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2 years ago

13

-8d > - 2 Inequality

1 answer:

Tasya [4]2 years ago

3 0

Answer:

D<1/4

hopefully it's correct

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Label the equations in number 5-7 as True or False:

kkurt [141]

Answer:

${ \rm{ {(7}^{1}) ^{9 } = 7 {}^{5} \times {7}^{5} : \: { \tt{ \red{false}}} }} \\{ \boxed{ \mathfrak{correction \dashrightarrow \: ( {7}^{1}) {}^{9} = {7}^{9} }}} \\ \\ { \rm{ \frac{ {y}^{13} }{ {( {y}^{2}) }^{6} } = {y}^{ - 10} \times {y}^{9} : \:{ \tt { \red{false}}}}} \\ { \boxed{ \mathfrak{correction \: \dashrightarrow \: { \rm{{y}^{1} }}}}} \\ \\ { \rm{ {b}^{4} \times {b}^{ - 6} = \frac{ {b}^{1} \times {b}^{3} }{ {b}^{6} } }} \: : \: { \tt{ \green{true}}}$

8 0

2 years ago

Read 2 more answers

Two numbers are 10 units away in different directions from their midpoint, m, on a number line. The product of the numbers is –9

irinina [24]

In my opinion the answer is <span>m2 – 100 = –99
proof
</span>m2 – 100 =<span>(m – 10)(m + 10)
for eg. m=1, </span>(-9)(11)= <span>–99</span>

4 0

2 years ago

Read 2 more answers

The loading ramp for a truck makes a 30° angle to the ground. The back of the truck is 4' above the ground.

Gnesinka [82]

correct answer is d i think so

8 0

2 years ago

Solve for n: c= 6n−5g /11t

qwelly [4]

Answer: $\frac{11tc+5g}{66t}=n$

Step-by-step explanation:

You have the equation $c=6n-\frac{5g}{11t}$ .

Then, to solve for the variable $n$ from the equation you need:

Make the subtraction of the right side of the equation:

(As the denominators are 1 and $11t$ , the least common denominator is $11t$ )

$c=\frac{(6n)(11t)-5g}{11t}\\\\c=\frac{66nt-5g}{11t}$

Multiply $11t$ to both sides:

$(11t)c=(\frac{66nt-5g}{11t})(11t)\\\\11tc=66nt-5g$

Add $5g$ to both sides:

$11tc+5g=66nt-5g+5g\\\\11tc+5g=66nt$

And finally divide both sides by $66t$ :

$\frac{11tc+5g}{66t}=\frac{66nt}{66t}\\\\\frac{11tc+5g}{66t}=n$

3 0

2 years ago

Find f ∘ g f∘g and g ∘ f g∘f. f ( x ) = x 2 / 3 , g ( x ) = x 9 f(x)=x2/3, g(x)=x9 (a) f ∘ g f∘g (b) g ∘ f g∘f Find the domain o

faltersainse [42]

I don’t know it’s soooo long:)

8 0

2 years ago