All categories

3 years ago

Solve: 13/19 - (-1/19) A. -12/19 B. -14/19 C. 14/19 D. 12/19

Mathematics

2 answers:

wel3 years ago

7 0

Answer:

Step-by-step explanation:Any time you have a subtraction and a negitive right next to eachother, it means to add. So, you would add 13 and 1 to get 14, then keep the denominator 19 to get 14/19.

yarga [219]3 years ago

3 0

Answer:

c.14/19

Step-by-step explanation:

because math exists

You might be interested in

Write an equation of the line that passes through (−5, −2)(−5, −2) and is parallel to y=23x+1y=23x+1.

Klio2033 [76]

Assuming you meant 2/3
y=2/3x+1 and (-5,-2)

paralell means same slope
y=mx+b
m=slope

y=2/3x+1
sllope=2/3

point slope
y-y1=m(x-x1)
for slope=m
a point is (x1,y1)

slope=2/3
point=(-5,-2)

y-(-2)=2/3(x-(-5))
y+2=2/3(x+5)
or
y=2/3x+16/3

7 0

3 years ago

Big Smiles Toy Company needs to pack marbles into a box. The company has two boxes they can use. Box A is 9 inches wide, 13 inch

shtirl [24]

Answer:

Big Smiles Toy Company should choose Box B as its volume is 82.5 $inch^3$ larger than Box A

Step-by-step explanation:

Given

Dimension of Box A= 9 inches width, 13 inches length, 11.5 inches height

Dimension of Box B =14 inches width, 8.5 inches length, 12 inches height

Volume of Box A= Length $\times$ width $\times$ height= $9\times13\times11.5= 1345.5 inches^3$

Volume of Box B =Length $\times$ width $\times$ height= $14\times8.5\times12= 1428 inches^3$

Difference = Volume of Box B - Volume of Box A = 1428 - 1345.5 = 82.5 $inches^3$

Box B has 82.5 $inch^3$ volume larger than Box A

4 0

3 years ago

Calculate the total hours Nicki worked last week.

belka [17]

Answer:

47 hours

$517

Step-by-step explanation:

M8 T9 W9 T8 F0 S8 S5

From the information above :

Total hours worked last week :

(8 + 9 + 9 + 8 + 0 + 8 + 5) = 47 hours

Rate = $11 / hour

Last week earning :

Total hours worked * rate per hour

47 * $11

= $517

6 0

3 years ago

What is the slope of the line?

Elan Coil [88]

Answer:

1/4

Step-by-step explanation:

The slope of a line is rise over run. We can see here rise is 1 and run is 4. 1 over 4 = 1/4.

5 0

2 years ago

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:

brainly.com/question/8959311

6 0

2 years ago