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3 years ago

What is the equation in point-slope form of a line that passes through the points (7, −8) and (−4, 6) ?

1 answer:

5 0

First you differentiate to find the gradient

dy/dx = (6 - - 8) / (-4 - 7)
= 14 / -11
= - 14/11

Then you use this formula
y - y1 = m (x - x1)

y1 being the y coordinate
x1 being the x coordinate
m being the gradient (or slope)

y - 6 = -14/11 (x - - 4)
y - 6 = -14/11x - 56/11
y = -14/11x + 10/11

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Help please!!!!!.....

GarryVolchara [31]

Answer:

294

Step-by-step explanation:

7*7*6=294

6 0

3 years ago

Read 2 more answers

The volume of this triangular prism is 247.5 cubic inches. What is the value of t?

elixir [45]

Volume of a triangle is 1/2 base x height x length.

Using the given information we can find t:

247.5 = 1/2(5)(t)(11)

Simplify:

247.5 = 2.5(t)(11)

247.5 = 27.5t

Divide both sides by 27.5:

t = 247.5 / 27.5

t = 9

8 0

3 years ago

Read 2 more answers

Melissa buys 212 pounds of salmon and 114 pounds of swordfish. She pays a total of $31.25, and the swordfish costs $0.20 per pou

eimsori [14]

Answer:

C. $16.60

Step-by-step explanation:

Let the cost per pound of salmon be represented by x

The cost per pound of swordfish be represented by y

Melissa buys 2 1/2 pounds of salmon and 1 1/4 pounds of swordfish. She pays a total of $31.25

2.5 × x + 1.25× y = $31.25

2.5x + 1.25y = 31.25

The swordfish costs $0.20 per pound less than the salmon.

y = x - 0.20

Hence, we substitute

2.5x + 1.25y = 31.25

2.5y + 1.25(x - 0.20) = 31.25

2.5y + 1.25x - 0.25 = 31.25

2.5x + 1.25x = 31.25 + 0.25

3.75x = 31.5

x = 31.5/3.75

x = $8.4

The cost per pound of salmon be represented by x = $8.4

y = x - 0.20

y = 8.4 - 0.20

y = $8.2

The cost per pound of swordfish be represented by y = $8.2

The cost of a pound of salmon and sword fish

= $8.4 + $8.2

=$16.60

3 0

3 years ago

The total resistance R of two resistors connected in parallel circuit is given by 1/R = 1/R_1 + 1/R_2. Approximate the change in

Naily [24]

Answer:

a) Approximate the change in R is 0.5 ohm.

b) The actual change in R is 0.04 ohm.

Step-by-step explanation:

Given : The total resistance R of two resistors connected in parallel circuit is given by $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

To find :

a) Approximate the change in R ?

b) Compute the actual change.

Solution :

a) Approximate the change in R

$R_1=12\ ohm$ and $R_2=10\ ohm$

$R_1$ is decreased from 12 ohms to 11 ohms.

i.e. $\triangle R_1=21-11=1\ ohm$

$R_2$ is increased from 10 ohms to 11 ohms.

i.e. $\triangle R_2=11-10=1\ ohm$

The change in R is given by,

$\frac{1}{\triangle R}=\frac{1}{\triangle R_1}+\frac{1}{\triangle R_2}$

$\frac{1}{\triangle R}=\frac{\triangle R_2+\triangle R_1}{(\triangle R_1)(\triangle R_2)}$

$\triangle R=\frac{(\triangle R_1)(\triangle R_2)}{\triangle R_2+\triangle R_1}$

$\triangle R=\frac{(1)(1)}{1+1}$

$\triangle R=\frac{1}{2}$

$\triangle R=0.5\ ohm$

b) The actual change in Resistance

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{1}{12}+\frac{1}{10}$

$R=\frac{10\times 12}{10+12}$

$R=\frac{120}{22}$

$R=5.46\ ohm$

When resistances are charged, $R_1=R_2=11$

$\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R'}=\frac{1}{11}+\frac{1}{11}$

$R'=\frac{11}{2}$

$R'=5.5\ ohm$

Change in resistance is given by,

$C=R'-R$

$C=5.5-5.46$

$C=0.04\ ohm$

3 0

2 years ago

What is a good word problem for 3+(3*12)

Agata [3.3K]

Twelve times 3 added to three is equal to?

4 0

3 years ago