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Gala2k [10]
3 years ago
13

Here is a net made of right triangles and rectangles ​

Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
3 0

not sure your question?

comment then i can edit my answer XD

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Explain why it might be important to be able to work with and manipulate fractions when working in construction.
nirvana33 [79]

Answer:

Addition and Subtraction with different denominators. If the denominators are different then a common denominator needs to be found.

2 pages.

hope this helps

4 0
2 years ago
If f (a.b) = f(a) + f(b) and f (2) = 3, then f (32) equals how can we get 15 after this
attashe74 [19]
Hello,

f(4)=f(2*2)=f(2)+f(2)=3+3=6
f(8)=f(2*4)=f(2)+f(4)=3+6=9
f(16)=f(2*8)=f(2)+f(8)=3+9=12
f(32)=f(2*16)=f(2)+f(16)=3+12=15



4 0
4 years ago
A young sumo wrestler goes on a special diet to gain weight. The variable w models the wrestlers weight in km after the wrestler
svetlana [45]

Answer:

10.8kg

Step-by-step explanation:

Before he was on the diet he already weighed 80 kilograms, so it is not important to determining how much weight we gains in two months.  Instead we need to focus on the 5.4t part of the equation.  t represents the number of months so we need to substitute 2 for t and you get

5.4t

5.4(2)

10.8kg

5 0
3 years ago
Find the value of x so that the function has the given value.<br><br> f(x) = 6x+9; f(x)=21
VashaNatasha [74]

Answer: =135

Step-by-step explanation:

f(x) = 6(21)+9

f(x)= 135

8 0
3 years ago
Read 2 more answers
If a translation maps ∠D onto ∠B, which of the following statements is true? triangle CAB, point E is on segment AC between poin
mrs_skeptik [129]

Answer:

The correct option is;

ΔCED ~ ΔCAB

Step-by-step explanation:

Given that the translation maps angle ∠D to angle ∠B, we have;

Angle ∠D is congruent to ∠B (Given)

Segment ED is parallel to segment AB (lines having similar angles to a common transversal)

Therefore, ∠A is congruent to ∠E, (Angle on the same side of a transversal to two parallel lines)

∠C is congruent to ∠C reflexive property

Therefore, we have;

∠C ≅ ∠C

∠E ≅ ∠A  

∠D ≅ ∠B

Which gives ΔCED is similar to ΔCAB (not ΔCBA)

4 0
3 years ago
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