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Kazeer [188]
3 years ago
13

HELP ME tysm :) Brainiest answer to the one that the website BELIEVES is correct

Mathematics
1 answer:
a_sh-v [17]3 years ago
6 0

Answer:

0.75CM^2 and 10 secs

Step-by-step explanation:

And then its 10 seconds for part 2

OH BTW U NEED TO DRAW A TANGENT

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Find the shaded area
grin007 [14]

Answer:

area of big rectangle=l×b=12×18=216cm²

area of smaller square=l²=8²=64cm²

area of shaded area =area of (bigger rectangle-smaller square)=216-64=152cm²

5 0
3 years ago
Joseph has started completing the square on the equation 3x^2 - 7x + 12 = 0. He has worked to the point where he has the express
ololo11 [35]
3x^2 - 7x + 12 = 0...move the 12 to the other side
3x^2 - 7x = -12....now divide by 3
x^2 - 7/3x = -4..this is where he made his mistake
3 0
4 years ago
Solve the equation. <br> 4y^3 – 7y^2 + 28 = 16y
Studentka2010 [4]

Answer: -2, 2, 7/4

***If the pictures is sideways, use the rotate button. I hope the explanation made sense!

4 0
3 years ago
Write the word sentence as an equation. Then solve the equation. A number multiplied by 2/3 is 3/20.
vaieri [72.5K]

Answer:

Equation = X*(2/3) = 3/20

Solve for X = 0.23

Step-by-step explanation:

Let, the number be "X"

According to the question,

X*(2/3) = 3/20..........(i)

From equation (i), we can get,

X = (3/20)/(2/3)

or, X = 0.15/0.66

or, X = 0.23

Alternative way,

Let, the number be "X"

According to the question,

X*(2/3) = 3/20..........(i)

From equation (i), we can get,

X = (3/20)/(2/3)

or, X = (3/20)*(3/2)

or, X = 9/40

or, X = 0.23

7 0
4 years ago
Help math question derivative!
atroni [7]
Let f(x)=\sec^{-1}x. Then \sec f(x)=x, and differentiating both sides with respect to x gives

(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1
f'(x)=\dfrac1{\sec f(x)\tan f(x)}

Now, when x=\sqrt2, you get

(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}

You have \sec^{-1}\sqrt2=\dfrac\pi4, so \sec\left(\sec^{-1}\sqrt2\right)=\sqrt2 and \tan\left(\sec^{-1}\sqrt2\right)=1. So (\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}
5 0
4 years ago
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