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1 year ago

Given that ABCD is a rhombus. Prove ∠1≅∠2.

Mathematics

1 answer:

natka813 [3]1 year ago

5 0

Answer:

1) ABCD is a rhombus

2) All sides of a rhombus are equal.

4) SSS

5) CPCTC

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Chelsey wants to join a fitness club. The fitness club charges an initial membership fee of $55 and a monthly fee of 19.50. she

Naddik [55]

Answer:

Chelsey can be a member of the fitness club for 10 months.

Step-by-step explanation:

Given:

Chelsey wants to join a fitness club.

The fitness club charges an initial membership fee of $55 and a monthly fee of $19.50.

She has $250 to spend on a membership at the fitness club.

Now, to find the number of months Chelsey can be a member of the fitness club.

Let the number of months Chelsey can be a member of the fitness club be $x.$

Monthly fee = $19.50.

Membership fee = $55.

Now, to get the number of months Chelsey can be a member of the fitness club we write and solve an equation:

$55+19.50(x)=250.$

$55+19.50\times x=250\\\\55+19.50x=250$

Subtracting both sides by 55 we get:

$19.50x=195$

Dividing both sides by 19.50 we get:

$x=10.$

Therefore, Chelsey can be a member of the fitness club for 10 months.

5 0

3 years ago

Can anybody help me with this

kondaur [170]

Answer:

c ≈ 9.1 m

Step-by-step explanation:

Using Pythagoras' identity in the right triangle.

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides , that is

c² = 4.1² + 8.1² = 16.81 + 65.61 = 82.42 ( take square root of both sides )

c = $\sqrt{82.42 }$ ≈ 9.1 m ( to the nearest tenth )

5 0

1 year ago

D -e/3d + e= e, for d

navik [9.2K]

Answer:

$\large\boxed{d=\dfrac{e^2+e}{1-3e}}$

Step-by-step explanation:

$\dfrac{d-e}{3d+e}=e\\\\\dfrac{d-e}{3d+e}=\dfrac{e}{1}\qquad\text{cross multiply}\\\\(1)(d-e)=(e)(3d+e)\qquad\text{use the distributive property}\\\\d-e=3de+e^2\qquad\text{add}\ e\ \text{to both sides}\\\\d=3de+e+e^2\qquad\text{subtract}\ 3de\ \text{from both sides}\\\\d-3de=e^2+e\qquad\text{distribute}\\\\d(1-3e)=e^2+e\qquad\text{divide both sides by}\ (1-3e)\\\\d=\dfrac{e^2+e}{1-3e}$