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GREYUIT [131]
2 years ago
9

Using a compass and a straightedge, a student constructed a triangle in which XY is one of the sides.

Mathematics
1 answer:
Savatey [412]2 years ago
8 0

The length of the compass so that ΔXYZ is isosceles and not equilateral could be;

Option C; between ¹/₂XY and XY

Option E; greater than XY

We are told that;

Student has drawn a line XY

Now the student opened the compass with a set length to draw two intersecting arcs with X and Y as centers.

Now, for these two arcs to meet, the compass has to be opened to a length that is more than half of segment XY.

Secondly, since the triangle is not equilateral, it means the compass cannot be opened to length AB.

The next step would be t open the compass to a length slightly bigger or slightly smaller than AB.

Finally, the options that apply are Options C and E.

Read more about construction of Isosceles triangles at; brainly.com/question/2520169

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What is: 7m+(-5)=180
Virty [35]

Answer:

m = 7 / 185 or 26.42857

Step-by-step explanation:

7m - 5 = 180 add 5

7m = 185 divide 7

m = 7 / 185 or 26.42857 (and so on; just estimate it)

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3 years ago
64,32, 16..1 missing number
kaheart [24]

Answer:

48

Step-by-step explanation:

1..16..32....48...64

8 0
3 years ago
Read 2 more answers
He range R and the maximum height H of a projectile fired at an inclination θ to the horizontal with initial speed v0 are given
ziro4ka [17]

Answer:

<h2>A. R=1098.7ft</h2><h2>B. H=274.7ft</h2>

Step-by-step explanation:

<em>"The question is not complete, here is the complete question</em>

<em> The range R and the maximum height H of a projectile fired at an inclination θ to the horizontal with initial speed v0 are given by the formulas​ below, where g≈32.2 feet per second per second is the acceleration due to gravity. Complete parts A and B.</em>

<em />R=\frac{2v_o^2sin \theta cos \theta}{g}<em />

<em />H=\frac{v_o^2sin^2 \theta}{2g}<em />

<em>A. Find the range R if the projectile is fired at an angle 45° to the horizontal with an initial speed of 190 ft per second</em>

<em>B. Find the maximum height H  if the projectile is fired at angle 45° to the horizontal with an initial speed of 190 ft per second"</em>

<em />

<u>Solution</u>

A. given the expression for Range R

substitute v= 190 ft/sec and ∅=45°

R=\frac{2(190)^2sin(45) cos(45)}{32.2}\\\\R=\frac{72200*0.7*0.7}{32.2}\\\\R=\frac{35378}{32.2} \\\\R=1098.7ft

B. . given the expression for maximum height

substitute v= 190 ft/sec and ∅=45°

H=\frac{v_o^2sin^2 \theta}{2g}\\\\H=\frac{190*190*0.7*0.7}{2*32.2}\\\\H=\frac{17689}{64.4}\\\\H=274.7ft

5 0
3 years ago
4. Solve -5x2 + 3x = 11<br> Quadratic
exis [7]

Answer:

X=7

Step-by-step explanation:

-5×2+3x=11

(-5×2)+3x=11

-10+3x=11

3x=11+10

3x=21

x=21÷3

x=7

4 0
3 years ago
HELP MEE PLEASE HELPPPP
MrRissso [65]
I would feel as if the answer would b 7
6 0
3 years ago
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