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pashok25 [27]
3 years ago
10

What is the slope of the line in the graph?

Mathematics
2 answers:
mr Goodwill [35]3 years ago
8 0

Answer:

2

Step-by-step explanation:

Slope is Y/x. In this case, you could do 4 /2 which would give you a slope of 2/1=2.

Sergeeva-Olga [200]3 years ago
3 0

Answer:

1

Step-by-step explanation:

rise 1 over run 1 1/1=1

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B. The area of another sheet of paper is 200 square inches. Write an equation that
nata0808 [166]

Answer:

200/3n

Step-by-step explanation:

It is a strangely worded question but you are dividing the paper in 3's n times so 3n will be part of the answer and you are dividing 200. So, the answer is 200/3n.

6 0
2 years ago
I need help finding if it’s either SOH, CAH or TOA and the indicated side of the triangle, thank you!
Brilliant_brown [7]

Answer:

\huge\boxed{IJ \approx 2.01}

Step-by-step explanation:

In order to find the side mentioned (IJ), we need to use SOH CAH TOA.

SOH CAH TOA is an acronym to help us remember what sin, cos, and tan mean. It stands for:

Sin = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tan = Opposite / Adjacent

Since we know the measure of angle K (42) and we know one of the sides, we can use this to find the missing length.

Since the side given to us is the hypotenuse, and we're looking for the side opposite of the angle (IJ), the only possible one to use would be SIN as it includes Opposite and Hypotenuse.

Our equation is now this: \text{sin(42)} = \frac{x}{3}

Let's now solve for x.

  • \text{sin(42)} = \frac{x}{3}
  • 3 \cdot \text{sin(42)} = x
  • \text{sin(42)} \approx 0.67
  • 3 \cdot 0.67 \approx 2.01

Therefore, the length of IJ will be around 2.01.

Hope this helped!

4 0
3 years ago
Please explain and ANSWER this question for BRAINLIEST!
pogonyaev

Answer:

155 in^{2}

Step-by-step explanation:

Hope this work helps

4 0
2 years ago
I need help on this it’s about trig identities
Cerrena [4.2K]

Answer:

The answer to your question is:

Step-by-step explanation:

1.-

\frac{1 + sin\alpha }{cos\alpha } + \frac{cos\alpha }{1 + sin\alpha } = 2 sec\alpha

\frac{(1 + sin\alpha)^{2} + cos^{2} \alpha  }{cos\alpha (1 + sin\alpha) }

\frac{1  + 2sin\alpha + sin^{2} \alpha+ cos^{2} \alpha  }{cos\alpha + sin\alphacos\alpha  }

\frac{2 + 2sin\alpha }{cos\alpha+ sin\alpha cos\alpha  }

\frac{2(1 + sin\alpha) }{cos\alpha(1 + sin\alpha ) }

\frac{2}{cos\alpha }

2sec\alpha

2.-

   sec²x - tanxsecx

\frac{1}{cos^{2}x } - \frac{sinx}{cosx} \frac{1}{cosx}

\frac{1}{cos^{2} x} - \frac{sinx}{cos^{2}x}

\frac{1 - sinx}{cos^{2}x }

\frac{1 - sinx}{1 - sin^{2}x }

\frac{1 - sinx}{(1 - sinx)(1 + sinx)}

\frac{1}{1 + sinx}

3 0
3 years ago
Units are squared when dealing with area but not when dealing with perimeter
Mashcka [7]
The answer is true I believe 
7 0
4 years ago
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