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soldi70 [24.7K]
3 years ago
12

What is the number of lines that can be drawn perpendicular to a given point on that line in a plane. A.1 B.0 C infinitely many

D.2
Mathematics
1 answer:
REY [17]3 years ago
4 0

Answer:

C. Infinitely many

Step-by-step explanation:

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Find the perimeter of a rectangle whose length is 15 cm and length of one diagonal is 17cm
Ulleksa [173]

Answer:

<h2>46 cm</h2>

Step-by-step explanation:

first ,we use the Pythagorean theorem to determine the width:

width = √(17^2-15^2) = 8

then

perimeter = 2(width+lengt) = 2(8 + 15) = 2×23 = 46

6 0
4 years ago
What is equivalent to 3^4/3^7
Misha Larkins [42]

i believe your answer will be 0.04 or 4/100

OR if you need an equivalent expression, that would be 81/2187

5 0
3 years ago
Assume that the weight loss for the first month of a diet program varies between 6 pounds and 12 pounds, and is
swat32

Answer:

1/3

Step-by-step explanation:

Range of data set: 12 - 6 = 6

Given that there is a uniform distribution:

P(6<x<7) = 1/6

P(7<x<8) = 1/6

P(8<x<9) = 1/6

P(9<x<10) = 1/6

P(10<x<11) = 1/6

P(11<x<12) = 1/6

where x is weight loss

Probability that weight loss is more than 10 pounds:

P(10<x<11) + P(11<x<12) = 1/6 + 1/6 = 1/3

3 0
4 years ago
Nancy's mother has a large collection of shoes. The table shows how many pairs of shoes she
SSSSS [86.1K]

Answer:

P(pumps) = 2/5

Step-by-step explanation:

(Pumps)/(Everything)

(10)/(9+6+10)

10/25

2/5

7 0
3 years ago
Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv
ch4aika [34]

Answer:

Part 1) The function of the First graph is f(x)=(x-3)(x+1)

Part 2) The function of the Second graph is f(x)=-2(x-1)(x+3)

Part 3) The function of the Third graph is f(x)=0.5(x-6)(x+2)

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

f(x)=a(x-c)(x-d)

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

f(x)=a(x-3)(x+1)

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

-4=a(1-3)(1+1)

-4=a(-2)(2)

a=1

therefore

The function is equal to

f(x)=(x-3)(x+1)

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

f(x)=a(x-1)(x+3)

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

8=a(-1-1)(-1+3)

8=a(-2)(2)

a=-2

therefore

The function is equal to

f(x)=-2(x-1)(x+3)

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

f(x)=a(x-6)(x+2)

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

-8=a(2-6)(2+2)

-8=a(-4)(4)

a=0.5

therefore

The function is equal to

f(x)=0.5(x-6)(x+2)

3 0
4 years ago
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