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3 years ago

(t^-4)^-9t^2 A)t^26 B)t^-26 C)t^38 D)t^-38

1 answer:

4 0

The relevant rules of exponents are
.. (t^a)^b = t^(a·b)
.. t^a·t^b = t^(a+b)

You have
.. (t^-4)^-9·t^2
.. = t^((-4)*(-9) +2)
.. = t^38 . . . . . . . . . . . selection C

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Determine two numbers that have a sum of 12 but a difference of 12

sveta [45]

Answer:

0 And 12

Step-by-step explanation:

0 + 12 = 12

But 0 is 12 away from 12. hope i helped :)

6 0

2 years ago

Read 2 more answers

Yati has a coin box containing some coins. 40% of the coins are 20-cent coins and the rest are 50-cent coins.

lana66690 [7]

The fraction that should be removed is 1/5.

<h3>How to calculate the fraction?</h3>

It should be noted that from the information, 40% of the coins are 20-cent coins and the rest are 50-cent coins. Those that are 50 cent will be:

= 100 - 40%

= 60%

The fraction that's should be removed will be:

= 2/5 - x = 1/5

x = 2/5 - 1/5

x = 1/5

The fraction that should be removed is 1/5.

Learn more about fractions on:

brainly.com/question/17220365

#SPJ1

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1 year ago

Step 3: Displaying numerical data in dot plots, and describing and analyzing

mylen [45]

Answer:

Step-by-step explanation:

i added it all

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2 years ago

Express the radical using the imaginary unit, i

Kipish [7]

Answer:

7i.

Step-by-step explanation:

Note : The given expression must be $\sqrt{-49}$ .

We have to write this radical value in the imaginary value.

The given value can be rewritten as

$\sqrt{-49}=\sqrt{49\times (-1)}$

$\sqrt{-49}=\sqrt{49}\times \sqrt{-1}$ $[\because \sqrt{ab}=\sqrt{a}\sqrt{b}]$

$\sqrt{-49}=7\times i$ $[\because \sqrt{-1}=i]$

$\sqrt{-49}=7i$

Therefore, the required expression is 7i.

8 0

3 years ago

Let a, b, c, and d be real numbers with a, c 6= 0. Prove that the lines y = ax+b and y = cx + d have the same x-intercept if and

Monica [59]

Step-by-step explanation:

We have got the lines :

$y=ax+b\\y=cx+d$

Both lines intercept the x-axis in the point :

$I = (i_{1} ,i_{2})$

In all point from x-axis the y-component is equal to 0.

$I=(i_{1},o)$

We replace the I point in the lines equations:

$0=a(i_{1})+b \\0=c(i_{1})+d$

From the first equation :

$0=a(i_{1})+b \\-b=a(i_{1})\\i_{1}=\frac{-b}{a}$

From the second equation :

$0=c(i_{1})+d\\ -d=c(i_{1})\\i_{1}=\frac{-d}{c}$

Then $i_{1}=i_{1}$

Finally :

$\frac{-b}{a}=\frac{-d}{c} \\\frac{b}{a}=\frac{d}{c} \\ad=bc$

y = ax + b and y = cx + d have the same x-intercept ⇔ad=bc

8 0

2 years ago