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1 year ago

Evaluate p-(m-4)(p-q) given m = -6, p=-5 and q=-5

Mathematics

1 answer:

Svetradugi [14.3K]1 year ago

5 0

Answer:

Step-by-step explanation:

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9) Which equation represents the line that passes through the point (3,4) and is parallel to the x-axis?

kobusy [5.1K]

Since the line is parallell to the x-axis we know the slope is zero: $m=0$

Using the point-slope formula we can substitute know values:
$y=mx+b$
$4=0(3)+b$
$b=4$
$y=0x+4$
$y=4$

You could also jsut do it by inspection: since it passes through (3,4) and is horizontal it must cross at (0,4) i.e. all y=4.

3 0

3 years ago

Combine like terms to create an equivalent expression.<br> -1/2(-3y+10)

Studentka2010 [4]

Answer: 3/2y - 5

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Help me please 25 points.

sveticcg [70]

The answer i got is VP<------------->

AKA the first option

i hope this is right

4 0

2 years ago

Read 2 more answers

Multiplying two digits<br>

Vika [28.1K]

Answer:

Get a calculator

Step-by-step explanation:

Most of the time for simple math like multiplying 2 digits, it is easiest if you just get a calculator. It saves a lot of time.

If you can't get a calculator then i recommend memorizing some common equations.

For example 10*15= 150

so that if you get 12*15=?, you can easily just add two extra 15's to your answer of 150.

15+15 is 30, and 30+150(from 10*15)=180.

Therefore 12*15=180.

:) Hope that helped!

4 0

2 years ago

If the mean weight of 4 backfield members on the football team is 221 lb and the mean weight of the 7 other players is 202 lb, w

MatroZZZ [7]

Let $x_1, x_2, x_3, x_4, x_5, x_6, x_7, x_8, x_9, x_{10}, x_{11}$ be the weight of i-th player.

1. If the mean weight of 4 backfield members on the football team is 221 lb, then

$\dfrac{x_1+x_2+x_3+x_4}{4}=221\ lb.$

2. If the mean weight of the 7 other players is 202 lb, then

$\dfrac{x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{7}=202\ lb.$

3. From the previous statements you have that

$x_1+x_2+x_3+x_4=221\cdot 4=884 \lb,\\ \\x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=202\cdot 7=1414\ lb.$

Add these two equalities and then divide by 11:

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}=884+1414=2298\ lb,\\ \\\dfrac{x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_{10}+x_{11}}{11}=\dfrac{2298}{11}=208\dfrac{10}{11}\ lb.$

Answer: the mean weight of the 11-person team is $208\dfrac{10}{11}\ lb.$

4 0

2 years ago