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Morgarella [4.7K]
3 years ago
5

The answer is 3.8 and I CANT FIGURE IT OUT

Mathematics
1 answer:
Alex17521 [72]3 years ago
8 0
3.8 is the answer you already figured it out
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What the expression has the factor of (H +2)
katen-ka-za [31]

Answer:

a - h^2 + h - 2

Step-by-step explanation:

To find the expression with factor h + 2, I solved it the usual way I would

6 0
3 years ago
The pie chart below illustrates the future plans of 200members of St.Thomas graduating class.
ruslelena [56]

Answer:

<h2>a) approximately 133 graduates</h2><h2>b) approximately 120°</h2>

Step-by-step explanation:

a) the number of graduates planning to continue studying :

= (37 1/2% + 12 1/2% + 16 2/3%) × 200

=\frac{\left( 37\frac{1}{2} +12\frac{1}{2} +16\frac{2}{3} \right)  }{100} \times200

= (37.5 + 12.5 + 16.666666666667)×2

= 133.333333333334

…………………………………

b) the measurement of the angle representing those who plan to work :

= (360× 33 1/2)÷100

= (360× 33.333333333333)÷100

=119.999999999999

8 0
2 years ago
Which of the following statements correctly illustrate the addition property of equality?
liubo4ka [24]

Answer:

x - y + z = 8

2x +3y -z= -2

3x - 2y -9z = 9

Step-by-step explanation:

[1 - 1  1  8]

[2  3  -1 -2]

[3 -2 -9  9]

4 0
3 years ago
I need assistance!!!!
erastovalidia [21]

Answer:

Step-by-step explanation:

Because midpoints were used to make similar triangles we can say

AC=2(DE)

5x+12=2(9x-20)

5x+12=18x-40

-13x+12=-40

-13x=-52

x=4

AC=5x+12

AC=5(4)+12

AC=20+12

AC=32

3 0
3 years ago
Read 2 more answers
Hiro bought a small carton of milk at lunch. If the approximate dimensions of the milk carton are shown what is the minimum amou
faltersainse [42]

Answer:

C.\ 104\ in^2

Step-by-step explanation:

At first, the question looks like an optimization problem, but since all the dimensions of the carton are given, we only have to compute the total area of the given figure.

Let's calculate the front (and back) areas, which are rectangles

A_1=(6.5)(3)=19.5\ in^2

Now with the lateral rectangles which happen to have the very same dimensions

A_2=19.5\ in^2

Next, we compute the front and back triangles of base 3 in and height 1.5 in

A_3=\frac{1}{2}(3)(1.5)=2.25\ in^2

Now, the lateral inclined rectangles of base 3 in and height 2 in

A_4=(3)(2)=6\ in^2

Finally, the base rectangle who happens to be a square of side 3 in

A_5=(3)(3)= 9\ in^2

This last area, unlike all others, is not doubled because its counterpart is inside the carton and is not part of the lateral area

Our total area of cardboard is

A_t=2(19.5)+2(19.5)+2(2.25)+2(6)+9=103.5\ in^2

The closest option to this answer is

C.\ 104\ in^2

5 0
3 years ago
Read 2 more answers
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