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Nadusha1986 [10]
2 years ago
14

What is the quotient of StartFraction 2 Superscript 4 Baseline Over 2 Superscript negative 4 Baseline EndFraction? StartFraction

1 Over 256 EndFraction One-half 1 256.
Mathematics
1 answer:
Lina20 [59]2 years ago
4 0

The quotient of StartFraction 2 Superscript 4 Baseline Over 2 Superscript negative 4 Baseline End Fraction  will be 256

<h3>What will be the quotient of  StartFraction 2 Superscript 4 Baseline Over 2 Superscript negative 4 Baseline EndFraction?</h3>

From the given data

=\dfrac{2^{4} }{2^{-4} }

Now it will become

2^{4} \times 2^{4} = 2^{8} =256  

Thus the quotient of StartFraction 2 Superscript 4 Baseline Over 2 Superscript negative 4 Baseline End Fraction  will be 256

To know more about Start and End fraction follow

brainly.com/question/1697242

 

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The sum of 9, 1​, and a number amounts to 23. Find the number.
maksim [4K]

Answer:

13

Step-by-step explanation:

9 + 1` = 10

23 - 10 = 13

4 0
3 years ago
Please Help Me!!! ASAP
FrozenT [24]

Answer:

x= 8

Below is the diagram.

Step-by-step explanation:

Given,

YW \ bisects\ \angle XYZ

Such that,

m\angle XYW=\ m\angle WYZ

Thus, substituting their values.

(3x-7)=(2x+1)\\3x-7=2x+1\\3x-2x=1+7\\x=8

Therefore on simplifying we get x= 8

7 0
4 years ago
A triangle has an area of 78 square inches. Its base is measured to be 13 inches. What is the height of this triangle in inches?
Fynjy0 [20]

Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{12 \: inches}}}}}

Step-by-step explanation:

Given,

Area of a triangle = 78 square inches

Base of a triangle = 13 inches

Height of a triangle = ?

<u>Finding</u><u> </u><u>the</u><u> </u><u>height</u><u> </u><u>of</u><u> </u><u>a</u><u> </u><u>triangle</u>

\boxed{ \sf{area \: of \: a \: triangle =  \frac{1}{2}  \times  \: base \:  \times  \: height}}

\dashrightarrow{ \sf{78 =  \frac{1}{2}  \times 13 \times h}}

\dashrightarrow{ \sf{78 = 6.5 \: h \: }}

\dashrightarrow{ \sf{6.5 \:  h = 78}}

\dashrightarrow{ \sf{ \frac{6.5h}{6.5}  =  \frac{78}{6.5}}}

\dashrightarrow{ \sf{h = 12 \: inches}}

Hope I helped!

Best regards! :D

4 0
3 years ago
An equation for loudness L in decibels is given by L= 10LogR where R is the sound’s relative intensity. An air-raid siren can re
bazaltina [42]
We are asked to find ratio of two relative intensities. Let's start by rearranging formula for R:
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For air-raid siren we have:
R_{1}= 10^{\frac{150}{10}} =10^{15}
For jet engine noise we have:
R_{2}= 10^{\frac{120}{10}} =10^{12}

To find out <span>many times greater is the relative intensity of the air-raid siren than that of the jet engine noise we need to divide these two numbers:
</span>tex] \frac{R_{1}}{R_{2}} = \frac{10^{15}}{10^{12}} =10^{3}=1000[/tex]
4 0
3 years ago
4.<br>Which is the best buy in the following cases.<br>a) 4 pens for 56 or 6 pens for rs 90<br>​
alukav5142 [94]

Answer:4 pens for 56

Step-by-step explanation:

4 0
3 years ago
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