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Rama09 [41]
3 years ago
5

A checker board has 64 squares and 32 of them are black.

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0
50% or 32/64 because 32 is half of 64
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Write an inequality to represent the graph.
Westkost [7]

Answer:

y ≥ −4x − 2

Step-by-step explanation:

8 0
3 years ago
There are 120 students in the seventh grade. Seventy percent of these students are involved in extracurricular activites. How ma
Tanzania [10]

First, turn 70% into a decimal by dividing 70 by 100.

70% ⇒ 0.70

Multiply 120 by 0.70.

120 × 0.70 = 84

Now we know that 84 students are involved in extracurricular activities. Subtract 84 from 120 to find the amount of students who are <em>not</em> involved in extracurricular activities.

120 - 84 = 36

<h2>Answer:</h2>

<u>36 seventh grade students are not involved in extracurricular activities.</u>

6 0
3 years ago
Read 2 more answers
For the given functions f and g, complete parts (a)-(h). For parts (a)-(d), also find the domain.
slamgirl [31]

The domain for (a) , (b) , (c) is all set of real numbers , for (d) (x ε R except x = 3/8.

The complete question is

a) (f + g)(x), (b) (f – g)(x), (c) (f × g)(x), (d) (f / g)(x), (e) (f + g)(7), (f) (f - g)(2), (g) (f.g)(3), (h) (f/g)(27)

<h3>What is a Function ?</h3>

A function is the law that defines the relation between independent and a dependent variable.

It is given that

f(x) = 3x+8

g(x) = 8x-3

(a)  (f + g)(x) = 3x+8 +8x-3 = 11x+5

(b)  (f – g)(x) = 3x+8 -8x+3 = -5x +11

(c) (f × g)(x) = (3x+8)(8x-3) = 24x²-9x+64x -24 = 24x² +55x-24

(d)(f / g)(x) = (3x+8)/(8x-3)

(e) (f + g)(7) = 11*7+5 = 77+5 = 82

(f) (f - g)(2)= -5*2 +11 = 1

(g) (f.g)(3) =24*9 +55*3 -24 = 357

(h)  (f/g)(27) = (3x+8)/(8x-3)  = (3 * 27 +8)/(8*27-3) = 89/213

The domain for (a) , (b) , (c) is all set of real numbers , for (d) (x ε R except x = 3/8)

To know more about Function

brainly.com/question/12431044

#SPJ1

3 0
2 years ago
Expression that equals 118
Andrei [34K]
Here are the following multiples:
59 * 2 = 118

and 118 * 1 = 118

The * means times.
5 0
3 years ago
Prove the identity (sina-cosatanb)/(cosa+sinatanb) = tan (a+b)
OLga [1]
Sina - (cosa)(tanb)/cosa + (sina)(tanb)
sina ≡ (tana)(cosa)

(tana)(cosa) - (cosa)(tanb)/cosa + (tana)(cosa)(tanb)
= cosa(tana - tanb)/cosa(1 + tanatanb)
(cosas cancel out)
= (tana - tanb)/(1 + tanatanb) ≡ tan(a-b)
7 0
3 years ago
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