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ddd [48]
3 years ago
13

I will give brainliest to the first one that does it correctly

Mathematics
1 answer:
Effectus [21]3 years ago
8 0

Answer:

180-138

=42, is the answer

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I really need help with this problem steps
melamori03 [73]
Solve for y by setting one equation =to X after that you can substitute the equation into the other one let's set the first one in X= FORM
X-7y=10
-2x+14y=-20

Add 7y to both sides of the first equation to let x stand alone
X=7y+10
Now you can substitute x on the second equation
-2 (7y+10)+14y=-20
Distribute
-14y-20+14y=-20
Add and simplify
Us cancel out =0
-20=-20
0=0
They are the same equations you can also divide the second equation by -2 which would make it look like this
X-7y=10
-2 (x-7y=10)
Let me know if I have answered your question
5 0
3 years ago
This question has two parts. First, answer Part A. Then, answer Part B.
masya89 [10]

Answer:

$10.12

Step-by-step explanation:

Step one:

given data

we are told that the cost of binoculars is $155.75

and also the sales tax is 7%

Step two:

We want to first solve for the sales tax which is 7% of  $155.75

=7/100* 155.75

=0.070*155.75

=10.12

Cassie will pay a tax of $10.12

Also Cassie will pay a total amount of

= 155.75+10.12

=$165.87

3 0
2 years ago
Choose the equation of the horizontal line that passes through the point (2, 3).
Marizza181 [45]
The equation for the horizontal line is
y=3
5 0
3 years ago
Which describes the solution of the inequality y > −15? Select all the points that are part of the solution.
Svetllana [295]

Step-by-step explanation:

y > -15

y is bigger than -15

= -14 , -13 , -12 , -11 , -10 , etc . . .

8 0
3 years ago
(1+cos2x)/(1-cos2x) = cot^2x
sesenic [268]

We will turn the left side into the right side.

\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x

Use the identity:

\cos 2x = \cos^2 x - \sin^2 x

\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x

\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x

Now use the identity

\sin^2 x + \cos^2 x = 1 solved for sin^2 x and for cos^2 x.

\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x

\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x

\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x

\cot^2 x = \cot^2 x


8 0
3 years ago
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