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3 years ago

Will make brainliest!

Mathematics

2 answers:

mestny [16]3 years ago

8 0

Answer:

∠4 and ∠12

A is correct

Step-by-step explanation:

Parallel lines r and s are cut by two transversals, parallel lines t and u

r || s and t || u

We are four parallel line. Two parallel line cuts two another parallel line.

Angle 8 is form by intersection of s and t line.

Corresponding angle: When two parallel lines are crossed by transversal line, the angles at matching corners are called corresponding angles.

For angle 8, t || u with s is transveral line.

Thus, ∠8 = ∠12 (By definition of corresponding angle)

For angle 8, s || r with t is transveral line.

Thus, ∠8 = ∠4 (By definition of corresponding angle)

Hence, ∠4 and ∠12 are corresponding angle of ∠8

Ahat [919]3 years ago

7 0

<h2>Answer:</h2><h2>A </h2>

Step-by-step explanation:

You might be interested in

Five weeks after the introduction of a new toy, the manufacturer found that the demand for the toy was modeled by f(x) = x^2 +5x

mamaluj [8]

Answer:

76 weeks

Step-by-step explanation:

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7 0

3 years ago

Please help me I don’t have that much time

Salsk061 [2.6K]

Answer:

$a \geq 18$

Step-by-step explanation:

4 0

3 years ago

A self-serve frozen yogurt restaurant charges its customers according to the weight of their filled frozen yogurt cup. So, the a

jeyben [28]

If the price of the yogurt is directly proportional to the weight of the purchase, then it can be expressed as:

Price of yogurt = weight of purchase

With the given above, it can be further expressed as:
$5.04 = 12 oz

These can be expressed in ratio form, which is:

5.04/12

Thus, if the cup weighs 8 oz and we don't know how much it costs, then the ratio should be:

n = 8 oz, or in fractions:
n/8

After this, simply equate the two and cross multiply.
5.04 / 12 = n/8
5.04(8) = 12n
40.32 = 12n

Then solve for n:
40.32 /12 = 12n/12
n = 3.36.

Thus, the 8 oz. cup of yogurt costs $3.36.

4 0

3 years ago

The perimeter of an ICU ward, that is rectangular in shape, is 258 feet. The width is 37 feet less than the length. Find the dim

zubka84 [21]

The dimensions of the rectangle is 83 feet x 46 feet

What is the Perimeter of a Rectangle?

The perimeter P of a rectangle is given by the formula, P = 2 ( L + W) , where L is the length and W is the width of the rectangle.

Perimeter P = 2 ( Length + Width )

Given data ,

The perimeter P of the rectangle = 258 feet

Let the length of the rectangle be = L

Let the width of the rectangle be = W

And , The width is 37 feet less than the length. ,

So , W = L - 37

Perimeter P = 2 ( Length + Width )

Substituting the values of L and W in the equation , we get

Perimeter P = 2 ( L + L - 37 )

2 ( L + L - 37 ) = 258

( L + L - 37 ) = 129

2L - 37 = 129

Adding 37 on both sides , we get

2L = 166

Divide by 2 on both sides , we get

Length L of the rectangle = 83 feet

Substituting the value of L in equation ,we get

W = L - 37

Width W of the rectangle = 46 feet

Therefore , Length = 83 feet and Width = 46 feet

Hence , the dimensions of the rectangle is 83 feet x 46 feet

To learn more about perimeter of rectangle click :

brainly.com/question/15725265

#SPJ1

3 0

2 years ago

Please help me to prove this!

Sophie [7]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: A + B = C → A = C - B

→ B = C - A

Use the Double Angle Identity: cos 2A = 2 cos² A - 1

→ (cos 2A + 1)/2 = cos² A

Use Sum to Product Identity: cos A + cos B = 2 cos [(A + B)/2] · 2 cos [(A - B)/2]

Use Even/Odd Identity: cos (-A) = cos (A)

<u>Proof LHS → RHS:</u>

LHS: cos² A + cos² B + cos² C

$\text{Double Angle:}\qquad \dfrac{\cos 2A+1}{2}+\dfrac{\cos 2B+1}{2}+\cos^2 C\\\\\\.\qquad \qquad \qquad =\dfrac{1}{2}\bigg(2+\cos 2A+\cos 2B\bigg)+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\dfrac{1}{2}\bigg(\cos 2A+\cos 2B\bigg)+\cos^2 C$

$\text{Sum to Product:}\quad 1+\dfrac{1}{2}\bigg[2\cos \bigg(\dfrac{2A+2B}{2}\bigg)\cdot \cos \bigg(\dfrac{2A-2B}{2}\bigg)\bigg]+\cos^2 C\\\\\\.\qquad \qquad \qquad =1+\cos (A+B)\cdot \cos (A-B)+\cos^2 C$

$\text{Given:}\qquad \qquad 1+\cos C\cdot \cos (A-B)+\cos^2C$

$\text{Factor:}\qquad \qquad 1+\cos C[\cos (A-B)+\cos C]$

$\text{Sum to Product:}\quad 1+\cos C\bigg[2\cos \bigg(\dfrac{A-B+C}{2}\bigg)\cdot \cos \bigg(\dfrac{A-B-C}{2}\bigg)\bigg]\\\\\\.\qquad \qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+(C-B)}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-(C-A)}{2}\bigg)$

$\text{Given:}\qquad \qquad =1+2\cos C\cdot \cos \bigg(\dfrac{A+A}{2}\bigg)\cdot \cos \bigg(\dfrac{-B-B}{2}\bigg)\\\\\\.\qquad \qquad \qquad =1+2\cos C \cdot \cos A\cdot \cos (-B)$

$\text{Even/Odd:}\qquad \qquad 1+2\cos C \cdot \cos A\cdot \cos B\\\\\\.\qquad \qquad \qquad \quad =1+2\cos A \cdot \cos B\cdot \cos C$

LHS = RHS: 1 + 2 cos A · cos B · cos C = 1 + 2 cos A · cos B · cos C $\checkmark$

5 0

3 years ago