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3 years ago

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A girl's feet are negative 2 over 4 yards from the surface of a pool. A boy's feet are negative 3 over 4 yards from the surface

of the pool. Determine whose feet are closer to the surface.

1 answer:

frozen [14]3 years ago

6 0

The girls are closer. Her feet are 1/2 the distance away from the surface pool, the boys are 3/4 away.

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HELP AGAIN PLEASE!!!!

daser333 [38]

Answer:

108 feet²

Step-by-step explanation:

Well we can just multiply the corresponding sides of Quadrilateral ABCD to fit Quadrilateral EFGH.

Line EF is 42 feet

Line FG is 18 feet

Line GH is 18 feet

Line HE is 30 feet

The perimeter of Quadrilateral EFGH is

30+18+18+42=

30+36+42=

66+42=

108 feet²

6 0

3 years ago

Rectangle A is a scale drawing of Rectangle B and has 25% of its area. If rectangle A has side lengths of 4 cm and 5 cm , what a

SpyIntel [72]

Answer:

8 cm and 10 cm

Step-by-step explanation:

Hello, <em> </em>I can help you with this.

Step 1

According to the question there are two rectangles A and B,

Rectangle A is a scale drawing of Rectangle B and has 25% of its area

in other words

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\$

Step 2

Let

Rectangle A

length (1)= 4 cm

length (2)= 5 cm

$Area_{A}=4\ cm * 5\ cm\\Area_{A}=20\ cm^{2}$

put this value into equation 1

$Area_{A}=0.25*Area_{B} (Equation\ 1)\\\\20\ cm^{2} =0.25*Area_{B} \\divide\ each\ side\ by\ 0.25\\\frac{20\ cm^{2} }{0.25}=\frac{0.25}{0.25}*Area_{B}\\ Area_{B}=80\ cm^{2}$

Now, we know the area of rectangle B, to know its length we need to formule other equation

Step 3

$Area_{B}=80\ cm^{2}\\length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\$

the ratio between the lengths must be constant, so the ratio of A must be equal to ratio in B, then

$\frac{length(1A)}{length(2A)}=\frac{length(1B)}{length(2B)} \\\\\\frac{4}{5}= \frac{length(1B)}{length(2B)}\\0.8=\frac{length(1B)}{length(2B)}\\length(1B)=0.8*length(2B) (Equation 3)$

Step three

using Eq 1 and Eq 2 find the lengths

put the value of length(1B) into equation (2)

$length (1B)*length (2B)=80\ cm^{2} (equation\ 2)\\\(0.8*length(2B)) (*length (2B)=80\ cm^{2} \\\\0.8*(length (2B))^{2} =80\ cm^{2}\\(length (2B))^{2} =\frac{80\ cm^{2}}{0.8} \\(length (2B))^{2}=100\\\sqrt{(length (2B))^{2}}=\sqrt{100\ cm^{2}} \\ length (2B)=10\ cm$

Now, put the value of length(2B) into equation 3 to know length (1B)

$length(1B)=0.8*length(2B)\\length(1B)=0.8*10\ cm\\length(1B)=8 cm$

I really hope this helps you, have a great day.

6 0

3 years ago

Evaluate the exponential function f(x)=(1/2)^x. Which value completes this table?

zheka24 [161]

0.5^-1=2
the answer is 2

5 0

3 years ago

Read 2 more answers

Wich system of equations would have NO solution? (Answer correctly with work for branliest)

siniylev [52]

Answer:

A and D both equations don't have any solution

Step-by-step explanation:

y = 3x + 2

so y ≠ - 3x - 2

if 3x + y = 8

then 3x + y ≠ 9

7 0

2 years ago

PLZZZZ HELP What is the slope of a line that is perpendicular to a line whose equation is 3y=−4x+2 ?

earnstyle [38]

$\bf 3y=-4x+2\implies y=\cfrac{-4x+2}{3} \\\\\\ y=\stackrel{\stackrel{slope}{\downarrow }}{-\cfrac{4}{3}}x+\cfrac{2}{3}\impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}$

so the slope of that line above is really -4/3, now

$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{4}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{3}{4}}\qquad \stackrel{negative~reciprocal}{+\cfrac{3}{4}\implies \blacktriangleright \cfrac{3}{4} \blacktriangleleft}}$

6 0

3 years ago