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3 years ago

Rewrite using a negative exponent. 1/a ^4

Mathematics

1 answer:

vesna_86 [32]3 years ago

8 0

Answer: a^-4

Step-by-step explanation:

$\frac{1-4}{a^4-4}$ = a^-4

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san4es73 [151]

Answer:

The answer is 2. 2 is the only number that can't change. There is no x or ^.

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3 years ago

Brad has 42 stamps to arrange on an album page

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3 years ago

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4x^2-5x-15=0 how to find the vertex and y intercept

mezya [45]

$ax^2+bx+c=0\\\\the\ vertex:\left(\frac{-b}{2a};\frac{-(b^2-4ac)}{4a}\right)\\\\y-intrcept=c\\=======================================\\\\4x^2-5x-15=0\\\\a=4;\ b=-5;\ c=-15\\\\\frac{-b}{2a}=\frac{-(-5)}{2\cdot4}=\frac{5}{8};\ \frac{-(b^2-4ac)}{4a}=\frac{-[(-5)^2-4\cdot4\cdot(-15)]}{4\cdot4}=-\frac{265}{16}\\\\\boxed{the\ vertex:\left(\frac{5}{8};-\frac{265}{16}\right)}\\\\\boxed{y-intercept:-15}$

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3 years ago

Please Help !!!

solong [7]

Best Answer: 2 LiCl = 2 Li + Cl2
mass Li = 56.8 mL x 0.534 g/mL=30.3 g
moles Li = 30.3 g / 6.941 g/mol=4.37
the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)

moles LiCl required = 4.37
mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g

Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag
the ratio between Cu and AgNO3 is 1 : 2
moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant
moles Cu reacted = 6.3 / 2 = 3.15
moles Cu in excess = 4.2 - 3.15 =1.05

N2 + 3 H2 = 2 NH3
moles N2 = 42.5 g / 28.0134 g/mol=1.52
the ratio between N2 and H2 is 1 : 3
moles H2 required = 1.52 x 3 =4.56
actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant
moles NH3 = 1.52 x 2 = 3.04
mass NH3 = 3.04 x 17.0337 g/mol=51.8 g
moles H2 in excess = 5.00 - 4.56 =0.44
mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g

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4 years ago

Anyone know how to do this

svlad2 [7]

Answer:

The area of the rectangle on the left side is

$9cm \: \times 4cm = 36 {cm}^{2}$

The area of the bottom rectangle is

$6cm \times 2cm = 12 {cm}^{2}$

The total area of the composite figure will be

$36 {cm}^{2} + 12 {cm}^{2} = 48 {cm}^{2}$

Step-by-step explanation:

The area of any given rectangle can be found by multiplying the length of that rectangle by its width. The rectangle on the left side has a length of 9cm but the width is unknown. To find the width, we subtract 6cm from the width of the bottom rectangle: 10cm. And that gives us 4cm.

Therefore, we can now calculate the area to be: length × width = 9cm × 4cm = 36cm²//

The area of the bottom rectangle can be found similarly by multiplying the length: 2cm by the width: 6cm of that rectangle. And the result gives us: 2cm × 6cm = 12cm²//

The total area of the composite figure is calculated by adding the results from the left and bottom rectangles together. And that gives us: 36cm² + 12cm² = 48cm²//

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3 years ago