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3 years ago

1. Given that ABCD is a rhombus with ∠DBC=49∘, what is ∠DAB?

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

5 0

Answer:

1). Option (A)

2). Option (D)

Step-by-step explanation:

1). We will use two properties of the rhombus,

- Diagonals of a rhombus are perpendicular and bisect each other.

- Diagonals of a rhombus bisect the opposite angles.

From the figure attached,

m∠DBC = 49° [Given]

m∠DBA = 49° [m∠DBC = m∠DBA]

In ΔAOB,

m∠AOB + m∠OBA + m∠BAO = 180° [Triangle sum theorem]

90°+ 49° + m∠BAO = 180°

m∠BAO = 41°

Since, m∠DAB = 2(m∠BAO)

= 2(41°)

= 82°

Option (A) will be the answer.

2). From the figure attached,

m∠JKF = (3y + 6)°

m∠KFG = (7y - 14)°

Since, m∠JKF ≅ m∠KFG

(3y + 6) = (7y - 14)

7y - 3y = 6 + 14

4y = 20 ⇒ y = 5

Option (D) will be the answer.

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allochka39001 [22]

Answer:

P ( 1.2 < X < 2.1 ) = 0.3

Step-by-step explanation:

Given:

Uniform distribution over interval (0,3) can be modeled by a probability density function f(x)

f(x) = 1 / (b - a)

Where a < x < b is the domain at which function is defined:

f(x) = 1 / (3) = 1 / 3

Where, X - U ( u , δ )

u = ( a + b ) / 2 = (0 +3) / 2 = 1.5

δ = ( b - a ) / sqrt (12) = (3 - 0) / sqrt (12) = 0.866

Hence,

X - U ( 1.5 , 0.866 )

There-fore calculating P ( 1.2 < X < 2.1 ):

$P ( 1.2 < X < 2.1 ) = \int\limits^b_a{f(x)} \, dx$

Where, a = 1.2 and b = 2.1

P ( 1.2 < X < 2.1 ) = x / 3 |

P ( 1.2 < X < 2.1 ) = 2.1 /3 - 1.2 / 3 = 0.3

Answer: P ( 1.2 < X < 2.1 ) = 0.3

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3 years ago

Please help as soon as possible

alexira [117]

I think that there’s around 16 squares.

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3 years ago

In parallelogram QRST QS RT Is QRST a rectangle ?

fomenos

Answer:

The correct option is A.

Step-by-step explanation:

It is given that the QRST is a parallelogram and the diagonals QS and RT are congruent.

According to the property of parallelogram if the length of diagonals are same, then the parallelogram is rectangle. We can also prove it.

Let QRST be an parallelogram and QS=RT.

In triangles QRT and RQS.

$RT=QS$ (Diagonals)

$QR=QR$ (Common side)

$QT=RS$ (Opposite sides of the parallelogram are same)

So by SSS rule of congruence, triangle QRT and triangle RQS are congruent.

By CPCT,

$\angle R=\angle Q$

Since the sum of two consecutive angles of a parallelogram is 180 degree.

$\angle R+\angle Q=180$

$2\angle R=180$

$\angle R=90$

Therefore the measure of angle R and angle Q are 90 degree.

Similarly in triangle QST and RTS,

$RT=QS$ (Diagonals)

$ST=ST$ (Common side)

$QT=RS$ (Opposite sides of the parallelogram are same)

So by SSS rule of congruence, triangle QST and RTS are congruent.

By CPCT,

$\angle S=\angle T$

Since the sum of two consecutive angles of a parallelogram is 180 degree.

Therefore the measure of angle S and angle T are 90 degree.

Since the measure of all interior angles of QRST are 90 degree, therefore the parallelogram QRST is a rectangle.

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