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ryzh [129]
3 years ago
13

I need help this is due Monday

Mathematics
2 answers:
ASHA 777 [7]3 years ago
8 0
1. 259 
2. 80 
3. 5316
4. 5 
5. 2524 
Korolek [52]3 years ago
8 0
1. 259 in
2. 80 ft.
3. 5,316 ft.
4. 5 ft.
5. 2,524 in.
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adoni [48]

Answer:

the answer is k=1

Step-by-step explanation:

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3 years ago
Which statement best describes the area of Triangle ABC shown below?
sergeinik [125]

Answer:

  It is one-half the area of a rectangle with sides 4 units × 3 units

Step-by-step explanation:

One side of the triangle is on the line y = 2 between points x=2 and x=6. So, that side has length 6-2 = 4.

The opposite vertex has y-value 5, so is 3 units away from the line y = 2.

The area of the triangle can be considered to have a base of 4 and a height of 3. In the formula ...

  A = (1/2)bh

we find the area to be ...

  A = (1/2)×(4 units)×(3 units) . . . . triangle area

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A rectangle's area is the product of its length and width. So, a rectangle that is 4 units by 3 units will have an area of ...

  A = (4 units)×(3 units) . . . . rectangle area

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3 years ago
Of the 25 sheep in a flock 34% are white. how many sheep are white
Iteru [2.4K]
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3 years ago
select the correct answer you're getting two side lengths of 10 cm and 8 centimeters the angle between the sides measure 40° how
Readme [11.4K]

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Step-by-step explanation:

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What is the function written in vertex form?
lys-0071 [83]

Answer:

The answer in the procedure

Step-by-step explanation:

The question does not present the graph, however it can be answered to help the student solve similar problems.

we know that

The equation of a vertical parabola into vertex form is equal to

f(x)=a(x-h)^{2}+k

where

a is a coefficient

(h,k) is the vertex

If the coefficient a is positive then the parabola open up and the vertex is a minimum

If the coefficient a is negative then the parabola open down and the vertex is a maximum

case A) we have

f(x)=3(x+4)^{2}-6

The vertex is the point (-4,-6)

a=3

therefore

The parabola open up, the vertex is a minimum

case B) we have

f(x)=3(x+4)^{2}-38

The vertex is the point (-4,-38)

a=3

therefore

The parabola open up, the vertex is a minimum

case C) we have

f(x)=3(x-4)^{2}-6

The vertex is the point (4,-6)

a=3

therefore

The parabola open up, the vertex is a minimum

case D) we have

f(x)=3(x-4)^{2}-38

The vertex is the point (4,-38)

a=3

therefore

The parabola open up, the vertex is a minimum

4 0
3 years ago
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