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Y_Kistochka [10]
3 years ago
11

Can anyone help me with this?

Mathematics
1 answer:
Zina [86]3 years ago
6 0

Answer:

  b = 15°

Step-by-step explanation:

The base angles (c = ∠QRP) of ΔQPR are such that the sum of angles of the triangle is 180°. Since the triangle isosceles, the base angles are equal:

  2c +60 = 180

  2c = 120

  c = 60 . . . . . . degrees; the base angle of ΔQPR

__

Likewise, the base angle (∠SRP) of ΔSPR is ...

  a = (180 -90)/2 = 45 . . . . . degrees

__

The measure of angle b is the difference between the larger and smaller base angles:

  b = c -a = 60° -45°

  b = 15°

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KATRIN_1 [288]
The answer is A. for this, you have to set up a system of equations. the first one will be the area equation. since you know area=length x width, your equation will be LxW=50. the next equation is L=2W, since the length is two times the width. then, plug in 2W for the L in the other equation and you get 2W^2=50. divide by 2 and get W^2=25. square root both sides and you get W=5. plug back into the other equation to find L=10. Then, add the sides of the rectangle for the perimeter. 10+5+10+5=30.
3 0
3 years ago
please help as soon as you see this because people keep answering my questions a day later it’s too late by then‍♂️ & if you
melisa1 [442]

Answer:

A. Cylinder, Shpere, Cone

Step-by-step explanation:

a cylander can be made by rotating a square.

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for a cone, a right triangle.

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7 0
3 years ago
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BabaBlast [244]

\underline{\bf{Given \:equation:-}}

\\ \sf{:}\dashrightarrow ax^2+by+c=0

\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.

\sf We\:know,

\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}

\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}

\underline{\large{\bf Identities\:used:-}}

\boxed{\sf (a+b)^2=a^2+2ab+b^2}

\boxed{\sf (√a)^2=a}

\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}

\boxed{\sf \sqrt{\sqrt{a}}=a}

\underline{\bf Final\: Solution:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}

\bull\sf Apply\: Squares

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}

\bull\sf Put\:values

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}

\bull\sf Simplify

\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}

\underline{\bf More\: simplification:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}

\underline{\Large{\bf Simplified\: Answer:-}}

\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}

5 0
2 years ago
Read 2 more answers
Can anybody help me please I'm stuck with this question.
tekilochka [14]

Answer: 18 units

Step-by-step explanation:since its is a horizontal line x2 - x1

7 -1 = 6

line 2:

since this is a vertical line y2 - y1

7 - 3 = 4

line 3:

since this is a horizontal line x2 - x1

7 - 4 = 3

line 4:

for this we need to use the distance formula which allows us to find the distance making a third point to form a right angle triangle

point 1: (1,3)

point 2: (4,7)

point 3 (new point) : (4,3)

now we can apply the pythogorean thereum (C squared = B squared + A squared) with the following lines.

line 1: (1,3) - (4,7)

line 2: (1,3) - (4,3)

line 3: (4,3) - (4,7)

line 1 squared = line2 squared + line 3 squared

calculate length of line 2 and 3

line 1 squared = (4 - 1) squared + (7 - 3) squared

line 1 squared = 3 squared + 4 squared

line 1 squared = 9 + 16

line 1 squared = 25

root both sides

line 1 = 5

add all the liens together

6 + 4 + 3 + 5 = 18

8 0
3 years ago
Simplify radical sign 0.08^12
vagabundo [1.1K]

Answer:

√0.08¹² = 0.08⁶

4 0
3 years ago
Read 2 more answers
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