Joanne is depositing money into a bank account. After 3 months there is $150 in the account. After 6 months there is $300 in the account. Determine the constant rate of change of the account.
Answer:
The constant rate of change of the account is $50 or Increasing by $50 per month.
Step-by-step explanation:
Consider the provided information.
Joanne is depositing money into a bank account. After 3 months there is $150 in the account. After 6 months there is $300 in the account.
Rate of change is known as how one quantity change in relation to other.
The rate of change can be calculated as:
[tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
Now use the above formula to calculated the rate of change.
[tex]\frac{300-150}{6-3}[/tex]
[tex]\frac{150}{3}[/tex]
[tex]50[/tex]
Hence, the constant rate of change of the account is $50 or Increasing by $50 per month.
Tim bought a pen for $2.25, a pencil for $0.59, a notebook for $6.49, and a highlighter for $1.49. He used a coupon that gave him $5.25 off his entire purchase. How much did he spend in total?
which number is greater 67.89 and 67.98
Kurt and Maria’s high school is having a newspaper drive.The goal is to collect 3,585 pounds of newspapers. So far, 21% of the goal has been reached. Kurt estimated the number of pounds of newspapers collected by finding 10% of 3,600 and then multiplying the result by 2. Maria estimated the number of pounds of newspapers collected by finding mc006-1.jpg of 3,600. Who is right, and why?
In order to start a business, a student takes out a simple interest loan for $7000.00 for 6 months at a rate of 8.00 %
the 4th and 13th terms of an AP are 5 and -1, find the 8th term of an AP
Is the subset w = {(x, y, z) | z = 1} ⊂ r 3 a vector subspace of r 3 ? explain why or why not?
If Mr. Khans buys 15 staplers, it would cost him $254.85. How would you write this using function notation?
Answer:
f(x) = don't spend more than 200 dollars on staplers
Step-by-step explanation:
Solve the equation using the Zero-Product Property. –8n(10n – 1) = 0
Answer: The solution is,
[tex]n = 0\text{ or }n = \frac{1}{10}[/tex]
Step-by-step explanation:
Since, Zero product property states that if the product of two numbers or expression is equal to zero then either of the numbers or expressions must be equal to zero.
That is, If a.b = 0 ⇒ a = 0 or b = 0
Here, the given expression is,
[tex]-8n(10n-1)=0[/tex]
[tex]\implies (-8n)(10n-1)=0[/tex]
Thus, by the above property,
[tex]-8n = 0\text{ or }(10n-1)=0[/tex]
[tex]\implies n = \frac{0}{-8}\text{ or }10n= 1[/tex]
[tex]\implies n = 0\text{ or }n = \frac{1}{10}[/tex]
One tablet contains 575 grams of muscle relaxing medication. How many grams are in 3 1/2 tablets
if y varies directly as x and y=6 when x =-7, find y when x is 4
When x is 4, y is approximately -3.43.
Explanation:To find y when x is 4, we can use the given information that y varies directly as x. This means that the ratio of y to x remains constant. We can set up a proportion using the initial values of y and x and solve for the unknown value:
6 / -7 = y / 4
By cross-multiplying, we get y = -24 / 7. Therefore, when x is 4, y is approximately -3.43.
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1.Find the coordinates of the midpoint of __ given that H (-1,3) and X (7,1).
HX
A.(3,1)
B (0,4)
C(-3,1)
D(-4,0)
2. Find the distance between the points R(0,5) and S(12,3). round the answer to the nearest tenth.
A 10.4
B 16
C 12.2
D 11.8
3. An airplane at T(80,20) needs to fly to both U(20,60) and V(110,85) what is the shortest possible distance for the trip?
A. 165 units
B. 170 units
C. 97 units
D. 169 units
Answer:
1. (3, 2)
2. Option C. 12.2
3. Option A. 165 units
Step-by-step explanation:
1. The midpoint of two coordinates (x₁, x₂) and (y₁, y₂) is calculate by,
[tex](x, y) = (\frac{x_{1} + x_{2}}{2},\frac{y_{1} + y_{2}}{2})[/tex]
⇒[tex](x, y) = (\frac{-1 + 7}{2},\frac{3 + 1}{2})[/tex]
Thus (x, y) = (3, 2)
Hence, none of given options are true.
2. The distance between two coordinates is calculate by,
[tex]Distance=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2[/tex]
⇒ Distance = 12.16 ≈ 12.2 unit
Hence, option (C) is correct.
3. The distance between T(80, 20) and V(110, 85) is comparatively smaller than T(80, 20) and U(20, 60).
Using the Distance formula,
[tex]Distance=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2[/tex]
Distance between T(80, 20) and V(110, 85) is 71.59 unit
and Distance between U(20, 60) and V(110, 85) is 93.41 unit
So, Airplane firstly go to point V from point T and then point U.
