Lailamg100idn
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Please help!!! giving 100 points for correct answer

(algebra 2)

|-3d-6|+|-5-d^2| for d =-3

please give a step by step too

Please Help Giving 100 Points For Correct Answeralgebra 2 3d65d2 For D 3please Give A Step By Step Too class=

Sagot :

Fieryanswererftidn

Answer:

17

Step-by-step explanation:

[tex]\sf |-3d-6|+|-5-d^2| \quad for\:\: d = -3[/tex]

Given expression:

[tex]\sf |-3d-6|+|-5-d^2|[/tex]

substitute d = -3

[tex]\sf |-3(-3)-6|+|-5-(-3)^2|[/tex]

simplify following inside parenthesis

[tex]\sf |9-6|+|-5-9|[/tex]

simplify by combining integers

[tex]\sf |3|+|-14|[/tex]

if |-a| or |a| then equals to a

[tex]\sf 3+14[/tex]

evaluate:

[tex]\sf 17[/tex]

Semsee45idn

Answer:

17

Step-by-step explanation:

Given expression:

[tex]|-3d-6|+|-5-d^2|[/tex]

To evaluate for d = -3, substitute d = -3 into the expression:

[tex]\implies |-3(-3)-6|+|-5-(-3)^2|[/tex]

[tex]\textsf{Apply exponent rule} \quad a\cdot a=a^2:[/tex]

[tex]\implies |(-3)^2-6|+|-5-(-3)^2|[/tex]

[tex]\textsf{Apply exponent rule} \quad (-a)^2=a^2:[/tex]

[tex]\implies |3^2-6|+|-5-3^2|[/tex]

[tex]\implies |9-6|+|-5-9|[/tex]

Subtract the numbers:

[tex]\implies |3|+|-14|[/tex]

[tex]\textsf{Apply absolute rule} \quad |a|=a, \quad a\geq 0[/tex]

[tex]\implies 3+|-14|[/tex]

[tex]\textsf{Apply absolute rule} \quad |-a|=a, \quad a > 0[/tex]

[tex]\implies 3+14[/tex]

Add the numbers:

[tex]\implies 17[/tex]

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