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3 years ago

11

What is the slope of a line that runs parallel to y =2x+5

2 answers:

Murljashka [212]3 years ago

5 0

2 since parallel lines have the same slope

Luden [163]3 years ago

5 0

A slope of something that is parallel is the same slope. It's 2 (2/1) and it can be anything for the y intercept but 5

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X - 6 = 8 what is x?

kondor19780726 [428]

Answer:

$x = 14$

Step-by-step explanation:

$x - 6 = 8$
Group like terms.
$x = 8 + 6$
$x = 14$

5 0

2 years ago

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A ship leaves port at 1:00 P.M. and travels S35°E at the rate of 27 mi/hr. Another ship leaves the same port at 1:30 P.M. and tr

Len [333]

To solve this problem you must apply the proccedure shown below:

1. You must apply the Law of Cosines, as you can see in the figure attached. Then:

- The first ship travels at $27 mi/hr$ in for two hours. Therefore, the side $a$ is:

$a=(27 mi/hr)(2 hr)=54mi$

- The second ship travels at $18 mi/hr$ for $1.5 hours$ . Therefore, the side $b$ is:

$b=(18mi/hr)(1.5hr)=27mi$

- Now, you can calculate $c$ :

$c=\sqrt{54^{2}+27^{2}-2(54)(27)Cos(55)}=44 mi$

The answer is: $44 miles$

8 0

3 years ago

25 Points! Show ALL Work! Image Attached.

SVEN [57.7K]

Answer:

a) x = 13

b) x = 7

Step-by-step explanation:

A) Since they have the same base of 2 so

x + 4 + (-5) = 12

x - 1 = 12

x = 13

b) Since they have the same base of -3 so

8 + x + 4 = 19

x + 12 = 19

x = 7

4 0

3 years ago

A motorist traveled 311 miles on 12 gallons of gas. To the nearest tenth, how many miles can the motorist travel on one gallon o

Alina [70]

D. 311 divided by 12 = 25.91
The nearest tenth would be 25.9

3 0

3 years ago

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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