Total shortest distance = 71.60 + 93.40 = 165 unit.
Hence, option (A) is correct.
Order the fraction from least to greatest.
2/3, 4/9, 5/6, 7/12
All of the following expressions represent the sum of n and itself, except _____.
n + n
2n
n²
Answer: [tex]n^{2}[/tex]
Step-by-step explanation:
(n + n) represents the definition of the sum of n and itself.
(2n) means two times n, wich is the same as adding n + n.
[tex]n^{2}[/tex] actually represents the multiplication of n two times: [tex]n*n[/tex]
For example, if n=3:
[tex]n+n=3+3=6[/tex]
[tex]2n=2(3)=6[/tex]
[tex]n^{2}=3^{2} =9[/tex]
what is the value of the 7 equal to (7×1/100)
This number represents the number of times the base is used as a factor
Compare methods of solving linear equations and methods of solving linear inequalities. what do they have in common? what is different?
if tanx=-4/3 and x is in quadrant 2 then cos2x=?
A. 7/25
b. -3/5
c. -7/25
d. 3/5
Trigonometric Identities are equalities that utilize trigonometry functions and hold true for all variables in the equation. The correct option is C, -7/25.
What are Trigonometric Identities?Trigonometric Identities are equalities that utilize trigonometry functions and hold true for all variables in the equation. There are several trigonometric identities relating to the side length and angle of a triangle.
Given that the value of tan(x) = -4/3 and x is in quadrant 2.
In order to find the value of cos(2x), we can use the trigonometric identity of cos(2x), therefore, for cos(2x) we can write,
[tex]\cos(2x) = \dfrac{1-\tan^2(x)}{1+\tan^2(x)}[/tex]
Since the value of tan(x) is known, substitute the value in the identity,
[tex]\cos(2x) = \dfrac{1-(-\frac{4}{3})^2}{1+(-\frac{4}{3})^2}\\\\\cos(2x) = \dfrac{1-(\frac{16}{9})}{1+(\frac{16}{9})}\\\\[/tex]
cos(2x) = (-7/9) × (9/25)
Cancelling 9 from the numerator and the denominator,
cos(2x) = -7/25
Hence, if tanx=-4/3 and x are in quadrant 2 then cos2x=-7/25.
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The correct answer is c. [tex]$\cos(2x) = -\frac{7}{25}$[/tex].
First, we find [tex]$\sin(x)$[/tex] and [tex]$\cos(x)$[/tex] using the given value of [tex]$\tan(x)$[/tex]:
[tex]$\tan(x) = \frac{\sin(x)}{\cos(x)} = -\frac{4}{3}$.[/tex]
We can use the Pythagorean identity [tex]$\sin^2(x) + \cos^2(x) = 1$[/tex] to find [tex]$\sin(x)$[/tex] and [tex]$\cos(x)$[/tex]. Let's assume [tex]$\sin(x) = \frac{4}{5}$[/tex] (since [tex]$\tan(x) = -\frac{4}{3}$[/tex], and we are looking for a positive value of [tex]$\sin(x)$[/tex] in quadrant 2). Using the Pythagorean identity:
[tex]$\left(\frac{4}{5}\right)^2 + \cos^2(x) = 1$ \\ $\frac{16}{25} + \cos^2(x) = 1$ \\ $\cos^2(x) = 1 - \frac{16}{25}$ \\ $\cos^2(x) = \frac{25}{25} - \frac{16}{25}$ \\ $\cos^2(x) = \frac{9}{25}$. \\[/tex]
Since [tex]$\cos(x)$[/tex] is negative in quadrant 2, we have [tex]$\cos(x) = -\frac{3}{5}$[/tex].
Now, we can use the double-angle formula for cosine:
[tex]$\cos(2x) = 2\cos^2(x) - 1$ \\ $\cos(2x) = 2\left(-\frac{3}{5}\right)^2 - 1$ \\ $\cos(2x) = 2\left(\frac{9}{25}\right) - 1$ \\ $\cos(2x) = \frac{18}{25} - 1$ \\ $\cos(2x) = \frac{18}{25} - \frac{25}{25}$ \\ $\cos(2x) = -\frac{7}{25}$.[/tex]
Therefore, the value of [tex]$\cos(2x)$[/tex] = [tex]$ -\frac{7}{25}$[/tex].
what fraction 2/3 is found between which pair of fraction on the number line?
The relationship between the number of $4 lunches you buy with a $100 school lunch card and the money remaining on the card
what is the prime factorization for 37
The prime factorization of 37 is 37
We have,
The prime factorization of a number involves expressing it as a product of prime numbers.
However, in the case of prime numbers themselves, their prime factorization is simply the number itself.
In this case,
The number 37 is a prime number because it is only divisible by 1 and itself.
