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Zanzabum
3 years ago
10

How do I solve absolute value inequalities?

Mathematics
1 answer:
evablogger [386]3 years ago
5 0
This link could be useful.
http://www.purplemath.com/modules/absineq.htm
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Which represents the solution(s) of the graphed system of equations, y = x2 + 2x – 3 and y = x – 1?
Sonja [21]

Answer:

Second option: (-2,-3) and (1,0)

Step-by-step explanation:

Given the system of equations \left \{ {{y = x^2 + 2x-3} \atop {y = x - 1}} \right., you can rewrite them in this form:

x^2 + 2x-3= x - 1

Simplify:

x^2 + 2x-3-x+1=0\\\\x^2+x-2=0

Factor the quadratic equation. Choose two number whose sum be 1 and whose product be -2. These are: 2 and -1, then:

(x+2)(x-1)=0\\\\x_1=-2\\\\x_2=1

Substitute each value of "x"  into any of the original equation to find the values of "y":

y_1= (-2) - 1=-3\\\\y_2=(1)-1=0

Then, the solutions are:

(-2,-3) and (1,0)

7 0
3 years ago
Read 2 more answers
Help with 33 and 34 please!!!
hjlf
33. 15x10=150x9
the answer is b

34. a
3 0
3 years ago
Read 2 more answers
Three cube-shaped boxes are stacked one above the other. The volumes of two of the boxes are 1,331 cubic meters each, and the vo
MAVERICK [17]
B.29. that's the answer
8 0
3 years ago
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In the supermarket a loaf of bread costs 37p how many loaves can david buy with a 2 coin ? How much money will be left over?
Rasek [7]

Answer: He can buy 5 loaves of bread.

After buying 5 loaves 15 p will be left.

Step-by-step explanation:

Given, In the supermarket a loaf of bread costs 37p .

To find: How many loaves can David buy with a \pounds 2 coin?

Since \pounds 1=100\text{ pence}

Then, \pounds 2=200\text{ pence}

Number of loaves he can buy = (Amount he has) ÷ (Cost of a loaf of bread)

= 200 p ÷ 37 p

=5\dfrac{15}{37}

i.e. he can buy 5 loaves of bread and 15 p will be left.

Hence, He can buy 5 loaves of bread.

After buying 5 loaves 15 p will be left.

6 0
3 years ago
Find the perimeter of triangle ABC with vertices A(6, 3), B(6, - 2) , and C(- 4, 3) .
anyanavicka [17]

Answer:

26.2 units

Step-by-step explanation:

We are given the points/vertices

A(6, 3),

B(6, - 2) , and

C(- 4, 3)

Step two

Let us find the distances between the given points/vertices

A-B =A(6, 3) to B(6,-2)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((6-6)²+(-2-3)²)

d=√(-2-3)²)

d=√(-5)²)

d=5 units

B-C=B(6, - 2) to C(-4, 3)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((-4-6)²+(3+2)²)

d=√(-10)²+(5)²)

d=√100+25

d=√125

d=11.2 units

C-A=C(-4, 3) to A(6, 3)

d=√((x2-x1)²+(y2-y1)²)

Substitute

d=√((6+4)²+(3-3)²)

d=√(10)²

d=√100

d=10 units

Hence the perimeter is 5+11.2+10

P=26.2 units

4 0
3 years ago
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