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ExtremeBDS [4]
3 years ago
12

During a pandemic, 135 students out of 750 who attend were absent. What percent of students were absent?

Mathematics
2 answers:
muminat3 years ago
8 0

Answer:

18%

Step-by-step explanation:

divide 135 by 750 getting  0.18. Then change it to percent form which is 18%

anzhelika [568]3 years ago
7 0

Answer:

18 percent of the students were absent

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Simplify the expression so there is only one positive power for the base, -5
s2008m [1.1K]

Answer:

C. -5 power of 5

Step-by-step explanation:

-5 power of 7 / -5 power of 2

Soo, you subtract the exponent 7 by 2 and you are left with 5.

You are thus left with -5 to the power of 5.

Please mark as brainliest if it helped!:)

6 0
3 years ago
A box of crackers weighs 11 1/4 ounces. The estimated serving is 3/4 ounce. How many servings are in the box?
eimsori [14]
15 servings because 11.25/0.75=15

3/4=.75
11 1/4=11.25
7 0
3 years ago
Read 2 more answers
Benny bought six new baseball trading cards to add to his collection . The next day his dog ate half of his collection. There ar
KIM [24]
90 because if his dog ate half his cards and benny had 45 left that means 45 times two equals 90 because the dog ate HALF of 90
3 0
3 years ago
What are the types of roots of the equation below?<br> - 81=0
Tju [1.3M]

Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0. This can be obtained by finding root of the equation using algebraic identity.    

<h3>What are the types of roots of the equation below?</h3>

Here in the question it is given that,

  • the equation x⁴ - 81 = 0

By using algebraic identity, (a + b)(a - b) = a² - b², we get,  

⇒ x⁴ - 81 = 0                      

⇒ (x² +  9)(x² - 9) = 0

⇒ (x² + 9)(x² - 9) = 0

  1. (x² -  9) = (x² - 3²) = (x - 3)(x + 3) [using algebraic identity, (a + b)(a - b) = a² - b²]
  2. x² + 9 = 0 ⇒ x² = -9 ⇒ x = √-9 ⇒ x= √-1√9 ⇒x = ± 3i

⇒ (x² + 9) = (x - 3i)(x + 3i)

Now the equation becomes,

[(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

Therefore x + 3, x - 3, x + 3i and x - 3i are the roots of the equation

To check whether the roots are correct multiply the roots with each other,

⇒ [(x - 3)(x + 3)][(x - 3i)(x + 3i)] = 0

⇒ [x² - 3x + 3x - 9][x² - 3xi + 3xi - 9i²] = 0

⇒ (x² +0x - 9)(x² +0xi - 9(- 1)) = 0

⇒ (x² - 9)(x² + 9) = 0

⇒ x⁴ - 9x² + 9x² - 81 = 0

⇒ x⁴ - 81 = 0

Hence Option B, that is Two Complex and Two Real which are x + 3, x - 3, x + 3i and x - 3i, are the types of roots of the equation x⁴ - 81 = 0.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: What are the types of roots of the equation below?

x⁴ - 81 = 0

A) Four Complex

B) Two Complex and Two Real

C) Four Real

Learn more about roots of equation here:

brainly.com/question/26926523

#SPJ9

5 0
1 year ago
Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =
Rainbow [258]

Answer:

\sin L = 0.60

tan\ N = 1.33

\cos L = 0.80

\sin N = 0.80

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

MN = 6

LN = 10

First, we need to calculate length LM,

Using Pythagoras theorem:

LN^2 = MN^2 + LM^2

10^2 = 6^2 + LM^2

100 = 36 + LM^2

Collect Like Terms

LM^2 = 100 - 36

LM^2 = 64

LM = 8

Solving (a): \sin L

\sin L = \frac{Opposite}{Hypotenuse}

\sin L = \frac{MN}{LN}

Substitute values for MN and LN

\sin L = \frac{6}{10}

\sin L = 0.60

Solving (b): tan\ N

tan\ N = \frac{Opposite}{Adjacent}

tan\ N = \frac{LM}{MN}

Substitute values for LM and MN

tan\ N = \frac{8}{6}

tan\ N = 1.33

Solving (c): \cos L

\cos L = \frac{Adjacent}{Hypotenuse}

\cos L = \frac{LM}{LN}

Substitute values for LN and LM

\cos L = \frac{8}{10}

\cos L = 0.80

Solving (d): \sin N

\sin N = \frac{Opposite}{Hypotenuse}

\sin N = \frac{LM}{LN}

Substitute values for LM and LN

\sin N = \frac{8}{10}

\sin N = 0.80

7 0
3 years ago
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