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3 years ago

9

What new development happened with some of the Soviet nuclear missiles in Cuba on October 25, 1962

1 answer:

Tanzania [10]3 years ago

5 0

The aircraft carrier USS Essex and the destroyer USS Gearing attempted to intercept the Soviet tanker Bucharest as it over the U.S. quarantine of Cuba.The Soviet ship failed to cooperate, but the U.S. Navy restrained itself from forcibly seizing the ship, deeming it unlikely that the tanker was carrying offensive weapons.

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Is this expression 6(2r+(−1)+1 equivalent to 12r-5

VARVARA [1.3K]

Answer:

no it's not

Step-by-step explanation:

6(2r+(-1)+1 = 12r + -6 + 6 = 12r+ 0 => not equivalent to 12r - 5

Hope this help :)

6 0

4 years ago

Read 2 more answers

The teacher has 5 cups or broth to share amoung 4 students. draw a model and explain your answer.

Mama L [17]

Answer:

1.25,but your teacher is prolly looking for a fraction so 1 1/4

Step-by-step explanation:

DO NOT PUT 20 THAT PERSON S WRONG

3 0

3 years ago

During a fundraiser, students in Math Club make a $0,50 profit for every candy bar

Dmitrij [34]

Answer:

profit(n) = n * .50

Profit(10) = 5.00

Step-by-step explanation:

Profit = amount of candy bars sold * profit per candy bar

profit(n) = n * .50

Let the number of candy bars sold be 10 ( n=10)

Profit(10) = 10 * .50

= 5.00

When you added you only added 9 .50 terms which is why you only got 4.50 for your answer.

4 0

3 years ago

Help!!! please and thank you :)

Andrews [41]

Answer:

c

d

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

An equilateral triangle is inscribed in a circle of radius 2r. Express the area A within the circle but outside the triangle as

defon

Answer:

$A=8\pi r^2-\dfrac{49\sqrt{3}x^2}{4}\ un^2.$

Step-by-step explanation:

An equilateral triangle with the side 7x units is inscribed in a circle of radius 2r units.

1. The area of the circle is

$A_{circle}=2\pi (2r)^2=2\pi \cdot 4r^2=8\pi r^2\ un^2.$

2. The area of the equilateral triangle is

$A_{\triangle }=\dfrac{(7x)^2\sqrt{3}}{4}=\dfrac{49\sqrt{3}x^2}{4}\ un^2.$

3. The area A within the circle but outside the triangle is

$A=A_{circle}-A_{\triangle}=8\pi r^2-\dfrac{49\sqrt{3}x^2}{4}\ un^2.$

7 0

3 years ago