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4 years ago

What is the absolute value of 9? -3? Show your work

Mathematics

1 answer:

myrzilka [38]4 years ago

7 0

|9|=9
|-3|=3

Basically absolute value is the posotive version of the number. Hope that makes sense!

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Solve for x:<br><br> x2 + 8x = -1

olga nikolaevna [1]

$x^2+8x=-1\\
x^2+8x+16-16=-1\\
(x+4)^2=15\\
x+4=\sqrt{15} \vee x+4=-\sqrt{15}\\
x=-4+\sqrt{15} \vee x=-4-\sqrt{15}$

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3 years ago

The price received for a bicycle is given by the equation b = 100 – 10x2, where x is the number of bicycles produced, in million

Alchen [17]

Answer:

I. B. P = – + 40x

II. 1.3

hope this helps! :o)

Step-by-step explanation:

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3 years ago

Read 2 more answers

The first side of a triangle is 3 inches shorter than the second side, and 2 inches longer than the third side. How long is each

ziro4ka [17]

Answer:

the length of the first side of the triangle is $\frac{28}{3}\ in$

the length of the second side of the triangle is $\frac{37}{3}\ in$

the length of the third side of the triangle is $\frac{19}{3}\ in$

Step-by-step explanation:

Let

x-----> the length of the first side of a triangle

y----> the length of the second side of a triangle

z---> the length of the third side of a triangle

we know that

$x=y-3$

$y=x+3$ -----> equation A

$x=z+3$

$z=x-3$ -----> equation B

The perimeter of the triangle is equal to

$P=x+y+z$

$P=28\ in$

$28=x+y+z$ -----> equation C

substitute equation A and equation B in equation C

$28=x+(x+3)+(x-3)$

solve for x

$28=3x$

$x=\frac{28}{3}\ in$

Find the value of each side

the first side of a triangle is x

$x=\frac{28}{3}\ in$

the second side of a triangle is y

$y=\frac{28}{3}+3=\frac{37}{3}\ in$

the third side of a triangle is z

$z=\frac{28}{3}-3=\frac{19}{3}\ in$

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3 years ago

Engh

Gennadij [26K]

Answer:

Step-by-step explanation: bc i said

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3 years ago

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Hunter-Best [27]

... because it is very difficult to prove, and because that notion is based on ... They may not make more commission to sell that product, but it is a ..... help, make sure you are working with a Registered Investment Advisor, not a broker.

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3 years ago