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2 years ago

Find the product. (3x2+6x-5)(-3x) PLEASE HELP!!! ASAP!!!

Mathematics

2 answers:

nekit [7.7K]2 years ago

8 0

Answer:

$\boxed{-9x^3-18x^2 + 15x}$

Step-by-step explanation:

(3x²+6x-5)(-3x)

Apply distributive law.

-3x(3x²)-3x(6x)-3x(-5)

Simplify.

-9x³ - 18x² + 15x

VashaNatasha [74]2 years ago

5 0

Answer:

$\boxed{\red{ - 9 {x}^{3} - 18 {x}^{2} + 15x}}$

Step-by-step explanation:

$( - 3x)(3 {x}^{2} + 6x - 5) \\ - 3x(3 {x}^{2} ) - 3x(6x) - 3x( - 5) \\ = - 9 {x}^{3} - 18 {x}^{2} + 15x$

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Order of Operations- -6+(-30) divided by 5-(-2). Explain step by step

Reil [10]

Answer:

-36/7 or -5.14287

Step-by-step explanation:

so I would get rid of parentheses you do not need them

$\frac{-6+-30}{5--2}$ = $\frac{-36}{7}$ (because when two negative come before two numbers like -x-y they are being added but negative is present to the result of the sum, when one negative sign comes after a number and the other negative sign comes before a number, like x--y they are being added as well but negative is not present to the result of the sum)

-36/7=-5.1428

hope this helps!

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Answer:

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How to estimate and multiply 83x367

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Assume that in a particular military exercise involving two units, Red and Blue, thereis a 60 % chance that the Red unit will su

Marysya12 [62]

Answer: a) There are 42% chances that both units will meet their objectives.

b) There are 88% chances that one or the other but not both of the units will be successful.

Step-by-step explanation:

Since we have given that

Probability that the Red unit will successfully meet its objectives = 60% = P(R)

Probability that Blue unit will successfully meet its objectives = 70% = P(B)

Probability that only Red unit will be successful = P(only Red) = 18%

As we know that

$P(only\ red)=P(R)-P(R\cap B)\\\\0.18=0.60-P(R\cap B)\\\\0.18-0.60=-P(R\cap B)\\\\-0.42=-P(R\cap B)\\\\P(R\cap B)=42\%$

Hence, there are 42% chances that both units will meet their objectives.

the probability that one or the other but not both of the units will be successful is given by

$P(R\cup B)=P(R)+P(B)-P(R\cap B)\\\\P(R\cup B)=0.60+0.70-0.42\\\\P(R\cup B)=0.88\\\\P(R\cup B)=88\%$

Hence, there are 88% chances that one or the other but not both of the units will be successful.

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