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stiv31 [10]
3 years ago
5

Guys this isn't hard please answer i need a good grade or im not gonna be able to play soccer

Mathematics
2 answers:
melamori03 [73]3 years ago
8 0

Answer:

0, 4.7, and 5.1

Step-by-step explanation:

Dvinal [7]3 years ago
3 0

Answer:

0, 4.7, 5.1, and 7.4

Step-by-step explanation:

≤ means that the number has to be either less than or equal to 7.4. 0, 4.7, and 5.1 are all smaller than 7.4, and 7.4 (D, the option), is of course, equal to 7.4.

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3q(p-q)+2r(p-q)-S(q-p)<br> Can u solve it step by step
Artyom0805 [142]

Answer:

-3q² + 3qp + 2rp - 2rq + Sq - Sp

Step-by-step explanation:

first part

3q(p-q) = 3qp - 3q²

second part

2r(p-q) = 2rp - 2rq

third part

S(q-p) = Sq - Sp

then we put it all together

3qp - 3q² + 2rp - 2rq + Sq - Sp

in the right place possibly

-3q² + 3qp + 2rp - 2rq + Sq - Sp

8 0
2 years ago
For the circle with equation (x-2)^2+(y+3)^2=9, answer each question. a) whats coords of the center? b) what are the radius and
MrRissso [65]

Answer:

<h2>a) center (2, -3)</h2><h2>b) radius r = 3</h2><h2>c) in the attachment</h2>

Step-by-step explanation:

The standard form of an equation of a circle:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

r - radius

We have:

(x-2)^2+(y+3)^2=9\\\\(x-2)^2+(y-(-3))^2=3^2

a) center (2, -3)

b) radius r = 3

c) in the attachment

4 0
3 years ago
Solve the equation <br><br> 14 = 9 - p
morpeh [17]

14 = 9 - p

Move 9 to the other side

Sign changes from +9 to -9

14-9= 9-9-p

14-9= -p

5=-p

Mutiply both sides by -1 to get + p

(5)(-1)= (-p)(-1)

p= -5

Answer: p= -5

3 0
3 years ago
Last week’s and this week’s low temperatures are shown in the table below. Low Temperatures for 5 Days This Week and Last Week L
STatiana [176]

Question:

Temperatures for 5 Days This Week and Last Week

Low Temperatures This Week (Degrees Fahrenheit) 4 10 6 9 6 Low Temperatures Last Week (Degrees Fahrenheit) 13 9 5 8 5 Which measures of center or variability are greater than 5 degrees? Select three choices.

a) the mean of this week’s temperatures

b) the mean of last week’s temperatures

c) the range of this week’s temperatures

d)the mean absolute deviation of this week’s temperatures

e) the mean absolute deviation of last week’s temperatures

Answer:

a) the mean of this week’s temperatures

b) the mean of last week’s temperatures

c) the range of this week’s temperatures

Step-by-step explanation:

I would be verifying the options a, b, and c in my answer above through calculations which are shown below.

We were given the following data:

Low Temperatures This Week (Degrees Fahrenheit)

4, 10, 6, 9, 6

Low Temperatures Last Week (Degrees Fahrenheit)

13, 9, 5, 8, 5

We are to find which measures of center or variability are greater than 5 degrees.

Option a

The mean of this week’s temperatures

(4+ 10+ 6+9+ 6) °F ÷ 5 = 35 °F ÷5 = 7°F

Option a is correct because it measures of center or variability which is 7 °F is higher than 5°F

Option b

The mean of last week’s temperatures

(13+ 9 + 5 + 8 + 5) °F = 40°F ÷ 5 = 8°F

Option b is correct , because its measure of variability which is 8°F is greater than 5°F.

Option c

the range of this week’s temperatures

This week's temperature is given as

(4, 10, 6, 9, 6) °F

Range is defined as the difference between the highest number and the lowest number

Range of this week's temperature = (10 - 4) °F = 6°F

Hence Option c is correct because it measures of center or variability which is 6°F is greater than 5°F

From the above calculations we can accurately confirm that options a, b, and c are correct because the measures of their center or variability is greater than 5°F

3 0
3 years ago
Read 2 more answers
The rate at which rain accumulates in a bucket is modeled by the function r given by r(t)=10t−t^2, where r(t) is measured in mil
mars1129 [50]

Answer:

36 milliliters of rain.

Step-by-step explanation:

The rate at which rain accumluated in a bucket is given by the function:

r(t)=10t-t^2

Where r(t) is measured in milliliters per minute.

We want to find the total accumulation of rain from <em>t</em> = 0 to <em>t</em> = 3.

We can use the Net Change Theorem. So, we will integrate function <em>r</em> from <em>t</em> = 0 to <em>t</em> = 3:

\displaystyle \int_0^3r(t)\, dt

Substitute:

=\displaystyle \int_0^3 10t-t^2\, dt

Integrate:

\displaystyle =5t^2-\frac{1}{3}t^3\Big|_0^3

Evaluate:

\displaystyle =(5(3)^2-\frac{1}{3}(3)^3)-(5(0)^2-\frac{1}{3}(0)^3)=36\text{ milliliters}

36 milliliters of rain accumulated in the bucket from time <em>t</em> = 0 to <em>t</em> = 3.

4 0
3 years ago
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