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RSB [31]
2 years ago
12

The quotient of a number and 5 is 10

Mathematics
2 answers:
RUDIKE [14]2 years ago
7 0

Answer:

The number could be 50.

Step-by-step explanation:

50/5 = 10

sergiy2304 [10]2 years ago
4 0

Answer:

50

Step-by-step explanation:

50 divided by 5 = 10

5×10=50

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Identify the similar triangles. Find each measure.​
boyakko [2]

Answer:

Example: these two triangles are similar: If two of their angles are equal, then the third angle must also be equal, because angles of a triangle always add to make 180°. So AA could also be called AAA (because when two angles are equal, all three angles must be equal)

5 0
3 years ago
X+ k = w + v for x ? Some one please explain this for me
Troyanec [42]

Answer:

x = w + v - k

Step-by-step explanation:

For this question, we would have to solve for the variable x.

Solve for x:

x + k = w + v

To solve it, we would have to get "x" by itself.

Subtract k from both sides.

x = w + v - k

Since "x" is by itself, we know what the value of "x" is.

Therefore, your answer is x = w + v - k

3 0
3 years ago
How to order -14.4, -14, -14 1/5 from least to greatest
ElenaW [278]
Think of  thermometer the negative degrees are the coldest so are the lowest on the vertical column

first convert -14 1/5 to decimals  =  -14.2

so the order is 

-14.4 , -14 1/5 , -14.

8 0
3 years ago
I WILL GIVE BRAINLIEST!
lorasvet [3.4K]
Q is 124 and 56 because it has to add up to 180 and sense r is also 56 degrees that means S is 68 degrees because the sum of each angle has to add up to 180 degrees
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Read 2 more answers
Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o
Brrunno [24]

Answer:

AB = \sqrt{a^2 + b^2-2abCos\ C}

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

Sin\ C = \frac{h}{a}

Make h the subject:

h = aSin\ C

Also, in BOC (Using Pythagoras)

a^2 = h^2 + x^2

Make x^2 the subject

x^2 = a^2 - h^2

Substitute aSin\ C for h

x^2 = a^2 - h^2 becomes

x^2 = a^2 - (aSin\ C)^2

x^2 = a^2 - a^2Sin^2\ C

Factorize

x^2 = a^2 (1 - Sin^2\ C)

In trigonometry:

Cos^2C = 1-Sin^2C

So, we have that:

x^2 = a^2 Cos^2\ C

Take square roots of both sides

x= aCos\ C

In triangle BOA, applying Pythagoras theorem, we have that:

AB^2 = h^2 + (b-x)^2

Open bracket

AB^2 = h^2 + b^2-2bx+x^2

Substitute x= aCos\ C and h = aSin\ C in AB^2 = h^2 + b^2-2bx+x^2

AB^2 = h^2 + b^2-2bx+x^2

AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2

Open Bracket

AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C

Reorder

AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C

Factorize:

AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C

In trigonometry:

Sin^2C + Cos^2 = 1

So, we have that:

AB^2 = a^2 * 1 + b^2-2abCos\ C

AB^2 = a^2 + b^2-2abCos\ C

Take square roots of both sides

AB = \sqrt{a^2 + b^2-2abCos\ C}

6 0
3 years ago
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