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g100num [7]
2 years ago
9

The line AB is shown on the grid.

Mathematics
1 answer:
Aleksandr [31]2 years ago
6 0

a) The gradient of a line perpendicular to AB is -2.

b) equation of line is y=-2x+6

<h3>What is gradient?</h3>

The gradient is the measure of the steepness of a straight line.

a)The gradient of line is

=(-2-2)/(7+1)

=-4/8

=-1/2

So, gradient of a line perpendicular to AB is  -2

b)The coordinates of mid point (3, 0)

m = -2

So,

0= -6+ b

b= 6

So, equation of line is y=-2x+6

Learn more about this concept here:

brainly.com/question/19465597

#SPJ1

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What is the value of h in the diagram below? If necessary, round your answer
Ostrovityanka [42]

ANSWER

10.9

EXPLANATION

We use the Altitude Theorem to determine the value of h.

According to the Altitude Theorem, the height, h is equal to the geometric mean of the two segment created by the leg of the altitude on the hypotenuse.

This implies that:

h=  \sqrt{MP \times PO}

That's,

h=  \sqrt{(24- 7)7}

h=  \sqrt{17 \times 7}

h=  \sqrt{119}

h= 10.9

to the nearest tenth.

3 0
4 years ago
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Olenka [21]
Haha very funny !!!!!!!!!!!!!
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3 years ago
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What is the area for the parallelogram/trapeziod in this image? (NOT THE AREA OF THE WHOLE SHAPE)
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Because the area of a trapezoid is A= ((a+b)/2)h, where a is the length of the top, b is the length of the base, and h is the height, you plug that in to find your answer.

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Now divide by 2 to get A = (8.5)h

Because the height is 9, A = (8.5)9 = 76.5

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3 0
3 years ago
Write an equation for the graph​ below, which represents an exponential function f with base 2 or​ 3, translated​ and/or reflect
evablogger [386]

Answer:

f(x) = 2*2^x + 3

Step-by-step explanation:

The model for an exponencial function is:

f(x) = a*b^x + c

'c' is the value of the asymptote, so we have c = 3.

To find the value of 'a', let's use the point (0,5):

5 = a*b^0 + 3

5 = a*1 + 3

a = 2

Now, to find the value of 'b', let's use the point (1, 7):

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7 = 2b + 3

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b = 2

So our function is:

f(x) = 2*2^x + 3

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4 = 2 * 2^{-1} + 3

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6 0
3 years ago
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Base angles.
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