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3 years ago

In the rhombus, m<1=8y-6. Find the value of y. Please help!!

Mathematics

2 answers:

agasfer [191]3 years ago

6 0

Answer: 12

The value of y= 12.

Step-by-step explanation:

We know that the diagonal of rhombus are perpendicular bisector of each other.

i.e. the angle made at the intersection of diagonal is 90 degrees.

Thus , for the given figure m∠1 = 90°

Since it is given that m∠1 = 8y-6

Thus , 8y-6= 90

⇒8y=96 [Adding 6 both sides]

⇒ y = 12 [Dividing both sides by 8]

Hence, the value of y= 12.

tia_tia [17]3 years ago

5 0

I think the answer is A.

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3 years ago

HELP! A semi circle of radius 6 is centered at the origin as shown. A rectangle has two of its vertices at (5,0) and (-5,0) and

Ugo [173]

The Area of rectangle is " $30 \ unit^2$ " and the equation of the semi circle is " $y = \sqrt{36-x^2}$ ".

According to the question,

The vertices of rectangle,

(5, 0) and (-5, 0)

Length,

l = 10 unit

Breadth,

b = 3 unit

Radius of semi circle,

r = 6

Centre of origin,

(0, 0)

As we know,

→ The Area of rectangle is:

= $Length\times Breadth$

= $10\times 3$

= $30 \ unit^2$

and,

→ The Equation of semi circle is,

$y = \sqrt{r^2-x^2}$

by substituting the values, we get

$=\sqrt{(6)^2-x^2}$

$= \sqrt{36-x^2}$

Thus the above is the correct answers.

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brainly.com/question/14383947

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2 years ago

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Artist 52 [7]

Answer:

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2 years ago

2. In the circle shown, secants PE and PF have been drawn such that mCE  82 , mDF  98 and the ratio 3. In circle O, tangent

Andrew [12]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The arc PR is $\theta _{PR} = 84$

Step-by-step explanation:

A descriptive diagram of the diagram given in the question is shown on the first uploaded image

In the circle tangent $\= {M \= Q}$ , secant $PE$ and chord $PF$ are drawn

given that $\theta _ {MPQ} = 102^o$ , $\theta _{M} = 60^o$

The objective is to determine PR

From the diagram we see that MQ is a straight which implies that the the angle

$\theta_{NPM} = 180 -102$

$\theta_{NPM} = 78^o$

Now looking at triangle NMP we see that

$\theta _{MNP} = 180 -[ \theta_{NMP} + \theta _{NPM}]$

$\theta _{MNP} = 180 - [60 + 78 ]$

$\theta _{MNP} = 42 ^o$

Now the measure of an inscribed angle is half the measure of its arc intercepted, this statement is the inscribed angle theorem

So with the knowledge

Then

$\theta_{RNP} = \frac{1}{2} \theta_{PR}$

Looking at the diagram we see that

$\theta _{MNP} = \theta_{RNP} = 42 ^o$

$42 = \frac{1}{2} \theta _{PR}$

$\theta _{PR} = 2 * 42$

$\theta _{PR} = 84$

7 0

2 years ago