A human baby weighs 7 lbs and 8 oz. at birth. After one month, she weighs 10 lbs and 3 oz. How much more does she weigh one mont
h after birth?
1 answer:
Answer:
she weighs 2lbs 11oz more one month after birth
Step-by-step explanation:
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Answer:
-26x + 94.8y + 16
Step-by-step explanation:
2x + 100y - 28x - 5.2y + 16
combine x's and y's
1/25 of a minute is 2,4 secs.
1/5 = 2,4s
5/5 = 12 secs
I think the answer is 179712ft
Answer:
- The longest side, which is the opposite side of a right angle is the hypotenuse ( h ).
- The opposite is the side opposite the angle involved and it is called the Perpendicular ( p ) .
- The adjacent is the side next to the angle involved ( but not the hypotenuse ) and it is called the base ( b )
![\large{ \tt{❃ \: TAKING \: \angle \: \: A \: AS \: A \: ANGLE \: OF \: REFERENCE : }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9D%83%20%5C%3A%20TAKING%20%20%5C%3A%20%5Cangle%20%5C%3A%20%5C%3A%20A%20%5C%3A%20AS%20%5C%3A%20A%20%5C%3A%20ANGLE%20%5C%3A%20OF%20%5C%3A%20REFERENCE%20%3A%20%20%7D%7D)
- Here , Perpendicular ( p ) = 6 , hypotenuse = y and now we're going to find the value of y first : We know :
![\large{ \tt{❁ \: sin \: 60 \degree = \frac{perpendicular}{hypotenuse} }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9D%81%20%5C%3A%20sin%20%5C%3A%20%2060%20%5Cdegree%20%3D%20%20%5Cfrac%7Bperpendicular%7D%7Bhypotenuse%7D%20%7D%7D)
![\large{ \tt{➝ \: \frac{ \sqrt{3} }{2} = \frac{6}{y} }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9E%9D%20%5C%3A%20%20%5Cfrac%7B%20%5Csqrt%7B3%7D%20%7D%7B2%7D%20%3D%20%20%5Cfrac%7B6%7D%7By%7D%20%20%7D%7D)
![\large{ \tt{➝ \sqrt{3}y = 6 \times 2 }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9E%9D%20%5Csqrt%7B3%7Dy%20%3D%206%20%5Ctimes%202%20%7D%7D)
![\large{ \tt{➝ \: \sqrt{3} \: y = 12 }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9E%9D%20%5C%3A%20%20%5Csqrt%7B3%7D%20%5C%3A%20y%20%3D%2012%20%7D%7D)
![\large{ \tt{➝ \: y= \frac{12}{ \sqrt{ 3} } }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9E%9D%20%5C%3A%20y%3D%20%20%5Cfrac%7B12%7D%7B%20%5Csqrt%7B%203%7D%20%7D%20%7D%7D)
![\boxed{ \large{ \tt{➝ y = \: 4 \sqrt{3} }}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%5Clarge%7B%20%5Ctt%7B%E2%9E%9D%20y%20%3D%20%5C%3A%204%20%5Csqrt%7B3%7D%20%7D%7D%7D)
![\large{ \tt{❇ \: TAKING \: \angle \: B \: AS \: THE \: ANGLE \: OF \: REFERENCE}} :](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9D%87%20%5C%3A%20TAKING%20%20%5C%3A%20%5Cangle%20%5C%3A%20B%20%5C%3A%20AS%20%5C%3A%20THE%20%5C%3A%20ANGLE%20%5C%3A%20OF%20%5C%3A%20REFERENCE%7D%7D%20%3A%20)
- Here - Perpendicular ( p ) = x , hypotenuse = y = 4 √ 3 and Now , we're gonna find the value of x :
![\large{ \tt{❊ \: sin \: 30 \degree = \frac{perpendicular}{hypotenuse} }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9D%8A%20%5C%3A%20sin%20%20%5C%3A%2030%20%5Cdegree%20%3D%20%20%5Cfrac%7Bperpendicular%7D%7Bhypotenuse%7D%20%7D%7D)
![\large{ \tt{⟶ \: \frac{1}{2} = \frac{x}{4 \sqrt{3} } }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9F%B6%20%5C%3A%20%20%5Cfrac%7B1%7D%7B2%7D%20%3D%20%20%5Cfrac%7Bx%7D%7B4%20%5Csqrt%7B3%7D%20%7D%20%20%7D%7D)
![\large{ \tt{⟶ \: 2x = 4 \sqrt{3} }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9F%B6%20%5C%3A%202x%20%3D%204%20%5Csqrt%7B3%7D%20%7D%7D)
![\large{ \tt{⟶ \: x = \frac{4 \sqrt{3} }{2} }}](https://tex.z-dn.net/?f=%20%5Clarge%7B%20%5Ctt%7B%E2%9F%B6%20%5C%3A%20x%20%3D%20%20%5Cfrac%7B4%20%5Csqrt%7B3%7D%20%7D%7B2%7D%20%7D%7D)
![\boxed{\large{ \tt{⟶ \: x = 2 \sqrt{3} }}}](https://tex.z-dn.net/?f=%20%20%5Cboxed%7B%5Clarge%7B%20%5Ctt%7B%E2%9F%B6%20%5C%3A%20x%20%3D%202%20%5Csqrt%7B3%7D%20%7D%7D%7D)
![\boxed{ \large{ \tt{❈ \: OUR \: FINAL \: ANSWER : \boxed{ \tt \: x = 2 \sqrt{3} \: \: y = 4 \sqrt{3} }✓}}}](https://tex.z-dn.net/?f=%20%5Cboxed%7B%20%20%5Clarge%7B%20%5Ctt%7B%E2%9D%88%20%5C%3A%20%20OUR%20%5C%3A%20FINAL%20%5C%3A%20ANSWER%20%3A%20%20%5Cboxed%7B%20%5Ctt%20%5C%3A%20x%20%3D%202%20%5Csqrt%7B3%7D%20%20%20%5C%3A%20%5C%3A%20y%20%3D%204%20%5Csqrt%7B3%7D%20%7D%E2%9C%93%7D%7D%7D)
- And we're done! Hope I helped! Let me know if you have any questions regarding my answer and also notify me , if you need any other help ! :)
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Answer:
RSQA ~ ZXYA
Step-by-step explanation: