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4 years ago

15

Solve each compound inequality and grahp the solutions. ( that is the question ) . Algebra 1

1 answer:

Gnesinka [82]4 years ago

3 0

1 < = x + 7 < 7....subtract 7
1 - 7 < = x + 7 - 7 < 7 - 7...simplify
-6 < = x < 0

closed circle on -6, open circle on 0, shading in between

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Solve whether the equation have 1 solution no solution or infinity solution

babymother [125]

Steps to solving equations:
Simplify/combine the like terms
Isolate the variable (Get the variable on one side
Solve for the variable

-2y - 7 + 5y = 13 - 2y Simplify/Combine terms
3y -7 = 13 - 2y Isolate variable by adding 2y to both sides
5y - 7 = 13 Add 7 to both sides to get y alone on one side
5y = 20 Divide by 5 on both sides to solve for y
y = 4
You can check your answer by replacing y with 4 in the original equation.

5 0

3 years ago

Read 2 more answers

3/5x + 2/5y + 1/8 - (3/5x + 1/5y + 1)

notka56 [123]

I just copied and pasted your question on google and a graph came up it may not be the answer but i tried...

4 0

3 years ago

76. Currently, your savings for each month is capped at $400. If you start investing all of this into a savings plan earning 1%

Alexxandr [17]

Answer:

10 years and 10 months.

Step-by-step explanation:

The annually interest rate (ia) can be converted by monthly (im) by the equation:

(1 + im)¹² = 1 + ia

(1 + im)¹² = 1 +0.01

(1 + im)¹² = 1.001 (putting ln in both sides)

ln(1 + im)¹² = ln1.001

12*ln(1 + im) = 1.0x10⁻³

ln(1 + im) = 8.33x10⁻⁵(applying "e in both sides)

$e^{ln(1 + im)} = e^{8.33x10^{-5}}$

1 + im = 1.00083

im = 0.00083 = 0.083%

For a investimenting, the final amount (A) can be calculated by:

$A = R*(\frac{(1+i)^n-1}{i} )$

Where R is the amount invested per month, i is the interest, and n the number of months:

160000 = 400 * $\frac{(1 +0.00083)^n - 1}{0.00083}$

$\frac{(1 +0.00083)^n - 1}{0.00083}$ = 400

1.00083ⁿ - 1 = 0.332

1.00083ⁿ = 1.332 (applying ln in both sides)

n*ln1.00083 = ln1.332

8.3x10⁻⁴n = 0.2867

n = 345.4 months

345.4 months *1 yea12 months = 10 years and 10 months.

7 0

3 years ago

Read 2 more answers

Express the negations of each of these statements so that all negation symbols immediately precede predicates. a) ∀x∃y∀zT(x, y,

Gnesinka [82]

Answer:

a) ∀x∃y ¬∀zT(x, y, z)

∀x∃y ∃z ¬T(x, y, z)

b) ∀x¬[∃y (P(x, y) ∨ Q(x, y))]

∀x∀y ¬ [P(x, y) ∨ Q(x, y)]

∀x∀y [¬P(x, y) ^ ¬Q(x, y)]

c) ∀x ¬∃y (P(x, y) ^ ∃zR(x, y, z))

∀x ∀y ¬(P(x, y) ^ ∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ¬∃zR(x, y, z))

∀x ∀y (¬P(x, y) v ∀z¬R(x, y, z))

d) ∀x¬∃y (P(x, y) → Q(x, y))

∀x∀y ¬(P(x, y) → Q(x, y))

∀x∀y (¬P(x, y) ^ Q(x, y))

3 0

3 years ago

Read 2 more answers

To test the effectiveness of a business school preparation course, 8 students took a general business test before and after the

Lisa [10]

Solution :

Group Before After

Mean 693.75 743.75

Sd 155.37 143.92

SEM 54.93 50.88

n 8 8

Null hypothesis : The preparation course not effective.

$H_0: \mu_d = 0$

Alternative hypothesis : The preparation course is effective in improving the exam scores.

$H_a : \mu_d>0$ (after - before)

4 0

3 years ago