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3 years ago

Y=-1/2x +7

Mathematics

1 answer:

yarga [219]3 years ago

4 0

Answer:

B; (2,-4)

Step-by-step explanation:

The general equation of a straight line is;

y = mx + b

where m is the slope and b is the y-intercept

So from the question, when we look at the given equation, its slope is -1/2

mathematically, when two lines are perpendicular, the product of their slopes is -1

thus;

m1 * m2 = -1

m2 * -1/2 = -1

-m2 = -2

m2 = 2

The general equation form is;

y-y1 = m(x-x1)

where (x1,y1) = (4,0)

y-0 = 2(x-4)

y = 2x - 8

So, now we look at the point that will work for this equation

For the line that will work, if we substitute its x-value, we get the y-value

Let us take a look at option B

y = 2(2) -8 = 4-8 = -4

we can see that (2,-4) works

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2 years ago

Work out the size of one of the exterior angles

hodyreva [135]

Given Information :-

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A polygon with 10 sides ( Decagon )

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To Find :-

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The value of one of the exterior angles

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Formula Used :-

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$\qquad \diamond \: \underline{ \boxed{ \pink{ \sf Exterior ~angle = \dfrac {360^\circ}{no. ~of~sides}}}} \: \star$

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Solution :-

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Putting the given values, we get,

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$\sf \dashrightarrow Exterior ~angle = \dfrac{360 ^\circ}{10} \: \: \\ \\ \\ \sf \dashrightarrow Exterior ~angle = \frac{36 \cancel{0}^\circ}{ \cancel{10}} \: \: \\ \\ \\ \sf \dashrightarrow Exterior ~angle = \underline{ \boxed{ \frak{ \red{36^\circ}}}} \: \star \\ \\$

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2 years ago

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Find the minimum and maximum value of the function on the given interval by comparing values at the critical points and endpoint

Kitty [74]

Answer:

maximum: y = 1

minimum: y = 0.

Step-by-step explanation:

Here we have the function:

y = f(x) = √(1 + x^2 - 2x)

we want to find the minimum and maximum in the segment [0, 1]

First, we evaluate in the endpoints, which are 0 and 1.

f(0) =√(1 + 0^2 - 2*0) = 1

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Now let's look at the critical points (the zeros of the first derivate)

To derivate our function, we can use the chain rule:

f(x) = h(g(x))

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f'(x) = h'(g(x))*g(x)

Here we can define:

h(x) = √x

g(x) = 1 + x^2 - 2x

Then:

f(x) = h(g(x))

f'(x) = 1/2*( 1 + x^2 - 2x)*(2x - 2)

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The maximum is 1 (when x = 0)

the minimum is 0 (when x = 1)

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