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3 years ago

What's the answer I'm not sure.

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

7 0

For the first one EF is 119 degrees now since that's made with two radii it makes a central angle so whatever that arc is the central angle will be the same. fro the second one since SRQ is 52 degrees you just take that and subtract it from 180 to get SQ. I'm not quite sure about the third one. the forth you are taking the measurement of the arc its connected to and dividing that by 2 so it would be 71 degrees. I'm pretty sure in right.

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Find the value of the following expression: (2^8 ⋅ 5^−5 ⋅ 19^0)^−2 ⋅ 5 to the power of negative 2 over 2 to the power of 3, whol

koban [17]

Answer:

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

Step-by-step explanation:

$\left(2^8\cdot5^{-5}\cdot19^0\right)^{-2}\cdot\left(\dfrac{5^{-2}}{2^3}\right)^4\cdot228\\\\\text{use}\ a^{-n}=\dfrac{1}{a^n}\ \text{and}\ a^0=1\ \text{and}\ (a^n)^m=a^{nm}\\\\=\left(2^8\cdot\dfrac{1}{5^5}\cdot1\right)^{-2}\cdot\left(\dfrac{\frac{1}{5^2}}{2^3}\right)^4\cdot228=\left(\dfrac{2^8}{5^5}\right)^{-2}\cdot\left(\dfrac{1}{2^35^2}\right)^4\cdot228$

$=\dfrac{(2^8)^{-2}}{(5^5)^{-2}}\cdot\dfrac{1^4}{(2^3)^4(5^2)^4}\cdot228=\dfrac{2^{-16}}{5^{-10}}\cdot\dfrac{1}{2^{12}5^8}\cdot228\\\\\text{use}\ a^n=\dfrac{1}{a^{-n}}\to\dfrac{1}{a^n}=a^{-n}\\\\=2^{-16}\cdot5^{10}\cdot2^{-12}\cdot5^{-8}\cdot228\\\\\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=2^{-16+(-12)}\cdot5^{10+(-8)}\cdot228=2^{-28}\cdot5^2\cdot228\\\\=2^{-28}\cdot5^2\cdot4\cdot57=2^{-28}\cdot5^2\cdot2^2\cdot57=2^{-28+2}\cdot5^2\cdot57\\\\=2^{-26}\cdot5^2\cdot57=\dfrac{5^2\cdot57}{2^{26}}$

$\large\boxed{\dfrac{5^2\cdot57}{2^{26}}=\dfrac{1425}{67108864}}$

4 0

3 years ago

I have included a picture of the equation

Alchen [17]

Answer:

it's c , x= 1.5 or -4

2x^2 + 5x - 12 = 0

First factor the left side of the equation

(2x-3)(x+4)=0

Second, set the factors equal to 0

2x-3=0 or x+4=0

+3 +3 -4 -4

2x=3 -4

divide both side by 2 and you get 3/2 which equals 1.5

x=1.5 or x= -4

3 0

2 years ago

In ΔDEF,{EF}\cong{DE} EF ≅ DE and m∠E = 104°. Find m∠D.

Alexeev081 [22]

The 3 inside angles need to equal 180.

Subtract E from 180 to find the sum of the other two angles:

180 - 104 = 76

Now because EF ≅ DE , angle D and F would be the same so divide 76 by 2

Angle D = 38

7 0

3 years ago

PLEASE HELP! SEE THE IMAGES BELOW! 30 POINTS

Mkey [24]

Answer:

d for the first and c for the second

Step-by-step explanation:

1. both are solid lines so it is or equal to something. the first line is going down by -3 and crosses the y axis at -3 so it is _>_ - 3x-3

the second line is going down at -1/2 and crosses the y axis at 2 so it is _<_ -1/2x+2

2. The point where the two lines cross is at (2.5, -6.5)

6 0

3 years ago

At 7:00 am, the temperature was -4℉. At 9:00 am, it was 8° warmer. What was the temperature at 9:00 am? *

Nana76 [90]

4℉.

What we know about Degrees is that there is a Positive type and a negative type.

(i.e: 30℉ is positive and -30℉ is negative.)

If the temperature was -4℉ at 7AM, then it is negative. If it goes up by an amount that is more than 4 then that negative will go up to a positive temperature. In this case: At 9AM it was 8° warmer.

Warmer is a keyword. If it is warmer by an amount, Negative temperatures will go up to a positive and positive temperature will just go up. If it gets cooler, negative temperatures will go down further and positive temperatures will go down to a negative.

So lets work out this problem with our newfound knowledge.

-4° F at 7AM

8° warmer at 9AM

-4 + 8 = 4.

The temperature was 4° at 9AM.

-Snooky

8 0

3 years ago