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snow_tiger [21]
3 years ago
13

Graph the line with the equation y=-x-4

Mathematics
1 answer:
Zielflug [23.3K]3 years ago
4 0

y=-x-4

Here is the picture of the graph:

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Is anyone good at geometry if so can someone help me please ?<br><br> NO LINKS PLEASE
irga5000 [103]

Answer:

d.

I and IV only.

Step-by-step explanation:

5 0
4 years ago
Which pairs of quadrilaterals can be shown to be congruent using
Zielflug [23.3K]

Rigid transformation do not alter the sides and angle measurements of a shape.

The true statements are:

  • <em>quadrilateral 1 and  quadrilateral 2 - Not congruent </em>
  • <em>quadrilateral 1 and  quadrilateral 3 - Not congruent </em>
  • <em>quadrilateral 1 and  quadrilateral 4 - Not congruent </em>
  • <em>quadrilateral 2 and  quadrilateral 3 - Congruent </em>

<em />

From the graph, we have the following highlights

  • Quadrilaterals 2, 3 and 4 are congruent
  • Quadrilateral 1 does not have equal measure with any of the other quadrilaterals.

Hence, the first three statements are not congruent, while the last statement is congruent.

Read more about congruent shapes at:

brainly.com/question/24430586

3 0
3 years ago
HELPPP FOR BRIANLEST!!!!
grandymaker [24]

Answer:

4/13

Step-by-step explanation:

NOT 2010s and NOT spruce wood (4) over total (13)

4/13 is in its reduced form

4 0
2 years ago
Read 2 more answers
Please help out , i’ll mark the brainliest answer
Len [333]

Answer:

Step-by-step explanation:

7) The tangent angles are 90° each.

The angles of a quadrilateral add up to 360°, so ? = 360°-90°-90°-73° = 107°.

8)  Solve this the same way as question 7). ? = 360°-90°-90°-57° = 123°.

Questions 9) and 10) are cut off, so I assume you don't need to know those.

4 0
3 years ago
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Evaluate the double integral. . ∫∫ y sqrt(x^2-y^2) dA, R={(x,y)|0≤y≤x, 0≤x≤1}. R. . Please explain
polet [3.4K]
First we will evaluate: ( substitution: u = x² - y²,  du = - 2 y dy )
\int\limits^x_0 {y \sqrt{ x^{2} - y^{2} } } \, dy= \\   \frac{-1}{2} \int\limits^x_0 { u^{1/2} } \, du  =
=\frac{-1}{3} \sqrt{ (x^{2} - y^{2} ) ^{3} } ( than plug in x and 0 )
=- \frac{1}{3} (  \sqrt{( x^{2} - x^{2}) ^{3}  }  -  \sqrt{ (x^{2} -0 ^{2} ) ^{3} } =
= 1/3 x³ ( then another integration )
1/3\int\limits^1_0 { x^{3} } \, dx = 1/3 (  x^{4}/4)}= 1/3 ( 1 ^{4}/4 - 0^{4} /4 )
= 1/3 * 1/4 = 1/12
4 0
3 years ago
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