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fgiga [73]
3 years ago
10

Write the equation of a line that passes through (3.5,0) and is perpendicular to 8y+4x=64.

Mathematics
1 answer:
pantera1 [17]3 years ago
7 0

Answer:

y = 2x + 3.5

Step-by-step explanation:

Step 1: find the slope

8y + 4x = 64

8y = 64 - 4x

Make y the subject of the formula

y = (64 - 4x)/8

y = ( -4x + 64)/8

Separate to get slope

y = -4x/8 + 64/8

y = -x/2 + 8

Slope is the coefficient of x

m = -1/2

Note: if two lines are perpendicular to the other , it is negative reciprocal to each other

m = 2

Using the point slope form equation

y - y1 = m(x - x1)

y - y1 = 2(x - x1)

Substitute the point

( 3.5 , 0)

x1 = 3.5

y1 = 0

y - 3.5 = 2( x - 0)

open the bracket

y - 3.5 = 2x - 0

y = 2x - 0 + 3.5

y = 2x + 3.5

The equation of the line is

y = 2x + 3.5

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If two angles are supplementary and one angle measures 90, then the other angle also measures 90 deg, and the two angles are congruent. If one angle measures less than 90 deg, then the other measure more than 90 deg and are not congruent. Therefore, these angles may or may not be congruent.

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a tank is 3/7 full of water.After removing 420 litres it became 12/35 full .how much can the tank hold when full? ​
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Answer:

4900 Litres

Step-by-step explanation:

First we need a common demoninator:

3/7=15/35

Then we subtract the two to figure out how much 420 Litres is:

15/35-12/35=3/35

3/35=420 litres

Divide amount by the numerater for thow much 1/x is.

1/35=140

and multiply by the denomenator to get a full number on your fraction, and therefore a full tank.

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Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95
Virty [35]

Answer:

L(x)=1+\dfrac{1}{3}x

\sqrt[3]{0.95} \approx 0.9833

\sqrt[3]{1.1} \approx 1.0333

Step-by-step explanation:

Given the function: g(x)=\sqrt[3]{1+x}

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial,L(x)=f(a)+f'(a)(x-a)

a=0

g(0)=\sqrt[3]{1+0}=1

g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}

Therefore:

L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x

(b)\sqrt[3]{0.95}= \sqrt[3]{1-0.05}

When x = - 0.05

L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833

\sqrt[3]{0.95} \approx 0.9833

(c)

(b)\sqrt[3]{1.1}= \sqrt[3]{1+0.1}

When x = 0.1

L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333

\sqrt[3]{1.1} \approx 1.0333

7 0
3 years ago
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