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LenaWriter [7]
3 years ago
11

Question 6 of 10

Mathematics
2 answers:
hram777 [196]3 years ago
7 0
True. mark brainliest
Katarina [22]3 years ago
6 0

Answer:

the answer is true.. I think

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Find all numbers with the given absolute value<br> 1)12<br> 2)1.7<br> 3)3/5<br> 4)3 1/6<br> 5)0
telo118 [61]
It would be the negative form of those bumbers and the non-negative form of those numbers. The absolute value applies to negative numbers the same. For example absolute value of -3 is 3.
5 0
3 years ago
Please help me with the equation<br><br> 6+z/10= -2
oksian1 [2.3K]

Answer:

z = -26

Step-by-step explanation:

6+z/10= -2

6 + z = -2 (10)

z = -20 - 6

z = -26

7 0
3 years ago
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The solution to this system of equations lies between the x-values -2 and -1.5. At which x-value are the two equations approxima
katovenus [111]
D) -1.8
-1.8
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3 years ago
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Difference of 10 and a number W is no more than eight
algol [13]
10 - nw is less than or equal to 8

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4 0
3 years ago
Solve (x + 4)2 – 3(x + 4) – 3 = 0 using substitution. u = Select the solution(s) of the original equation.
bezimeni [28]

Answer:

\dfrac{-5-\sqrt{21}}{2},\ \dfrac{-5+\sqrt{21}}{2}

Step-by-step explanation:

For the equation (x+4)^2-3(x+4)-3=0 use the substitution u=x+4. Then the equation will take look

u^2-3u-3=0.

Solve this quadratic equation:

D=(-3)^2-4\cdot 1\cdot (-3)=9+12=21,\\ \\u_{1,2}=\dfrac{-(-3)\pm \sqrt{21}}{2}=\dfrac{3\pm\sqrt{21}}{2}.

Thus,

x+4=\dfrac{3-\sqrt{21}}{2}\text{ or }x+4=\dfrac{3+\sqrt{21}}{2},\\ \\x_1=\dfrac{3-\sqrt{21}}{2}-4=\dfrac{-5-\sqrt{21}}{2}\text{ or }x_2=\dfrac{3+\sqrt{21}}{2}-4=\dfrac{-5+\sqrt{21}}{2}.

8 0
3 years ago
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