Ask question

Ask question

All categories

3 years ago

15

At the beach, Pancho and his sister both built sandcastles and then measured their heights. Pancho's sandcastle was 3/5 of a foo

t tall and his sister's was 2/5 of a foot tall. How much taller was Pancho's sandcastle than his sister's?

1 answer:

iVinArrow [24]3 years ago

8 0

Answer: 1/5 a foot taller

Step-by-step explanation: Basically you just subtract 2/5 from 3/5. That gives you 1/5.

You might be interested in

Can you explain what went wrong and the actual answers these 3

vlada-n [284]

For the first one, they got rid of the 6x when they shouldn’t have, and the correct answer is x = 21.5

For the second one, 47 would be subtracted by 180, not 90, and the correct answer is 133

For the third one, 5x - 10 doesn’t equal 115, 5x - 10 equals 180 - 115, and the correct answer is x = 17

7 0

2 years ago

Read 2 more answers

Write the mixed number as an improper fraction.<br> -4 1/5

Sophie [7]

-(4 1/5)
5(4)+1 = 21
Thus the answer would be -21/5

4 0

3 years ago

Read 2 more answers

Joe wants to put a fence around his rectangular garden. his garden measures 33 ft by 39 ft. the garden has a path around it that

sukhopar [10]

The perimeter of a rectangle is given by:
P = 2W + 2L
Where,
W: width
L: long
Substituting the values we have to:
P = 2 * (33 + 2 * (3)) + 2 * (39 + 2 * (3))
P = 168 feet
Answer:
joe needs to enclose the garden and path 168 feet of fencing material

4 0

3 years ago

Read 2 more answers

-7/6 or -11/15<br>which is the greatest?

Lelechka [254]

Answer:

-11/15 is the greatest because ita in negative

6 0

3 years ago

Read 2 more answers

Solve for x:a(a²+b²)x²+b²x-a

m_a_m_a [10]

Answer:

x = a/(a² + b²) or x = -1/a

Step-by-step explanation:

a(a²+ b²)x² + b²x - a =0

Use the quadratic equation formula:

$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a} =\dfrac{-b\pm\sqrt{D}}{2a}$

1. Evaluate the discriminant D

D = b² - 4ac = b⁴ - 4a(a² + b²)(-a) = b⁴ + 4a⁴ + 4a²b² = (b² + 2a²)²

2. Solve for x

$\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{D}}{2a}\\\\ & = & \dfrac{-b^{2}\pm\sqrt{(b^{2}+2a^{2})^{2}}}{2a(a^{2} + b^{2})}\\\\ & = & \dfrac{-b^{2}\pm(b^{2}+ 2a^{2})}{2a(a^{2} + b^{2})}\\\\x = \dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}\\\\x =\dfrac{-b^{2}+(b^{2} + 2a^{2})}{2a(a^{2} + b^{2})}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\\end{array}$

$\begin{array}{rcl}x = \large \boxed{\mathbf{\dfrac{a}{a^{2} + b^{2}}}}&\qquad& x =\dfrac{-b^{2}-(b^{2} +2a^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2b^{2}- 2a^{2}}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\dfrac{-2(a^{2}+ b^{2})}{2a(a^{2} + b^{2})}\\\\&\qquad& x =\large \boxed{\mathbf{-\dfrac{1}{a}}}\\\\\end{array}$

5 0

3 years ago