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S_A_V [24]
3 years ago
9

Someone pls help me with this it’s not cut off this time but pls help me

Mathematics
1 answer:
stealth61 [152]3 years ago
5 0
It’s A “9/14”
Explanation:
3/4 is bigger than 9/14
If you graph those two fractions, you can clearly see that 9/14 is one number/block less than 3/4. Hope this helped!
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What is the order of the steps to solve for x in the equation 3x-6x-7=29
dexar [7]
First, you should combine the x terms:

-3x - 7 = 29

Then, get rid of the negative 7, by adding 7 to both sides.

-3x = 22

Finally divide 22 by -3, to solve for x.

x = -22/3 or -7.33
7 0
3 years ago
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-1/5y+7=7<br> What is the value of y?
liq [111]
The answer is y=0

See the attached photo for details

Hope this helps! Please make me the brainliest, it’s not necessary but appreciated, I put a lot of effort and research into my answers. Have a good day, stay safe and stay healthy.

8 0
3 years ago
Consider angle θ in Quadrant II, where sinθ = 4/5 . What is the value of cosθ?
Kay [80]

Answer:

3/5

Step-by-step explanation:

sin theta=opposite/hypotenuse

4/5=opposite/hypotenuse

therefore opposite=4 and hypotenuse=5

for adjacent

using pythagoras theorem

a^2+b^2=c^2

opposite^2 + adjacent^2 =hypotenuse^2

4^2 + adjacent^2 =5^2

16 + adjacent^2 =25

adjacent^2 =25-16

adjacent =\sqrt{9

adjacent=3

cos theta=adjacent/hypotenuse

=3/5

therefore the value of cos theta is 3/5

8 0
3 years ago
78 * 40
Debora [2.8K]

Answer:

3120

Step-by-step explanation:

step 1:multiply 8 with 0 and 7 with 0

step 2:add a 0 in the end and multiply 8 with 4 . you get 32. take 2 and carry over 3. multiply 7 with 4. you get 28 . add 3 you get 31. now add. you get 3120

3 0
3 years ago
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Using function notation, describe the transformation. x^2+y^2=1 --&gt; (x-7)^2 + (y+4)^2 = 1
Fiesta28 [93]

Answer:

f(x - 7) - 4

Translation: < 7 , -4 >

Step-by-step explanation:

x^2+y^2=1

Centre: (0,0)

(x-7)^2 + (y+4)^2 = 1

Centre: (7,-4)

Translation: < 7 , -4 >

If the first function is f(x), the transformed function is:

f(x - 7) - 4

3 0
3 years ago
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