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Stells [14]
3 years ago
11

How do i solve this problem ?

Mathematics
1 answer:
Usimov [2.4K]3 years ago
8 0
N is 7.5 because prq is 3x the size of abc
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Help meeeee<br> plsssssssss
marysya [2.9K]

Answer:

18

Step-by-step explanation:

7 0
3 years ago
How many lines of symmetry does this figure have?
Sergeeva-Olga [200]

Answer:

2

Step-by-step explanation:

There are two lines of symmetry and here I list them:

1) The first is a horizontal line that divides the square in to even parts such that the top part is the projection of the down one trough the symmetry line (and vice versa).

2) The second one is the vertical line that divides the square in two even sides. Note that this line will also divide both stars at half. The left side will be projected on the right one (and vice versa) trough the symmetry line.

A third line could be thought to be a diagonal between opposite vertices, but notice that the stars projection won't by symmetric in this case.

So, we only have 2 symmetry lines.

4 0
3 years ago
Graph a line that is perpendicular to y=7/8x+3
FromTheMoon [43]
One such line is y = -8/7x + 3 (The perpendicular slope will be the negative reciprocal of the given slope. You can graph it to make sure.)

Hope this helps!
Brainliest Please!
4 0
3 years ago
ILL MARK BRAINIEST IF YOU DO THIS RIGHT!!!
Arlecino [84]
First let’s find the equation:
a = (42-21)/(26-13)
a= 21/13
y-y1 = a(x-x1)
y-21=(21/13)(x-13)
y-21=(21/13)x - 21
y=(21/13)x-21+21
y=(21/13)x

Now we will find the other point:
X = 2 >> y =(21/13)*2 = 3.23 (not)
X = 3 >> y =(21/13)*3 = 4.84 (not)
X = 4 >> y =(21/13)*4 = 6.46 (not)

X = 5 >> y =(21/13)*5 = 8.07 (yes) find the answer
7 0
2 years ago
I need help solving this problem. <img src="https://tex.z-dn.net/?f=4x%20cos%20%5E%7B-1%7D%20%282x%2B4%29-%20%5Csqrt%7B3-3%20x%5
Gnesinka [82]
Given:
f(x)=4xcos^{-1}(2x+4)-\sqrt{3-3x^2}

Using
\frac{d}{dx}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}
we derive
\frac{d}{dx}4xcos^{-1}(2x+4)
=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}

Similarly, using
\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}
we derive
\frac{d}{dx}(-\sqrt{3-3x^2})
=\frac{3x}{\sqrt{3-3x^2}}

Therefore, the derivative is
f'(x)=\frac{d}{dx}(4xcos^{-1}(2x+4)-\sqrt{3-3x^2})
=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}+\frac{3x}{\sqrt{3-3x^2}}
3 0
3 years ago
Read 2 more answers
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