Answer:
30 strawberry sundaes.
Step-by-step explanation:
We know there is a total of 70 sundaes.
3 sundaes will be strawberry.
4 sundaes will be choclate.
When you add 3 and 4 together, you get 7. This is your total ratio.
So 3/7 of the sundaes are strawberry.
And 4/7 of sundaes are choclate.
We know there is a total of 70 sundaes.
The total ratio we figured out is 7.
If the total sundaes is 70, and the total ration is 7. Then the total sundaes is 10 times bigger than the ratio.
This means we need to multiply the 3 strawberry sundaes and 4 choclate sundaes by 10.
3*10=30
4*10=40
So there are 30 strawberry sundaes and 40 choclate sundaes.
We need to find the strawberry sundae amount, so the answer is 30.
Hope this helps!
Answer:
VY = 28 units
Step-by-step explanation:
the perimeter (P) of a rectangle is calculated as
P = 2 ( length + breadth )
given P = 130 , then
2(5y + 2 + 4y) = 130 ( divide both sides by 2 )
9y + 2 = 65 ( subtract 2 from both sides )
9y = 63 ( divide both sides by 9 )
y = 7
Then
VY = 4y = 4 × 7 = 28
Answer:
A. 5:6
Step-by-step explanation:
The scale factor of the smaller figure to the larger = the ratio of a side of the smaller figure to the corresponding side of the larger figure.
Thus, side length of 25 units of the smaller figure corresponds to the side length of 30 units.
Therefore:
Scale factor = 25:30
Simplify
Scale factor of the smaller figure to the larger = 5:6
Answer:
31.438
Step-by-step explanation:
yan ppo ang tama sagot
<h2>
Answer:</h2>
∠LMN is a right angle
<h2>
Step-by-step explanation:</h2>
If we want to prove that two right triangles are congruent by knowing that the corresponding hypotenuses and one leg are congruent, we begin as follows:
- Since two legs are congruent and we know this by the hash marks, then the triangle ΔLKN is isosceles.
- By definition LN ≅ NK
- If ∠LMN is a right angle, then MN is the altitude of triangle ΔLKN
- Also MN is the bisector of LK, so KM ≅ ML
- So we have two right triangles ΔLMN and ΔKM having the same lengths of corresponding sides
- In conclusion, ΔLMN ≅ ΔKMN