Since it has no other factors, its prime factorization is simply 37.
Thus,
The prime factorization of 37 is 37.
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Ken watches a marching band. He sees 2 rows of flute players. Six people are in each row. He sees 8 trombone players. How many flute or trombone players does Ken see
The area of a square garden is 98 meters squared. How long is the diagonal
When mary began her trip from san jose to la, she filled her car's tank with gas and reset its trip meter to zero. after traveling 324 miles, she stopped at a gas station to refuel; the gas tank required 17 gallons. mary wants a program that calculates and displays her car's gas mileage at any time during the trip. the gas mileage is the number of miles her car was driven per gallon of gas?
To calculate Mary's car's gas mileage, divide the number of miles driven by the number of gallons of gas used. In this scenario, her car's gas mileage is approximately 19.06 miles per gallon.
Explanation:To calculate Mary's car's gas mileage, we need to divide the number of miles driven by the number of gallons of gas used. In this scenario, Mary traveled 324 miles and used 17 gallons of gas. Therefore, her car's gas mileage can be calculated as:
Gas mileage = Miles driven / Gallons of gas used
Substituting the values:
Gas mileage = 324 miles / 17 gallons
Simplifying the equation:
Gas mileage = 19.06 miles per gallon (rounded to two decimal places)
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Let f(x,y)=x2 −y2. find the gradient of f at the point (√2,1). sketch the level curve of f through this point, together with the gradient at that point. g
Answer:
[tex]\displaystyle \nabla f(\sqrt{2}, 1) = 2\sqrt{2} \hat{\i} - 2 \hat{\j}[/tex]
General Formulas and Concepts:
Calculus
Differentiation
DerivativesDerivative NotationDerivative Rule [Basic Power Rule]:
f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Multivariable Calculus
Differentiation
Partial DerivativesDerivative NotationGradient: [tex]\displaystyle \nabla f(x, y, z) = \frac{\partial f}{\partial x} \hat{\i} + \frac{\partial f}{\partial y} \hat{\j} + \frac{\partial f}{\partial z} \hat{\text{k}}[/tex]
Gradient Property [Addition/Subtraction]: [tex]\displaystyle \nabla \big[ f(x) + g(x) \big] = \nabla f(x) + \nabla g(x)[/tex]
Step-by-step explanation:
Step 1: Define
Identify.
[tex]\displaystyle f(x, y) = x^2 - y^2[/tex]
[tex]\displaystyle P(\sqrt{2}, 1)[/tex]
Step 2: Find Gradient
[Function] Differentiate [Gradient]: [tex]\displaystyle \nabla f(x, y) = \frac{\partial f}{\partial x} \bigg[ x^2 - y^2 \bigg] \hat{\i} + \frac{\partial f}{\partial y} \bigg[ x^2 - y^2 \bigg] \hat{\j}[/tex][Gradient] Rewrite [Gradient Property - Addition/Subtraction]: [tex]\displaystyle \nabla f(x, y) = \bigg[ \frac{\partial f}{\partial x}(x^2) - \frac{\partial f}{\partial x}(y^2) \bigg] \hat{\i} + \bigg[ \frac{\partial f}{\partial y}(x^2) - \frac{\partial f}{\partial y}(y^2) \bigg] \hat{\j}[/tex][Gradient] Differentiate [Derivative Rule - Basic Power Rule]: [tex]\displaystyle \nabla f(x, y) = 2x \hat{\i} - 2y \hat{\j}[/tex][Gradient] Substitute in point: [tex]\displaystyle \nabla f(\sqrt{2}, 1) = 2\sqrt{2} \hat{\i} - 2(1) \hat{\j}[/tex][Gradient] Simplify: [tex]\displaystyle \nabla f(\sqrt{2}, 1) = 2\sqrt{2} \hat{\i} - 2 \hat{\j}[/tex]∴ the gradient of the given f(x, y) function is equal to <2√2, -2>.
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Topic: Multivariable Calculus
Unit: Directional Derivatives
Find the distance between points M(6,16) and Z(-1,14) to the nearest tenth.
The equation of a line is y=3x+7. Change the equation so that it is proportional.
Answer:
y=3x
Step-by-step explanation:
For a line to be proportional it must have the form y=mx where the y-intercept is (0,0) through the origin. To change y=3x+7 to be proportional, write it as y=3x.
3x + 5x = 10 Which problem requires the same strategy (combining like terms)?
Determine the effective rate for $1 for 1 year at 5.9% compounded quarterly.
Solve: 5x - 7x + 6 = -2(x - 3).
A)
0
B)
3
4
C)
2
D)
infinitely many solutions
Answer:
D) infinitely many solutions
Step-by-step explanation:
The equation reduces to ...
-2x +6 = -2x +6
This is true for all possible values of x, so there are infinitely many solutions.