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3 years ago

Simplify the expression: 8(1 + 8z) = Submit

Mathematics

1 answer:

Firlakuza [10]3 years ago

7 0

Answer: My answer is in the bottom :)

Step-by-step explanation:8(1 + 8z) do the distributive property

8+64z=72z or 8=64z

64 64

1/8=z

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2.5=25% is it true or false

Gnom [1K]

Answer: false

explanation;

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3 years ago

Read 2 more answers

(4x³-4+7x)-(2x³-x-8)

Ket [755]

(4x³-4+7x)-(2x³-x-8) = 4x³-4+7x - 2x³ + x+8 = (4x³-2x³) + (7x+x) -4+8 =

= 2x³ + 8x +4

(4x³-4+7x)-(2x³-x-8) = 2x³ + 8x +4

4 0

3 years ago

A+b/c=d/4 solve for a

RUDIKE [14]

Answer:

a = d/4 - b/c

Step-by-step explanation:

Subtract b/c from both sides. We can subtract a fraction

a = d/4 - b/c

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3 years ago

When f(x) is replaced with f(x)+k, the graph that results is shown below K=__?

serious [3.7K]

Answer:

k = 6

Step-by-step explanation:

Parent function of the given graph → f(x) = (x - 1)²

If the parent function is shifted 6 units up.

Equation of the translated function will be → g(x) = (x - 1)² + 6

g(x) = f(x) + 6

Therefore, k = 6 will be the answer.

6 0

3 years ago

Twenty identical looking packets of white power are such that 15 contain cocaine and 5 do not. Four packets were randomly select

xxMikexx [17]

Answer:

The probability of selecting randomly 6 packets such that the first 4 contain cocaine and 2 not is 0.3522.

Step-by-step explanation:

Probabilities of selecting 4 packets with the ilegal substance:

$(\frac{15}{4})\\$ = $\frac{15!}{4!(15-4)!}$

Combinassions possible= 1365

Probabilities of selecting 2 packets with white powder:

$(\frac{5}{2})$ = $\frac{5!}{2!(5-2)!}$

Combinations possible= 10

Probabilities of selecting 6 packets from the totality of them:

$(\frac{20}{6})$ = $\frac{20!}{6!(20-6)!}$

Combinations possible= 38760

The probability of picking 4 with the substance and 2 with only white powder is:

$\frac{(15ncr4)(5ncr2)}{(20ncr6)}$ = 0.3522

I hope this answer helps you.

6 0

3 years ago