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Kryger [21]
3 years ago
11

Make q the subject of the formula 2(q+p) = 1+5q. Give your answer in the form ap-b/c where a,b and c are all positive integers

Mathematics
1 answer:
musickatia [10]3 years ago
8 0

Answer:

\frac{2p-1}{3}

Step-by-step explanation:

Since you want to get q only and q appears in both side of the equation. Try to isolate q to one side.

1) Expand 2(q+p)

2q + 2p = 1 + 5q

2) Move all q terms to one side

5q - 2q = 2p - 1

3q = 2p - 1

3) Divide 3 on both side (to isolate q)

q = \frac{2p-1}{3}

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Simplify the given equation. 6 - (3x 10) 4(2 - x) = 15
mamaluj [8]

Answer:

6 − 30 x ,  8 − 4 x

Step-by-step explanation:

hopefully this helps

4 0
3 years ago
Read 2 more answers
What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S
Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is y=\frac{1}{3} x+2

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3  

And as the line passing through (-3,1) and is  perpendicular to 3x + y = -8, product of slopes of two line will be -1  as lies are perpendicular.

Let required slope = x  

\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}

So we need to find the equation of a line whose slope is \frac{1}{3} and passing through (-3,1)

Equation of line passing through (x_1 , y_1) and having lope of m is given by

\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)

\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}

Substituting the values we get,

\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}

Hence the required equation of line is found using slope intercept form

4 0
3 years ago
A professor determined the relationship between the time spent studying (in hours) and performance on an exam.
kykrilka [37]
Step 1: Insert 2.6 hours for the variable of time and solve.
70.443 + 4.885(2.6)
70.443+ 12.701
= 83.144
This is Ann's EXPECTED exam score.  Because of the concert, Ann scored 16 points lower than this number.
Step 2: Subtract 16 from Ann's expected score.
83.144 - 16
= 67.144
3 0
3 years ago
a farmer hires 110 workers to harvest apples from his trees.it takes the 12 days to harvest the farmers entire crop.how long doe
bazaltina [42]

Answer:

110 = 12

110 + 110 + 110 + 110 = 440

so 110 × 4 = 440

4 × 12 =<em><u> 48 days</u></em>

hope this helps

5 0
3 years ago
Suppose the object moving is Dave, who hasamassofmo =66kgatrest. Whatis Daveâs mass at 90% of the speed of light? At 99% of the
nignag [31]

Step-by-step explanation:

Formula that relates the mass of an object at rest and its mass when it is moving at a speed v:

m=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}

Where :

m = mass of the oebjct in motion

m_o = mass of the object when at rest

v = velocity of a moving object

c = speed of the light = 3\times 0^8 m/s

We have :

1) Mass of the Dave = m_o=66 kg

Velocity of Dave ,v= 90% of speed of light = 0.90c

Mass of the Dave when moving at 90% of the speed of light:

m=\frac{66 kg}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}

m = 151.41 kg

Mass of Dave when when moving at 90% of the speed of light is 151.41 kg.

2) Velocity of Dave ,v= 99% of speed of light = 0.99c

Mass of the Dave when moving at 99% of the speed of light:

m=\frac{66 kg}{\sqrt{1-\frac{(0.99c)^2}{c^2}}}

m = 467.86 kg

Mass of Dave when when moving at 99% of the speed of light is 467.86 kg.

3) Velocity of Dave ,v= 99.9% of speed of light = 0.999c

Mass of the Dave when moving at 99.9% of the speed of light:

m=\frac{66 kg}{\sqrt{1-\frac{(0.999c)^2}{c^2}}}

m = 1,467.17 kg

Mass of Dave when when moving at 99.9% of the speed of light is 1,467.17 kg.

4) Mass of the Dave = m_o=66 kg

Velocity of Dave,v=?

Mass of the Dave when moving at v speed of light: 500

500 kg=\frac{66 kg}{\sqrt{1-\frac{(v)^2}{(3\times 10^8 m/s)^2}}}

v=2.973\times 10^8 m/s

Dave should be moving at speed of 2.973\times 10^8 m/s.

4 0
3 years ago
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