise 14 Answer
Given: The equation V=1/3πr2 h.
To find The value of h.
Evaluate to get the final answer.
We need to isolate the variable h from other variables, V and we get the result.
The value of h can be evaluated as V=1/3πr2h
3V=πr2h
h=3V/π.r2
The value of h found after evaluating is h=3V/π.r2
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 26 Exercise 15 Answer
Given: The equationI=V/R.
To find The value of R.
Evaluate to get the final answer.
We need to isolate the variable R from other variable, V and we get the result.
The value of R can be evaluated asI=V/R
IR=V
R=V/I
The value of R found after evaluating is V/I.
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 26 Exercise 16 Answer
Given: The equation PV=nRT.
To find The value of R.
Evaluate to get the final answer.
We need to isolate the variable R from other variables, P, V, and T and we get the result.
The value of R can be evaluated as PV=nRT
R=PV/nT.
The value of R found after evaluating is PV/nT.
Exercise 1.5 Big Ideas Math Algebra 1 Guide Informational
Big Ideas Math Algebra 1 Student Journal 1st Edition Chapter 1 Solving Linear Equations Exercise 1.5 Page 26 Exercise 17 Answer
Given: The equation A=P+Prt.
To find The value of r.
Evaluate to get the final answer.
We need to isolate the variable from other variables A, P, and t and we get the result.
The value of r can be evaluated as A=P+Prt
Prt=A−P
rt=A−P/P
r=A−P/Pt
The value of r found after evaluating is A−P/Pt.
Given: The amount after interest earned is $1080, and time is 2 years.
To find The annual interest rate.
Evaluate to get the final answer.
The interest for 2 years can be evaluated as:
$1080−1000=80.
The annual interest for 1 year is 80/2=40.
The annual interest rate is $40.
Given: The equation A=P+Prt.
To find The formula for P.
Evaluate to get the final answer.
We need to isolate the variable P from other variables, A,r, and t and we get the result.
The value of P can be evaluated as A=P+Prt
A=P(1+rt)
P=A/1+rt
The value of P found after evaluating is A/1+rt.
How To Solve Exercise 1.5 Big Ideas Math