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3 years ago

The sum of the number and three times its reciprocal is 79/10. Find the number.

Mathematics

1 answer:

luda_lava [24]3 years ago

5 0

so x + 3 (-x) = 79/10

x-3x = 79/10

-2x = 79/10

x = 79/10 / -2

X = 79/ 10 * -1/2

x = - 79/20

x = - 3.95

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Pls help me

Vladimir [108]

<h3>Learning Task 1</h3>

Correct responses;

$\displaystyle 1. \hspace{0.3 cm} 7^{-1} = \frac{1}{7}$

2. (14·a·b·c)⁰ = 1

$\displaystyle 3. \hspace{0.3 cm} 10^{-9} = \frac{1}{10^9}$

4. 5·(x·y)⁰ = 5

5. 0¹⁵ = 0

$\displaystyle 6. \hspace{0.3 cm} \frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4} = 4 \cdot x^3$

$\displaystyle 7. \hspace{0.3 cm} \left(\frac{5 \cdot x^0}{y} \right)^{-1} = \frac{y}{5}$

$8. \hspace{0.3 cm}\displaystyle \frac{120 \cdot a^5 \cdot b^6 \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0 \cdot c^{2}} = \frac{10 \cdot a^7 \cdot b^6}{c^7}$

$\displaystyle 9. \hspace{0.3 cm} \frac{(4 \cdot x)^0 \cdot y^{-5} \cdot z^{-2} }{(234 \cdot x \cdot y \cdot z)^0} = \frac{1}{y^{5} \cdot z^2}$

$\displaystyle 10. \hspace{0.3 cm} \left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2} = 100$

$11. \displaystyle \hspace{0.3 cm} \left(\frac{(a \cdot b)}{9} \right)^{-1} = \frac{9}{a \cdot b}$

$\displaystyle 12. \hspace{0.3 cm} \frac{9}{x^{-2}} = 9 \cdot x^2$

$13. \displaystyle \hspace{0.3 cm} \frac{12 \cdot b^{-3}}{a^{-4}} = \frac{12 \cdot a^4}{b^3}$

$14. \displaystyle \hspace{0.3 cm} \frac{1}{a^{-5 \cdot n}} =a^{5 \cdot n}$

$15. \displaystyle \hspace{0.3 cm} \left(\frac{1}{2} \right)^{-5} = 32$

<h3>Methods by which the above expressions are simplified</h3>

The given expressions expressed into non zero and non negative exponents are;

$\displaystyle 1. \hspace{0.3 cm} \mathbf{7^{-1}} = \underline{\frac{1}{7}}$

2. (14·a·b·c)⁰ = 14⁰ × a⁰ × b⁰ × c⁰ = 1 × 1 × 1 × 1 =<u> 1</u>

$\displaystyle 3. \hspace{0.3 cm} \mathbf{ 10^{-9}} = \underline{ \frac{1}{10^9}}$

4. 5·(x·y)⁰ = 5 × x⁰ × y⁰ = 5 × 1 × 1 = <u>5</u>

5. 0¹⁵ = <u>0</u>

$\displaystyle 6. \hspace{0.3 cm} \mathbf{\frac{24 \cdot x^8 \cdot y^4}{6 \cdot x^5 \cdot y^4}} = \frac{ 4 \times 6 \times x^5 \times x^3 \times y^4}{6 \times x^5 \times y^4} = \underline{4 \cdot x^3}$

$\displaystyle 7. \hspace{0.3 cm} \mathbf{\left(\frac{5 \cdot x^0}{y} \right)^{-1} }= \frac{1}{\frac{5 \times 1}{y} } = \underline{\frac{y}{5}}$

$8. \hspace{0.3 cm}\displaystyle \mathbf{\frac{120 \cdot a^5 \cdot b^6 \cdot c^{-5} }{12 \cdot a^{-2} \cdot b^0 \cdot c^{2}}} = 10 \cdot a^{5 - (-2)} \cdot b^{6 - 0} \cdot c^{-5 -2} = \underline{\frac{10 \cdot a^7 \cdot b^6}{c^7}}$

$\displaystyle 9. \hspace{0.3 cm} \mathbf{\frac{(4 \cdot x)^0 \cdot y^{-5} \cdot z^{-2} }{(234 \cdot x \cdot y \cdot z)^0}} = y^{-5} \cdot z^{-2} = \underline{\frac{1}{y^{5} \cdot z^2}}$

$\displaystyle 10. \hspace{0.3 cm} \mathbf{\left(\frac{(5 \cdot x \cdot y)^0}{10} \right)^{-2}} = \left(\frac{1}{10} \right)^{-2} = \frac{1}{\left(\frac{1}{10} \right)^2 } = \frac{1}{\frac{1}{100} } = \underline{100}$

$11. \displaystyle \hspace{0.3 cm} \mathbf{\left(\frac{(a \cdot b)}{9} \right)^{-1}} = \frac{1}{\left(\frac{(a \cdot b)}{9} \right) } = \underline{\frac{9}{a \cdot b}}$

$\displaystyle 12. \hspace{0.3 cm} \mathbf{\frac{9}{x^{-2}}} = \frac{9}{\frac{1}{x^2} } = \underline{9 \cdot x^2}$

$13. \displaystyle \hspace{0.3 cm} \mathbf{\frac{12 \cdot b^{-3}}{a^{-4}}} = 12 \cdot \frac{b^{-3}}{a^{-4}} =12 \cdot \frac{1}{\frac{b^3}{a^4} } = 12 \cdot \frac{a^4}{b^3} = \underline{\frac{12 \cdot a^4}{b^3}}$

$14. \displaystyle \hspace{0.3 cm} \mathbf{\frac{1}{a^{-5 \cdot n}}} = \frac{1}{\frac{1}{a^{5 \cdot n}} } = \underline{a^{5 \cdot n}}$

$15. \displaystyle \hspace{0.3 cm} \mathbf{\left(\frac{1}{2} \right)^{-5}} = \frac{1}{\left(\frac{1}{2} \right)^5} = 2^5 = \underline{32}$

Learn more about the laws of indices here:

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6 0

2 years ago

PLEASE HELP!

miskamm [114]

Answer:

340 square feet

Step-by-step explanation:

If we "unwrap" the painted surface from the shed, it will have the shape shown in the attachment. It is essentially a 40' by 8' rectangle with two 10' wide by 2' high triangles added.

The rectangle area is ...

A = LW = (40 ft)(8 ft) = 320 ft²

The total area of the two triangles is ...

A = 2(1/2)bh = (10 ft)(2 ft) = 20 ft²

Then the painted area is ...

total area = 320 ft² +20 ft²

total area = 340 ft²

8 0

4 years ago

Find the area of the regular polygon.

Neporo4naja [7]

Answer:

Then, the area is 222,48 in

Step-by-step explanation:

The area of the regular polygon is: 1/2 x perimeter x apothem.

The perimeter 6 times the number of sides of the polygon which is 9*6in = 54 in.

The apothem is: s/2tan(180/n) where "s" is the lenght of the side and "n" the number of sides.

apothem = s/2tan(180/n) = 6in/(2tan(180/9)) = 8.24 in

Then, the area is: 1/2 x 54 in x 8.24 in = 222,48 in

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4 years ago

How many quarters if she delivers 21 newspapers

evablogger [386]

525...............;0;0

8 0

3 years ago

-18 + (-6)<br> nton vann answer below:<br> idk what this is

EastWind [94]

Answer:

-24

Step-by-step explanation:

you add them like you would in addition but since they are both negative you keep the sign, making it -24

7 0

3 years ago