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3 years ago

−2/3(3 - 1/2)(-1) = ?

Mathematics

1 answer:

chubhunter [2.5K]3 years ago

8 0

The correct answer would be B. 2 - 1/3

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8. If f(x) = 5x - 8, what is the value of f(3)?

Igoryamba

Answer: f(3) = 7

Step-by-step explanation:

f(3) = 5(3) - 8

15 - 8

f(3) = 7

7 0

3 years ago

Read 2 more answers

B = 54°, b=15, c= c = ? (Round answers to the nearest hundredth.)

zaharov [31]

Answer:

18.54

Step-by-step explanation:

sine is opposite/hypotenuse, and you have opposite (b=15), so use that

sin(54) = 15/c

c = 15/sin54

then use calculator

c = 18.54

8 0

2 years ago

Find the midpoint of points A(-2,-8) and B(7,-4) graphically.

k0ka [10]

Answer:

The midpoint is (2.5,-6) or written as a fraction ( $\frac{5}{2}$ , -6)

Step-by-step explanation:

$MP= (\frac{x_1+x_2}{2}),(\frac{y_1+y_2}{2} ) \\MP= (\frac{-2+7}{2}),(\frac{-8+-4}{2} ) \\MP= (\frac{5}{2}),(\frac{-12}{2} ) \\MP= (2.5,-6 )$

7 0

3 years ago

Eliza is two years younger than half her bro Juan age write an equation to represent Eliza’s age ; if Juan is 6 how old is Eliza

cricket20 [7]

Answer:

Eliza is 1

Step-by-step explanation:

If Eliza is 2 years younger than 1/2 her brothers age and Juan is 6.

- divide 6 by 2 (6÷2=3)

- Take 3 away from 2 ( 3-2=1)

So that means Eliza would be 1

6 0

3 years ago

One Sunday, 120 days before Christmas, Aldsworth store publishes an advertisement saying ‘120 shopping days until Christmas'. Al

Lena [83]

Answer:

(a)18

(b)1089

(c)Sunday

Step-by-step explanation:

The problem presented is an arithmetic sequence where:

First Sunday, a=1
Common Difference (Every subsequent Sunday), d=7

We want to determine the number of Sundays in the 120 days before Christmas.

(a)In an arithmetic sequence:

$\text{The nth term}, T_n=a+(n-1)d\\T_n \leq 120\\$Therefore:$\\1+7(n-1) \leq 120\\1+7n-7\leq 120\\7n-6\leq 120\\7n\leq 120+6\\7n\leq 126\\$Divide both sides by 7$\\n\leq 18$

Since the result is a whole number, there are 18 Sundays in which Aldsworth advertises.

Therefore, Aldsworth advertised 18 times.

(b)Next, we want to determine the sum of the first 18 terms of the sequence

1,8,15,...

$\text{Sum of a sequence}, S_n=\frac{n}{2}( 2a+(n-1)d)\\S_{18}=\frac{18}{2}( 2*1+(18-1)*7)\\=9(2+17*7)\\=9(2+119)\\=9*121\\S_{18}=1089$

The sum of the numbers of days published in all the advertisements is 1089.

(c)SInce the 120th day is the 18th Sunday, Christmas is on Sunday.

6 0

3 years ago