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3 years ago

106 percent of $__ is 7.95

1 answer:

3 0

$Method\ 1:\\\\p\%=\dfrac{p}{100}\\\\106\%=\dfrac{106}{100}=1.06\\\\106\%\ of\ \$x\ is\ \$7.95\to1.06x=7.95\ \ \ |divide\ both\ sides\ by\ 1.06\\\\\boxed{x=\$7.50}\\\\Method\ 2:\\\\106\%\ -\ \$7.95\\100\%\ \ -\ \$x\\\\cross\ multiply\\\\106\cdot x=100\cdot\$7.95\\106x=\$795\ \ \ \ |divide\ both\ sides\ by\ 106\\\boxed{x=\$7.50}$

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What is the equation of a line that contains the point (2,-5) and is parallel to the line y=3x-4

aev [14]

Answer:

$\large\boxed{y=3x-11}$

Step-by-step explanation:

$\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\===================================\\\\y=3x-4\to m_1=3\\\\\text{Therefore}\ m_2=3.\\\\\text{We have the equation:}\ y=3x+b.\\\\\text{Put the coorsdinates of the point (2, -5) to the equation of the line}\\\\-5=3(2)+b\\-5=6+b\qquad\text{subtract 6 from both sides}\\-11=b\to b=-11$

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3 years ago

How many times greater is the value of the 4 In 42,672 than the value of the 4 in 37,426

Olenka [21]

Step-by-step explanation:

it 100 more than the 4 in 37,426

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3 years ago

Find the average rate of change, in people per year, for the population between the years 1900 and 1920.

elena-s [515]

Step-by-step explanation:

the average change rate of a function f(x) in an interval [x1 .. x2] is

(f(x2) - f(x1)) / (x2 - x1)

in our case that is

(7000 - 2500) / (20 - 0) = 4500 / 20 = 225

the population changed in average by 225 people per year from 1900 to 1920.

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2 years ago

What perfect square that is between 1 and 100 has 27 as one of its factors?

lesya692 [45]

81 because 81 is a perfect square and 27 times 3 =81

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3 years ago

Estimates of the number of cells in the human body vary over an order of magnitude. Indeed, the precise number varies from one

Amiraneli [1.4K]

Answer:

a) 2.00 × 10¹⁰; b) 2.00 × 10¹³

Step-by-step explanation:

a) No. of cells in 1 cm³

$\text{No. of cells} =\text{1 cm}^{3} \times\left( \dfrac{\text{1 m}}{\text{100 cm}}\right )^{3} \times \left (\dfrac{1 \, \mu \text{m} }{10^{-6}\,\text{ m}}\right )^{3} \times \dfrac{\text{1 cell}}{\text{50 $\,\mu$m}^{3}} = \mathbf{2.00 \times 10^{10}} \textbf{ cells}$

b) No. of cells in 1 L

$\text{No. of cells} = \text{1 L} \times \dfrac{\text{1000 cm}^{3}}{\text{1 L}} \times \dfrac{2 \times 10^{10} \text{ cells}}{\text{1 cm}^{3}} = \mathbf{2.00 \times 10^{13}} \textbf{ cells}$

7 0

3 years ago