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pantera1 [17]
2 years ago
14

Solve for x. Please look at the picture and answer it. Thank you.

Mathematics
2 answers:
Tatiana [17]2 years ago
7 0
<h3>Answer :- </h3>

  • x = 24

<h3>Solution :- </h3>

  • 2x + 42 = 90
  • 2x = 90 - 42
  • 2x = 48
  • x = 48/2
  • x = 24

<em>Hope</em><em> it</em><em> helps</em><em> </em><em>~</em>

mr Goodwill [35]2 years ago
3 0

Answer:

x=24

Step-by-step explanation:

i cant quite explain how i did it

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Graph triangle RST with vertices R(3, 7), S(-5, -2), and T(3, -5) and its image after a reflection over x = -3.​
kirill115 [55]

Given:

The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).

To find:

The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.

Solution:

If a figure reflected across the line x=a, then

(x,y)\to (-(x-a)+a,y)

(x,y)\to (-x+a+a,y)

(x,y)\to (2a-x,y)

The triangle after a reflection over x = -3. So, the rule of reflection is

(x,y)\to (2(-3)-x,y)

(x,y)\to (-6-x,y)

The vertices of triangle after reflection are

R(3,7)\to R'(-6-3,7)

R(3,7)\to R'(-9,7)

Similarly,

S(-5,-2)\to S'(-6-(-5),-2)

S(-5,-2)\to S'(-6+5,-2)

S(-5,-2)\to S'(-1,-2)

And,

T(3,-5)\to T'(-6-3,-5)

T(3,-5)\to T'(-9,-5)

Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).

3 0
3 years ago
PLEASE HELP!!! YOU GET 10 BRAINLIST!!!! PLEASE SHOW ALL OF YOUR WORK/HOW YOU GOT THE ANSWER!!!
Mrrafil [7]
The answer is x< 9 so yeah

5 0
3 years ago
Read 2 more answers
#11. A) A person on the top of a burning building looks down at a firetruck at an angle of 35°. If
tamaranim1 [39]

Answer:

Location of a Fire Fire tower A is 30 kilometers due west of fire tower B. A fire is spotted from the towers, and the bearings from A and B are N76°W and N56°W respectively (see figure). Find the distance d of the fire from the line segment AB.

Hope this helped :)

Step-by-step explanation:

3 0
2 years ago
Use the chain rule calculate dw/dr , dw/ds and dw/dt
Liono4ka [1.6K]

Answer:

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Step-by-step explanation:

We proceed to derive each expression by rule of chain. Let be w = \ln (x+2\cdot y + 3\cdot z), x = r^{2}+t^{2}, y = s^{2}-t^{2} and z = r^{2}+s^{2}:

\frac{dw}{dr} = \frac{\frac{dx}{dr}+2\cdot \frac{dy}{dr} +3\cdot \frac{dz}{dr}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{dr} = 2\cdot r

\frac{dy}{dr} = 0

\frac{dz}{dr} = 2\cdot r

\frac{dw}{dr} = \frac{8\cdot r}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

\frac{dw}{dr} = \frac{8\cdot r}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (1)

\frac{dw}{ds} = \frac{\frac{dx}{ds}+2\cdot \frac{dy}{ds} +3\cdot \frac{dz}{ds}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{ds} = 0

\frac{dy}{ds} = 2\cdot s

\frac{dz}{ds} = 2\cdot s

\frac{dw}{ds} = \frac{10\cdot s}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

\frac{dw}{ds} = \frac{10\cdot s}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (2)

\frac{dw}{dt} = \frac{\frac{dx}{dt}+2\cdot \frac{dy}{dt} +3\cdot \frac{dz}{dt}  }{x+2\cdot y + 3\cdot z}

\frac{dx}{dt} = 2\cdot t

\frac{dy}{dt} = -2\cdot t

\frac{dz}{dt} = 0

\frac{dw}{dt} = -\frac{2\cdot t}{(r^{2}+t^{2})+2\cdot (s^{2}-t^{2})+3\cdot (r^{2}+s^{2})}

\frac{dw}{dt} = -\frac{2\cdot t}{4\cdot r^{2}-t^{2}+5\cdot s^{2}} (3)

7 0
3 years ago
Pls the real answer
Alexxandr [17]

Answer:

79.25 in.

Step-by-step explanation:

multiply 11 and 7.25 to get the actual size

3 0
3 years ago
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