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Serga [27]
3 years ago
12

Solve Diego's equation: x + 35 =180 x =

Mathematics
2 answers:
Irina18 [472]3 years ago
3 0

Answer:

x=145

Step-by-step explanation:

simply subtract 35 from both sides, so you can isolate the variable. once done, you will get that x=145

Bingel [31]3 years ago
3 0

Answer:

The answer is 215.

Step-by-step explanation:

When you add it, it equals 215

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Someone please help me with this geometry proof question ASAP!
vivado [14]

Answer:

if ∠Q=∠R

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now we have two triangle : QPS and RSP

in which angle : QPS=RPS ( PS is angular bisector)

and we have angle Q and angle R are equal

and PS is common side

this makes triangle QPS and RSP are congruent by AAS(angle angle side)

which also means that pQ=PR ( proved)

8 0
3 years ago
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2 years ago
Two ships leave a harbor together, traveling on courses that have an angle of 135°40' between them. If they each travel 402 mile
mario62 [17]

Answer:

Therefore they are 734.106 miles apart.

Step-by-step explanation:

Given that ,

Two ships have a harbor together. The angle between two ships  is  135°40'. Each of two ships travel 402 miles.

It forms a isosceles triangle whose two sides are 402 miles and one angle is 135°40'. Since it is isosceles triangle then other two angles of the triangle is equal.

Let ∠B= 135°40', and AB = 402 miles , BC =  402 miles

Then the distance between the ships = AC

We know

The sum of all angles = 180°

⇒∠A+∠B+∠C=180°

⇒∠A+135°40'+∠C=180°

⇒2∠A= 180°- 135°40'      [ since ∠A=∠C]

⇒2∠A=44°60'

⇒∠A= 22°30'

Again we know that,

\frac{AB}{sin\angle C}=\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Taking last two ratio,

\frac{BC}{sin \angle A}=\frac{AC}{sin \angle B}

Putting the value of BC , AC ,∠A,∠B

\frac{402}{sin 22^\circ30'}=\frac{AC}{sin 135^\circ40'}

\Rightarrow AC=\frac{402 \times sin135^\circ40'}{sin 22^\circ30'}

         ≈734.106 miles

Therefore they are 734.106 miles apart.

3 0
3 years ago
Can someone help with this pls it’s due today
KiRa [710]
D, hope this is right! :)
7 0
3 years ago
The weight of an object on the moon varies directly as its weight on Earth. With all of his gear on, Neil Armstrong weighed 360
babymother [125]

amistre64 Medals 0
e = k.m
360 = k.60
k = 360/60 = 36/6 = 6
108 = 6.m
108/6 = m = 18 
8 0
3 years ago
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