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laiz [17]
3 years ago
14

Find the missing number 1:2=3:

Mathematics
2 answers:
Bad White [126]3 years ago
8 0
The missing number is 6 because 3 is half of 6.
zheka24 [161]3 years ago
4 0
I think it is  4 thank you


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Convert the mixed number to a fraction greater than 1. 3 2/5
vitfil [10]

3 \frac{2}{5} = \frac{5(3) + 2}{5} = \frac{17}{5}

8 0
4 years ago
Can someone evaluate? And please show work.<br> 8z - 3 z=4
Umnica [9.8K]

Answer:

putting the value of z

8(4) - 3

32 - 3

=29

Step-by-step explanation:

8 0
3 years ago
I really need this help
IRISSAK [1]

Answer:

area: 454.2 sq.cm

perimeter: 77.7 cm

Step-by-step explanation:

to find area, find the area of all the separate shapes and add them together:

semi-circle:

radius is 12 cm

area = \frac{\pi (12)^2\\}{2} = 226.2 sq.cm

separate the trapezoid into two right triangles and a rectangle.

area for triangles:

\frac{1}{2}(5)(12) = 30

30 x 2 = 60 sq.cm (since there are two triangles)

area for rectangle:

12 x 14 = 168 sq.cm

area of whole shape: 226.2 + 60 + 168 = 454.2 sq.cm

for perimeter, just add the length of the exterior sides together.

we have some unknown sides, such as the semicircle and the sides of the trapezoid

we can find the hypotenuse of the triangles we used earlier to find the sides of the trapezoid with pythagorean theorem

5^2 + 12^2 = c^2

25 + 144 = c^2

169 = c^2

\sqrt{169} =\sqrt{c^2}

c = 13 cm

to find the outside perimeter of the semi-circle, solve as if you were finding the circumference of a normal circle and divide it in half

\frac{2\pi12}{2} = 37.7 cm

now add it all up:

14 + 13 + 13 +37.7 = 77.7

3 0
3 years ago
Help please!!!!!!!!!!
adell [148]

Answer:

1/2?

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Evaluate the expression when a=-15 and b=-5. b^2 / a + 5​
DENIUS [597]
Answer: 20/3

Solution:

-5^2/-15 + 5
Determine the sign

5^2/15 + 5
Calculate

25/15 + 5
Simplify

5/3 + 5
Transform the expression

5 + 15/3
Calculate

= 20/3

(But there’s alternative forms too!: 6.6, 6 2/3)
3 0
2 years ago
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