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3 years ago

For his long distance phone service, Dale pays a

1 answer:

3 0

Equation;
y=6+0.08x
Last month, Dale's long distance bill was
$13.92
so,
$13.92=6+0.08x
0.08x=$13.92-6
0.08x=$7.92
x=99 minutes

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Can someone help me please....

zhannawk [14.2K]

Answer:

B. 350 minutes

Step-by-step explanation:

7 0

3 years ago

Ebi, Jose, Derell, and Asami measured their heights. Ebi's height was 2.5 cm greater than Jose's height. Jose's height was 3.1 c

irga5000 [103]

Start with assigning each person with a variable to represent their height

Ebi: e
Jose: j
Derell: d
Asami: a

Ebi'd height was 2.5 cm greater than Jose's height

j + 2.5 = e

Jose's height was 3.1 cm greater than Derell's

d + 3.1 = j

Derell's height is 0.4 cm less than Asami's height

a - 0.4 = d

Ebi is 162.5 cm tall

e = 162.5

So, plug in 162.5 into any of the above equations were there is a variable of e

j + 2.5 = e

j + 2.5 = 162.5

Subtract 2.5 from both sides of the equation

j = 160 cm

Jose's height is 160 cm

Now, plug in 160 into any of the above equations where there is a j

d + 3.1 = j

d + 3.1 = 160

Subtract 3.1 from both sides of the equation

d = 156.9 cm

Derell's height 156.9 cm

so, plug in 156.9 into any of the above equations where there is a d

a - 0.4 = d

a - 0.4 = 156.9

Add 0.4 on both sides of the equation

a = 157.3 cm

Asami's height is 157.3 cm

7 0

3 years ago

Read 2 more answers

120÷(18-(20-3(9-5)simplify

Ilya [14]

Answer:

20÷[18-{20-3(4)}]

120÷[18-[20-12]]

120÷[18-8]]

120÷10

Step-by-step explanation:

6 0

3 years ago

How many times does 0.75 fit into 9 wholly I need help

nexus9112 [7]

Answer:

12. 9÷0.75.

Thank you. please mark me the braniliest.

3 0

1 year ago

Find the 2th term of the expansion of (a-b)^4.

vladimir1956 [14]

The second term of the expansion is $-4a^3b$ .

Solution:

Given expression:

$(a-b)^4$

To find the second term of the expansion.

$(a-b)^4$

Using Binomial theorem,

$(a+b)^{n}=\sum_{i=0}^{n}\left(\begin{array}{l}n \\i\end{array}\right) a^{(n-i)} b^{i}$

Here, a = a and b = –b

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

Substitute i = 0, we get

$$\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}=1 \cdot \frac{4 !}{0 !(4-0) !} a^{4}=a^4$

Substitute i = 1, we get

$$\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}=\frac{4 !}{3!} a^{3}(-b)=-4 a^{3} b$

Substitute i = 2, we get

$$\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}=\frac{12}{2 !} a^{2}(-b)^{2}=6 a^{2} b^{2}$

Substitute i = 3, we get

$$\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}=\frac{4}{1 !} a(-b)^{3}=-4 a b^{3}$

Substitute i = 4, we get

$$\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}=1 \cdot \frac{(-b)^{4}}{(4-4) !}=b^{4}$

Therefore,

$$(a-b)^4=\sum_{i=0}^{4}\left(\begin{array}{l}4 \\i\end{array}\right) a^{(4-i)}(-b)^{i}$

$=\frac{4 !}{0 !(4-0) !} a^{4}(-b)^{0}+\frac{4 !}{1 !(4-1) !} a^{3}(-b)^{1}+\frac{4 !}{2 !(4-2) !} a^{2}(-b)^{2}+\frac{4 !}{3 !(4-3) !} a^{1}(-b)^{3}+\frac{4 !}{4 !(4-4) !} a^{0}(-b)^{4}$ $=a^{4}-4 a^{3} b+6 a^{2} b^{2}-4 a b^{3}+b^{4}$

Hence the second term of the expansion is $-4a^3b$ .

3 0

3 years ago