1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
ExtremeBDS [4]
2 years ago
9

6- 4(2r - 5) < 26 - 8r This is algebra

Mathematics
1 answer:
RSB [31]2 years ago
4 0

Answer:

r>1

Step-by-step explanation:

6-4(2r-5)<26-8r do () first

6-8r+20<26-8r subtract 6 from both sides

-8r+20<20-8r Since this is the same on both sides, r>1 to make the equation not come out as 0

You might be interested in
Joe ate 1/3 of a pizza and jane ate 1/5 of the pizza. How much of the pizza was eaten
NikAS [45]

Answer:

7/15

Step-by-step explanation:

If we're seeing how much pizza they both ate, we would add them together.

\frac{1}{3}+\frac{1}{5}

Find the common denominator

\frac{5}{15}+ \frac{3}{15}

add

\frac{7}{15}

7/15 of the pizza was eaten

5 0
3 years ago
Please help, will mark you as brainleist!!!
svet-max [94.6K]

Answer:

(1.) 50 (2). 112.5

Step-by-step explanation:

multiply the both sides and then divide ( for each one)

8 0
2 years ago
C.) of a total of $10,000, part was invested at 10% simple interest and the remainder at
nlexa [21]

Answer:

$3,500 at 10% and $6,500 at 7%.

Step-by-step explanation:

M * 10% + (10000 - M) * 7% = 805

0.1M + 700 - 0.07M = 805

0.03M = 105

M = 3500

4 0
2 years ago
Find the value of x, the problem is below.
Svetradugi [14.3K]
Angles are complementary and x is 15
7 0
3 years ago
Read 2 more answers
Find a solution of x dy dx = y2 − y that passes through the indicated points. (a) (0, 1) y = (b) (0, 0) y = (c) 1 6 , 1 6 y = (d
Leni [432]
Answers: 

(a) y = \frac{1}{1 - Cx}, for any constant C

(b) Solution does not exist

(c) y = \frac{256}{256 - 15x}

(d) y = \frac{64}{64 - 15x}

Explanations:

(a) To solve the differential equation in the problem, we need to manipulate the equation such that the expression that involves y is on the left side of the equation and the expression that involves x is on the right side equation.

Note that

 x\frac{dy}{dx} = y^2 - y&#10;\\&#10;\\ \indent xdy = \left ( y^2 - y \right )dx&#10;\\&#10;\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}&#10;\\&#10;\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} &#10;\\&#10;\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln x + C_1}      (1)

Now, we need to evaluate the indefinite integral on the left side of equation (1). Note that the denominator y² - y = y(y - 1). So, the denominator can be written as product of two polynomials. In this case, we can solve the indefinite integral using partial fractions.

Using partial fractions:

\frac{1}{y^2 - y} = \frac{1}{y(y - 1)} = \frac{A}{y - 1} + \frac{B}{y}&#10;\\&#10;\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} &#10;\\&#10;\\ \indent \Rightarrow \boxed{\frac{1}{y^2 - y} = \frac{(A+B)y - B}{y^2 - y} }      (2)

Since equation (2) has the same denominator, the numerator has to be equal. So,

1 = (A+B)y - B&#10;\\&#10;\\ \indent \Rightarrow (A+B)y - B = 0y + 1&#10;\\&#10;\\ \indent \Rightarrow \begin{cases}&#10; A + B = 0&#10;& \text{(3)}\\-B = 1&#10; & \text{(4)}   \end{cases}

Based on equation (4), B = -1. By replacing this value to equation (3), we have

A + B = 0
A + (-1) = 0
A + (-1) + 1 = 0 + 1
A = 1 

Hence, 

\frac{1}{y^2 - y} = \frac{1}{y - 1} - \frac{1}{y}

So,

\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} &#10;\\&#10;\\ \indent \indent \indent \indent = \ln (y-1) - \ln y&#10;\\&#10;\\ \indent  \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}

Now, equation (1) becomes

\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1&#10;\\&#10;\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2&#10;\\&#10;\\ \indent  \frac{y-1}{y} = e^{C_1 - C_2}x&#10;\\&#10;\\ \indent  \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}&#10;\\&#10;\\ \indent  1 - \frac{1}{y} = Cx&#10;\\&#10;\\ \indent \frac{1}{y} = 1 - Cx&#10;\\&#10;\\ \indent \boxed{y = \frac{1}{1 - Cx}}&#10;       (5)

At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1&#10;&#10;

Hence, for any constant C, the following solution will pass thru (0, 1):

\boxed{y = \frac{1}{1 - Cx}}

(b) Using equation (5) in problem (a),

y = \frac{1}{1 - Cx}   (6)

for any constant C.

Note that equation (6) is called the general solution. So, we just replace values of x and y in the equation and solve for constant C.

At point (0,0), x = 0, y =0. Then, we replace these values in equation (6) so that 

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent 0 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1

Note that 0 = 1 is false. Hence, for any constant C, the solution that passes thru (0,0) does not exist.

(c) We use equation (6) in problem (b) and because equation (6) is the general solution, we just need to plug in the value of x and y to the equation and solve for constant C. 

At point (16, 16), x = 16, y = 16 and by replacing these values to the general solution, we have

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent 16 = \frac{1}{1 - C(16)} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - 16C}&#10;\\&#10;\\ \indent 16(1 - 16C) = 1&#10;\\ \indent 16 - 256C = 1&#10;\\ \indent - 256C = -15&#10;\\ \indent \boxed{C = \frac{15}{256}}&#10;&#10;&#10;

By replacing this value of C, the general solution becomes

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent y = \frac{1}{1 - \frac{15}{256}x} &#10;\\ &#10;\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}&#10;\\&#10;\\&#10;\\ \indent \boxed{y = \frac{256}{256 - 15x}}&#10;&#10;&#10;&#10;

This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
        - Substitute the values of x and y to the general solution.
        - Solve for constant C

At point (4, 16), x = 4, y = 16. First, we replace x and y using these values so that 

y = \frac{1}{1 - Cx} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - C(4)} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - 4C} &#10;\\ &#10;\\ \indent 16(1 - 4C) = 1 &#10;\\ \indent 16 - 64C = 1 &#10;\\ \indent - 64C = -15 &#10;\\ \indent \boxed{C = \frac{15}{64}}

Now, we replace C using the derived value in the general solution. Then,

y = \frac{1}{1 - Cx} \\ \\ \indent y = \frac{1}{1 - \frac{15}{64}x} \\ \\ \indent y = \frac{1}{\frac{64 - 15x}{64}} \\ \\ \\ \indent \boxed{y = \frac{64}{64 - 15x}}
5 0
3 years ago
Other questions:
  • Let f(x)=(x+4)^2−3.
    12·1 answer
  • a basketball court is a rectangle with a perimeter of 1040 feet. The length is 200 feet more than the width. What is the width a
    14·2 answers
  • What is the answer to <br>3n-5=7n+11
    12·2 answers
  • -X/3=9 linear equations
    5·2 answers
  • In 4 regular season basketball games, James Harden scored 141 points. Write an equation to represent the number of points scored
    11·2 answers
  • Subtract the mixed numbers: 3 5/6 - 1 2/3 *
    6·1 answer
  • No entiendo matemáticas alguien me podría ayudar​
    14·1 answer
  • Anyone mind helping ?
    12·2 answers
  • Please help! Insta: bri.editzz0
    15·2 answers
  • PLS PLS PLS HELP QUICKKKK<br><br> Find the value of x in each case
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!