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Aleksandr [31]
2 years ago
10

The maximum number of students in a classroom is 26. if there are 16 students sign up for the art class,how many more students c

an join the class without exceeding the maximum? ues ">=" for more than and "<=" for less than.
and thanks for answering​
Mathematics
2 answers:
Pavlova-9 [17]2 years ago
6 0

Answer:

x = 16 ≥ 26

Step-by-step explanation:

x = 10

10 + 16 = 26

Sorry for the delay, my man!

Marizza181 [45]2 years ago
5 0
Maximum 12 fr longgg
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I need this really fast can someone help me outMalik jogged 2 miles in 20 minutes. What was his rate in miles per hour? ​
Blababa [14]
It would be 6 miles per hour.

mph is a unit for speed. And the formula for speed is s=distance /time

So the distance is 2 miles
And the time is 20 minutes

*for this question you need to convert the minutes to hours to get the right units

20mins =0.3 hours (approx) [20/60 to get it in hours]

2 miles/0.3 hours ≈ 6 mph
7 0
3 years ago
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Solve for y.<br> 8+4y= -8
Andrew [12]

Answer:

-4

Step-by-step explanation:

8+4y= -8

4y= -8 -8

4y= -16

divide both sides by the coeffient of the variable. in this case divide both sides by 4

y=-4

7 0
3 years ago
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If logb2=x and logb3=y, evaluate the following in terms of x and y:
Alja [10]

log_b{162} = x + 4y\\\\log_b324 = 2x+4y\\\\log_b\frac{8}{9} = 3x-2y\\\\\frac{log_b27}{log_b16} = 3y-4x

<em><u>Solution:</u></em>

Given that,

log_b2 = x\\\\log_b3 = y --------(i)

<em><u>Use the following log rules</u></em>

Rule 1: log_b(ac) = log_ba + log_bc

Rule 2: log_b\frac{a}{c} = log_ba - log_bc

Rule 3: log_ba^c = clog_ba

a) log_b{162}

Break 162 down to primes:

162 = 2^1 \times 3^4

log_b{162} =log_b 2^1. 3^4\\\\By\ rule\ 1\\\\ log_b{162} = log_b 2^1 +log_b 3^4\\\\By\ rule\ 3\\\\1log_b2 + 4log_b3\\\\1x+4y\\\\x+4y

Thus we get,

log_b162 = x + 4y

Next

b) log_b 324

Break 324 down to primes:

324 = 2^2 \times 3^4

log_b324 = log_b 2^2.3^4\\\\By\ rule\ 1\\\\log_b324 = log_b2^2 + log_b3^4\\\\By\ rule\ 3\\\\log_b324 = 2log_b2 + 4log_b3\\\\From\ (i)\\\\log_b324 = 2x + 4y

Next

c) log_b\frac{8}{9}

By rule 2

log_b\frac{8}{9} = log_b8 - log_b9\\\\log_b\frac{8}{9} = log_b 2^3 - log_b3^2\\\\By\ rule\ 3\\\\log_b\frac{8}{9} =  3 log_b2 - 2log_b3\\\\From\ (i)\\\\log_b\frac{8}{9} =  3x - 2y

Next

d) \frac{log_b27}{log_b16}

By rule 2

\frac{log_b27}{log_b16} = log_b27 - log_b16\\\\ \frac{log_b27}{log_b16} = log_b3^3 - log_b2^4\\\\By\ rule\ 2\\\\ \frac{log_b27}{log_b16} = 3log_b3 - 4log_b2 \\\\From\ (i)\\\\\frac{log_b27}{log_b16} =  3y - 4x

Thus the given are evaluated in terms of x and y

3 0
2 years ago
If line segment BH = 3, line segment BF = 10, line segment GH= 2, find HD 15 11 10.5 ?
pav-90 [236]

HD = 10.5

Step-by-step explanation:

Given BH = 3, GH = 2, BF = 10

Step 1: To find HF:

HF = BF – BH

HF = 10 – 3

HF = 7

Step 2: To find HD:

We know that if two chords intersects inside a circle, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord.

⇒ GH × HD = BH × HF

⇒ 2 × HD = 3 × 7

⇒ HD = 10.5

Hence, the value of HD = 10.5.

5 0
3 years ago
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Need help please don’t know what to do here
kotykmax [81]

Answer:

Down below

Step-by-step explanation:

a. The range of y = sinθ is [-1,1]

b. The period of y = cosθ is 2π

c. The asymptotes of y = tanθ are -π2, π2, πn

d. The amplitude of y = sinθ is 1

e. The period of y = tanθ is π

f. The max value of y = cosθ is 1

4 0
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