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pochemuha
3 years ago
5

Can you find the simplest form of this math problem.​

Mathematics
2 answers:
vagabundo [1.1K]3 years ago
8 0

Answer:

-7y^2

Step-by-step explanation:

(0.01)^(-1/2) = (1/0.01)^(1/2) = 100^(1/2) = 10

(-343 × y^6 ÷ 1000)^(1/3) =

(-343)^(1/3) × (y^6)^(1/3) ÷ 1000^(1/3)

= -7 × y^2 ÷ 10 = -0.7y^2

10 × -0.7y^2

-7y^2

Wittaler [7]3 years ago
7 0

Answer:

- 7y²

Step-by-step explanation:

(0,01)⁻¹/² × ∛(- 343y⁶/1000) =

= -(1/100)⁻¹/² × ∛( - 7³y⁶/10³)

= - 100¹/² × 7y²/10

= -(10²)¹/² × 7y²/10

= - 10 × 7y²/10

= - 7y²

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pychu [463]

Answer:

C: one triangle

Step-by-step explanation:

well a triangle = 180°

45+45=90

90+90=180

7 0
3 years ago
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Write an equation of the line that passes through the points.<br><br> (0,1),(−2,−5)
Arada [10]

Answer:

y=4/-2

Step-by-step explanation:

I think that is the answer

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2 years ago
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What is the value of -2(x+5)?​
bezimeni [28]
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So -2x=-10
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7 0
3 years ago
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one days discharge at its mouth, 3 trillion gallons, could supply all of country A’s household for 5 months. how much water an a
serious [3.7K]

let's use engineering notation for the sake of brevity.

1 trillion is 1,000,000,000,000, or just 1E12, twelve zeros.

1 million is then 1E6, six zeros.

we know the discharge for one day is 3E12 gallons, and that'd do just fine for country A for 5 months.  How many gallons in 1 month only?

\bf \begin{array}{ccll} gallons&months\\ \cline{1-2} 3E12&5\\ x&1 \end{array}\implies \cfrac{3E12}{x}=\cfrac{5}{1}\implies 3E12=5x \\\\\\ \cfrac{3E12}{5}=x\implies 6E11=x\implies 600000000000=x

if there are 200million inhabitants in A, namely 200E6 or 2E8 inhabitants, how many gallons per inhabitant from all those 6E11 gallons?

\bf \begin{array}{ccll} gallons&households\\ \cline{1-2} 6E11&2E8\\ x&1 \end{array}\implies \cfrac{6E11}{x}=\cfrac{2E8}{1}\implies 6E11=2E8x \\\\\\ \cfrac{6E11}{2E8}=x\implies \cfrac{600000000000}{200000000}=x\implies 3000=x

5 0
3 years ago
The mean sat verbal score is 486, with a standard deviation of 95. use the empirical rule to determine what percent of the score
Pavel [41]
Find the z-scores for the two scores in the given interval.

z=\frac{x-\mu}{\sigma}

For the score x =391, z=\frac{391-486}{95}=\frac{-95}{95}=-1.

For the score x = 486, z=\frac{486-486}{95}=0

Now you want the area (proportion of data) under the normal distribution from z = -1 to z = 0. The Empirical Rule says that 68% of the data falls between z = -1 to z = 1. But the curve is symmetrical around the vertical axis at z = 0, so the answer you want is HALF of 68%.

8 0
3 years ago
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