The name of place value of 9 in 199 is ones.
The second 9 has a value of tens.
The number 1 has a place value of hundreds.
Answer:
The equation for area of a triangle is height multiplied by the base length, divided by 2. So, if the height is 2, and the base is 4, then you multiply the two together, making 8, and then divide it by 2, making the final area 4. So if the dimensions are doubled, then the equation would be 4 times 8 divided by 2. Which is 16. If we do this again with 16 times 8, we will get 128, which when divided by two is 64. This is 4 times 16, just as 16 is 4 times the original answer 4. This means that every time you double the dimensions you will get the original answer multiplied by 4.
Step-by-step explanation:
40 fifth graders take a bus to school. 8x8= 64 so they have you a good fraction to start with to find out how many kids are walking or taking their bike (3/8) all you have to do is multiply 3 and 8 which equals 24 to find the fraction of the amount of kids who take the bus you subtract 3 from 8 which gives 5/8 and to get the answer you multiply 5 and 8 which is 40.
Answer:
21.98 is the circumference because is it 7 pi 7 times 3.14. For the area though it is 38.48 because you need to get the radius of the circle and you divided 7 by 2 which is 3.5 and here is the work A=πr2=π·3.52≈38.48451
Step-by-step explanation:
Hope this helped and you understood if you want it would help a lot if you put it brainliest.
Answer:
The correct answer is - 1/2 or 50% for first and second child to be affected.
Step-by-step explanation:
Achondroplasia is an autosomal dominant disorder. Autosomal dominant disorder refers to the presence of a single copy of the defective gene that is enough to lead to dwarfness.
A cross of achondroplasia (Aa) parent to a person of normal height (aa) result in half of their children will be affected with dwarfism and the other half will be normal.
a cross between affected or dwarf and normal parent
Aa X aa
Punnett square:
a a
A Aa Aa
a aa aa
Aa- dwarfness
aa- normal height
The probability that both their first child and second child would have achondroplasia is
2/4 =1/2 or 50%.
