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gayaneshka [121]
3 years ago
10

A grasshopper has a length of (5x - 2) inches. A spider has a length of (2x - 1) inches. How much longer is the grasshopper?

Mathematics
1 answer:
shutvik [7]3 years ago
3 0
The grasshopper is 3x-1 inches longer

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Use graph to answer question
slamgirl [31]

Answer:

cos(θ) = 3/5

Step-by-step explanation:

We can think of this situation as a triangle rectangle (you can see it in the image below).

Here, we have a triangle rectangle with an angle θ, such that the adjacent cathetus to θ is 3 units long, and the cathetus opposite to θ is 4 units long.

Here we want to find cos(θ).

You should remember:

cos(θ) = (adjacent cathetus)/(hypotenuse)

We already know that the adjacent cathetus is equal to 3.

And for the hypotenuse, we can use the Pythagorean's theorem, which says that the sum of the squares of the cathetus is equal to the square of the hypotenuse, this is:

3^2 + 4^2 = H^2

We can solve this for H, to get:

H = √( 3^2 + 4^2) = √(9 + 16) = √25 = 5

The hypotenuse is 5 units long.

Then we have:

cos(θ) = (adjacent cathetus)/(hypotenuse)

cos(θ) = 3/5

5 0
3 years ago
What is 1 tenth of 3.0. as a decimal
tangare [24]
Answer=0.3

1/10 of 3.0 is 0.3

3.0*(1/10)=0.3
4 0
3 years ago
Help? I'd Appreciate it!
kondor19780726 [428]

<u>Answer</u>

y = (1/2)x - 1


<u>Explanation</u>

The first step is to get the gradient of the line.

The two points in the line are; (2,0) and (-2, -2).

Gradient = (-2 - 0)/(-2, - 2)

                = -2/-4

                 = 1/2

To get the function we use one of the point (2,0) and a general point (x,y).

1/2 = (y - 0)/(x - 2)

1/2 = y/(x - 2)

(x - 2) = 2y

2y = x - 2

y = (1/2)x - 1

8 0
3 years ago
What is 2a+b
Vinil7 [7]
The answer is -8. Do you need an explanation?
4 0
3 years ago
Read 2 more answers
Pls help meh pls<br> Find the area of the figure. (Sides meet at right angles.)
Vlad [161]
We find the area of the total square which is 8in • 8in=64in^2 . Now we find the area of the small quadrilateral which is 4in•4in=16 in^2.
Now we subtract the area of the small quadrilateral from the big quadrilateral’s area 64in^2 -16in^2 leaving us with the are of 48in^2 of the figure
3 0
3 years ago
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