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Jlenok [28]
3 years ago
7

Is 0.02 larger than 0.0013

Mathematics
1 answer:
ch4aika [34]3 years ago
5 0
Yes 0.02 is larger than 0.0013
You might be interested in
Is 11/16 less than 3/8
Brilliant_brown [7]
\dfrac{11}{16} \ \textless \  \dfrac{3}{8}-FALSE\\\\because\\\\\dfrac{3}{8}=\dfrac{3\cdot2}{8\cdot2}=\dfrac{6}{16} \ \textless \  \dfrac{11}{16}
5 0
4 years ago
Pls help me with this
astra-53 [7]

Answer:

the third one!

Step-by-step explanation:

Because it says, the SUM of the number, x (or z I can't really see it well) and 1/2, you must add those two together:

1/2 + x

Then it says those two numbers equal to 4, so it would look like this:

1/2 + x = 4

Because you are trying to isolate, and solve for x, you must SUBTRACT 1/2 from both sides of the equal sign:

1/2  +  x  =  4

-1/2            -1/2

1/2 minus 1/2 would result 0, so there would be x by itself. and 4 minus 1/2 would be 3 and a half ( 3  1/2), so x would be  3  1/2

x = 3 1/2

Hope this helped. If this is incorrect please specify so I can help :)

7 0
3 years ago
Write a real-world problem for 7.50y+9.<br><br> PLZ HURRY!!!
prisoha [69]

Answer:

You are going to school in a week and you are out to get school supplies. You need a backpack and it costs 9 dollars. Each pack of mechanical pencils cost $7.50. How much do you spend if you buy "y" packs of pencils?

Step-by-step explanation:

9 is initial fee (y-intercept) and 7.50 is the slope.

8 0
4 years ago
Three point one four times eight squared
Margaret [11]

Answer:

200.96

Step-by-step explanation:

Hope it helps! Good luck with math.

8 0
3 years ago
How can you determine which quadrant an equation falls in?
Sunny_sXe [5.5K]
\bf 2sec^2(\theta )-tan(\theta )-3=0&#10;\\\\\\&#10;2[1+tan^2(\theta )]-tan(\theta )-3=0\implies 2+2tan^2(\theta )-tan(\theta )-3=0&#10;\\\\\\&#10;2tan^2(\theta )-tan(\theta )-1=0\implies [2tan(\theta )+1][tan(\theta )-1]=0\\\\&#10;-------------------------------\\\\

\bf 2tan(\theta )+1=0\implies 2tan(\theta )=-1\implies tan(\theta )=-\cfrac{1}{2}&#10;\\\\\\&#10;\measuredangle \theta \approx&#10;\begin{cases}&#10;5.82\ radians\\&#10;2.68\ radians&#10;\end{cases}\qquad &#10;\begin{array}{llll}&#10;\textit{the tangent is negative}\\&#10;\textit{tha means the II and IV quadrants}&#10;\end{array}\\\\&#10;-------------------------------\\\\

\bf tan(\theta )-1=0\implies tan(\theta )=1\implies \measuredangle \theta =&#10;\begin{cases}&#10;\frac{\pi }{4}\\&#10;\frac{7\pi }{4}&#10;\end{cases}&#10;\\\\\\&#10;\textit{the tangent is positive}\\&#10;\textit{that means the I and III quadrants}


bear in mind that tangent is sine/cosine or y/x

for the tangent to be negative, the signs of "y" and "x" must differ, and that happens only on the II and IV quadrants

and for the tangent to be positive, the signs must be same, and that's only on I and III quadrants.
8 0
3 years ago
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