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3 years ago

Mixed Applications: Choose the best strategy

Mathematics

1 answer:

romanna [79]3 years ago

5 0

Where is the problem

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Is there sufficient evidence at the \alpha=0.10α=0.10alpha, equals, 0, point, 10 level to conclude that the mean weight of the e

andrew11 [14]

Answer:

hmfujdyjdyjfjydym

Step-by-step explanation:

yicgncgnxyjdyjdukfukfukfukfukfukfukf

3 0

3 years ago

Given m || n, find the value of x and y.

Ann [662]

Answer:

x = 41 , y = 139

Step-by-step explanation:

x and 41° are alternate exterior angles and are congruent , so

x = 41

x and y are a linear pair and sum to 180° , that is

x + y = 180

41 + y = 180 ( subtract 41 from both sides )

y = 139

3 0

2 years ago

Around the car park, cylindrical concrete bollards are going to go placed .

Llana [10]

Answer:

653.120 cm³

Step-by-step explanation:

diameter/2 = 40 cm = radius

Height = 130 cm

Volume?

Volume = π r² h

= 3,14 . 1600 . 130

= 653.120 cm³

CMIIW

8 0

3 years ago

Read 2 more answers

Identify the interval(s) on which the function y = 2x^2 - 8x - 10 is positive.

Serga [27]

Given:

$y=2x^2-8x-10$

To find:

The interval(s) on which the given function is positive.

Solution:

We have,

$y=2x^2-8x-10$

$y=2(x^2-4x-5)$

$y=2(x^2-5x+x-5)$

$y=2(x(x-5)+1(x-5))$

$y=2(x+1)(x-5)$

For zeroes, $y=0$

$2(x+1)(x-5)=0$

$x=-1,x=5$

-1 and 5 divide the number line in three parts $(-\infty , -1),(-1,5),(5,\infty )$ .

Interval Selected Value $y=2(x+1)(x-5)$ Sign

$(-\infty,-1)$ -2 $y=2(-2+1)(-2-5)=14$ +

$(-1,5)$ 0 $y=2(0+1)(0-5)=-10$ -

$(5,\infty)$ 6 $y=2(6+1)(6-5)=14$ +

So, the function is positive for interval $(-\infty,-1)$ and $(5,\infty)$ .

We can say that the function is positive for interval x < -1 and x > 5.

Therefore, the correct option is B.

8 0

3 years ago

Which of these points lie on the circumference of the circle X^2 + Y^2 = 25? Circle your answer. (-3, 4) (6.25, 6.25) (9, 16) (-

Bad White [126]

Answer:

A. (-2, 4)

Step-by-step explanation:

Given the circle:

$x^2+y^2=25$

$x^2+y^2=5^2$

According to the equation, any point with one of the coordinate > 5 lies outside of the circle since the radius is 5 units.

The points 2, 3 and 4 have at least one of the coordinates greater than 5.

Let's verify the point 1:

$(-3)^2+4^2=9+16=25\\$
$25=25$

3 0

3 years ago

Read 2 more answers