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vovangra [49]
2 years ago
11

What is (12 − 4) − (6 ÷ 2) − (2 x 2) =

Mathematics
1 answer:
Tanya [424]2 years ago
5 0

Answer:

1

Step-by-step explanation:

(12-4) - (6/2) - (2 x 2)

(8) - (3) - (4)

1

First you solve the expressions in the perenthesis, then solve the final part

Hope this helps!

:)

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Sophia downloaded 600 vacation pictures. Of the pictures, 63% show the beach. About how many pictures show the beach?
Mashcka [7]

Answer:

378 pictures

Step-by-step explanation:

600 multiplied .63

4 0
2 years ago
A tennis ball machine serves a ball vertically into the air from a height of 2 feet, with an initial speed of 120 feet per secon
Makovka662 [10]

Answer:

The maximum height is reached after 3.75 seconds.

Step-by-step explanation:

Assuming the deceleration due to gravity is unaided by air resistance, gravity causes the ball to lose vertical speed at the rate of 32 ft/s every second. The initial vertical speed of 120 ft/s will decline to zero when t satisfies ...

... 120 ft/s - (32 ft/s²)t = 0

... (120 ft/s)/(32 ft/s²) = t = 3.75 s

The ball will not go any higher after its vertical speed is zero.

6 0
3 years ago
Which of the following is closest to 1/2 + 1/3
andre [41]

Answer:

.83

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Verify that y1(t) = 1 and y2(t) = t ^1/2 are solutions of the differential equation:
Papessa [141]

Answer: it is verified that:

* y1 and y2 are solutions to the differential equation,

* c1 + c2t^(1/2) is not a solution.

Step-by-step explanation:

Given the differential equation

yy'' + (y')² = 0

To verify that y1 solutions to the DE, differentiate y1 twice and substitute the values of y1'' for y'', y1' for y', and y1 for y into the DE. If it is equal to 0, then it is a solution. Do this for y2 as well.

Now,

y1 = 1

y1' = 0

y'' = 0

So,

y1y1'' + (y1')² = (1)(0) + (0)² = 0

Hence, y1 is a solution.

y2 = t^(1/2)

y2' = (1/2)t^(-1/2)

y2'' = (-1/4)t^(-3/2)

So,

y2y2'' + (y2')² = t^(1/2)×(-1/4)t^(-3/2) + [(1/2)t^(-1/2)]² = (-1/4)t^(-1) + (1/4)t^(-1) = 0

Hence, y2 is a solution.

Now, for some nonzero constants, c1 and c2, suppose c1 + c2t^(1/2) is a solution, then y = c1 + c2t^(1/2) satisfies the differential equation.

Let us differentiate this twice, and verify if it satisfies the differential equation.

y = c1 + c2t^(1/2)

y' = (1/2)c2t^(-1/2)

y'' = (-1/4)c2t(-3/2)

yy'' + (y')² = [c1 + c2t^(1/2)][(-1/4)c2t(-3/2)] + [(1/2)c2t^(-1/2)]²

= (-1/4)c1c2t(-3/2) + (-1/4)(c2)²t(-3/2) + (1/4)(c2)²t^(-1)

= (-1/4)c1c2t(-3/2)

≠ 0

This clearly doesn't satisfy the differential equation, hence, it is not a solution.

6 0
3 years ago
Given the function f(x)=|x+5|,find each of the following<br> F(10),f(-5),f(0)
kirza4 [7]

Answer:

In attachment

Step-by-step explanation:

|a|=a\qquad\text{if}\ a>0\\\\|a|=-a\qquad\text{if}\ a

7 0
2 years ago
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