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Yuliya22 [10]
2 years ago
7

Find the sum and express it in simplest form. (-6b - 2b - 1) + (-2b + 4b)

Mathematics
1 answer:
viva [34]2 years ago
5 0
-4b+2b-1
2b-1
this should be the answer
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Give the inequality. PLEASE HELP ASAP, THANK YOU
Reil [10]

Answer:

-3\le x\le 1

or

|x+1|\le2


5 0
2 years ago
Write 19/25 as a decimal and a percent
bixtya [17]

Answer:

19/25

Decimal: 0.76

Percent: 76%

Step-by-step explanation:

5 0
3 years ago
PLEASE HELP!! (Give reasonable answer)
gladu [14]

Answer: B

Step-by-step explanation: You just multiplying 25 with 10 and w which can be simplified as 24(w+10) and adding with the 13.5 x 10 and 13.5 x w, which can be simplified as 13.5(10+w)

7 0
2 years ago
A vertical number line is given.
Ad libitum [116K]

The absolute value of the number plotted on the number line is; 0.5

<h3>How to find Absolute Value?</h3>

The absolute value (or modulus) | x | of a real number x is the non-negative value of x without regard to its sign. For example, the absolute value of 5 is 5, and the absolute value of −5 is also 5. The absolute value of a number may be thought of as its distance from zero along real number line

Thus, if 0.5 is marked on the number line, from the given options, we can say that if we apply the definition above that |0.5| = 0.5

Read more about Absolute Value at; brainly.com/question/24368848

#SPJ1

4 0
1 year ago
Read 2 more answers
X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?
cluponka [151]

Answer:

x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

Step-by-step explanation:

I'm assuming the system is \left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.:

x^2+y^2=2

x^2+(2x^2-3)^2=2

x^2+(4x^4-12x^2+9)=2

x^2+4x^4-12x^2+9=2

4x^4-11x^2+9=2

4x^4-11x^2+7=0

x^4-11x^2+28=0

(x^2-7)(x^2-4)=0

(4x^2-7)(x^2-1)=0

4x^2-7=0

4x^2=7

x^2=\frac{7}{4}

x=\pm\sqrt{\frac{7}{4}}

x^2-1=0

x^2=1

x=\pm1

y=2x^2-3

y=2(\pm\sqrt{\frac{7}{4}})^2-3

y=2({\frac{7}{4}})-3

y=\frac{7}{2}-3

y=\frac{1}{2}

y=2x^2-3

y=2(\pm1)^2-3

y=2(1)-3

y=2-3

y=-1

Therefore, x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}} and y=-1,\frac{1}{2}

The ordered pair solutions are (-\sqrt{\frac{7}{4}},0.5), (\sqrt{\frac{7}{4}},0.5), (-1,-1), and (1,-1).

4 0
2 years ago
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