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3 years ago

What is an equation of the line that passes through the points (-6, -3) and (-3, -5)?

Mathematics

1 answer:

RUDIKE [14]3 years ago

7 0

find the gradient = change in y/change in X

-5 - -3/-3 - -6

= -2/3

take any two point say -3,-5 and equate to the gradient

Y - 5 /X - 3 = -2/3

cross multiply

3y - 15 = -2x + 6

3y = -2x +6 + 15

3y = -2x + 21

y = -2/3x +7

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Find the area of an equilateral triangle whose value is 18 cm

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Answer:

$18 + 18 + 18 = 18 \times 3 = 54$

Step-by-step explanation:

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2 years ago

Please help me with this math!

STALIN [3.7K]

1. Alternate exterior angles

So, C

2.
180 = 76 + 2x
104 = 2x
52 = x

So, B

4.

4a. No
4b. Yes
4c. No
4d. Yes

5.
180 = 35 + 35 + 2x
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7.

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9.

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2 years ago

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For the following demand equation compute the elasticity of demand and determine whether the demand is elastic, unitary, or inel

adelina 88 [10]

Answer:

Demand is inelastic at p = 9 and therefore revenue will increase with

an increase in price.

Step-by-step explanation:

Given a demand function that gives <em>q</em> in terms of <em>p</em>, the elasticity of demand is

$E=|\frac{p}{q}\cdot \frac{dq}{dp} |$

If E < 1, we say demand is inelastic. In this case, raising prices increases revenue.
If E > 1, we say demand is elastic. In this case, raising prices decreases revenue.
If E = 1, we say demand is unitary.

We have the following demand equation $D(p)=-\frac{3}{4}p+29$ ; p = 9

Applying the above definition of elasticity of demand we get:

$E(p)=\frac{p}{q}\cdot \frac{dq}{dp}$

where

p = 9
q = $-\frac{3}{4}p+29$
$\frac{dq}{dp}=\frac{d}{dp}(-\frac{3}{4}p+29)$

$\frac{d}{dp}\left(-\frac{3}{4}p+29\right)=-\frac{d}{dp}\left(\frac{3}{4}p\right)+\frac{d}{dp}\left(29\right)\\\\\frac{d}{dp}\left(-\frac{3}{4}p+29\right)=-\frac{3}{4}$