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Ilya [14]
3 years ago
13

Write an inequality, in slope-intercept form, for the graph below. If necessary.

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
5 0

Answer:

y = 2x + -2

Step-by-step explanation:

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Multi step equation <br><br> —20=—4x—6x<br> Answer please
Vikki [24]

Answer:

Solve for x by simplifying both sides of the equation, then isolating the variable  

So your Answer is x = 2 Hope this helps :)

Step-by-step explanation:


7 0
4 years ago
A triangle has sides with lengths of 6 inches, 13 inches, and 17 inches. Is it a right triangle?
Tems11 [23]

Answer:

Nope

Step-by-step explanation:

To now if a triangle is a right angle, use Pythagoras Theorem,

a^2 + b^2 = c^2

C usually be the longest side of the triangle, C = 17

Then also let a and b be 6 and 13 respectively.

6^2 + 13^2 =205

C^2 = 205

C = sqrt(205) = 14.32

therefore no this is not a right angle triangle

4 0
3 years ago
Find the ratio to its lowest 9cm to 36cm​
abruzzese [7]

Answer:

1:4 or 1/4

Step-by-step explanation:

9 cm : 36 cm

They both can be divided by 9 so,

9 ÷ 9 : 36 ÷ 9

= 1 : 4

Hope this helps

3 0
3 years ago
Read 2 more answers
Bye guys take some points its my last day
Rzqust [24]

Answer:

B. 2

Step-by-step explanation:

5 0
3 years ago
A radar operator on a ship discovers a large sunken vessel lying parallel to the ocean surface, 120 m directly below the ship. T
valentinak56 [21]

Answer:

The length of the sunken vessel, to the nearest tenth is 200 m

Step-by-step explanation:

The given information are;

The depth of the sunken vessel below the ship = 120 m

The angle of depression to the front of the sunken vessel = 55°

The angle of depression to the back of the sunken vessel = 46°

Therefore, we have;

The length, A, of the from directly below the ship to the front of the sunken vessel given as follows;

Tan(55°) = ((120 m)/A)

A = 120 / Tan(55°) ≈ 84.025 m

The length, B, of the from directly below the ship to the back of the sunken vessel given as follows;

tan(46°) = ((120 m)/B)

B = 120 / tan(46°) ≈ 115.88 m

The length, L,  of the sunken vessel = A + B

L = A + B = 120 / Tan(55°) + 120 / tan(46°) ≈ 84.025 m + 115.88 m ≈  199.91 m

∴ The length, L, of the sunken vessel, to the nearest tenth = 200 m.

3 0
3 years ago
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