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jonny [76]
3 years ago
7

Are the two triangles similar?

Mathematics
2 answers:
Marysya12 [62]3 years ago
5 0

Answer:

I am thinking c, but I do not know

Step-by-step explanation:

I am sure but if not c then b

sergiy2304 [10]3 years ago
3 0
B ,because is a vertical angle
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5x 2 +5x x 2 −1 ​ What is an equivalent expression in lowest terms?
arsen [322]

Answer:

Step-by-step explanation:

Explanation:

Start by writing out your starting expression

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(

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(

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−

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x

−

2

)

(

x

−

2

)

⋅

(

x

+

7

)

This will get you

x

2

−

5

−

(

x

+

3

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(

x

−

2

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(

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5 0
2 years ago
To find out how many wolves lived in the valley, scientists put tracking devices on 68 wolves. Later that month, the scientists
morpeh [17]

Answer:

b. 2,924

Step-by-step explanation:

Write and solve a proportion.

24 / 1032 = 68 / x

24x = 70176

x = 2924

5 0
3 years ago
1.2.8 -0.9+y<br> Your answer
Annette [7]
0.06+y simplify the expression
8 0
3 years ago
Find the value of n so that the expression is a perfect square trinomial and then factor the trinomial.
Anna007 [38]

The correct expression is (a) n = 100; (x + 10)^2

<h3>How to determine the value of n?</h3>

The expression is given as:

x^2 + 20x + n

Take the coefficient of x

k = 20

Divide by 2

k/2= 10

Square both sides

(k/2)^2 = 100

The above value represents the value of n

i.e.

n = 100

So, we have:

x^2 + 20x + 100

Expand

x^2 + 10x + 10x + 100

Factorize

x(x + 10) + 10(x + 10)

Factor out x + 10

(x + 10)^2

Hence, the correct expression is (a) n = 100; (x + 10)^2

Read more about trinomial at:

brainly.com/question/1538726

#SPJ1

7 0
2 years ago
Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
sleet_krkn [62]

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

8 0
3 years ago
Read 2 more answers
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