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2 years ago

The length of a swimming pool is 8m longer than its width and the area is 105m2

2 answers:

8 0

$width : w \\ length : \ l = w + 8 \\ A = 105 \ m^2 \\ \\A= w \cdot l \\ \\w(w+l)=105 \\ \\w^2+w = 105 \\ \\w^2+w-105 =0 \\ \\(w+15)(w-7)=0 \\ \\w+15 =0 \ \ or \ \ w-7 = 0 \\ \\w = -15 \ \ or \ \ w=7 \\ \\ width \ cant \ be \ negative, \ so \\ \\ w = 7 \ and \ l = w+8 = 7 +8 =15 \\ \\chek : \\ \\ A=w\cdot l \\ \\A=7\cdot 15=105 \ m^2$

Talja [164]2 years ago

5 0

$l\ \rightarrow\ the\ length\\w\ \rightarrow\ the\ width \\\\l=w+8\ \ \ and\ \ \ l\cdot w=105\ \ \ and\ \ \ l>0,\ w>0\\\\(w+8)\cdot w=105\ \ \ \Rightarrow\ \ \ w^2+8w=105\\\\ w^2+2\cdot4w+4^2=105+4^2\\\\ (w+4)^2=105+16\ \ \ \Rightarrow\ \ \ (w+4)^2=121\\\\w+4=11\ \ \ or\ \ \ w+4=-11\\\\w=7\ \ \ \ \ \ \ \ \ \ or\ \ \ w=-15$

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2 years ago

There were 340,000 cattle placed on feed. Write an equivalent ratio that could be used to find how many of these cattle were bet

lutik1710 [3]

There were 340,000 cattle placed on feed

How many of the 340,000 cattle placed on feed were between 700 and 799 pounds?

Given the fraction of total cattle for 700 - 799 pounds is 2/5

Let x be the number of cattle between 700 - 799 pounds

We make a proportion using the fraction

$\frac{2}{5} = \frac{x}{340,000}$

Cross multiply it and solve for x

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680000 = 5x

Divide by 5 on both sides

So x= 136,000

There were 136,000 cattle between 700 and 799 pounds

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2 years ago

Read 2 more answers

A gymnast dismounts the uneven parallel bars. Her height, h, depends on the time, t, that she is in the air h = −162 + 24t + 16.

dimulka [17.4K]

That is when h=0
so assuming yo meant
h(t)=-16t²+24t+16
solve for t such that h(t)=0
because when height=0, the gymnast hits the ground

so
0=-16t²+24t+16
using math (imma complete the square
0=-16(t²-3/2t)+16
0=-16(t²-3/2t+9/16-9/16)+16
0=-16((t-3/4)²-9/16)+16
0=-16(t-3/4)²+9+16
0=-16(t-3/4)²+25
-25=-16(t-3/4)²
25/16=(t-3/4)²
sqrt both sides
+/-5/4=t-3/4
3/4+/-5/4=t
if we do plus (because minus would give us negative height)
8/4=t
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2 years ago

There are 200 students in eleventh grade high school class. There are 40 students in the soccer team and 50 students in the bask

bekas [8.4K]

Answer:

P(A) = 0.2

P(B) = 0.25

P(A&B) = 0.05

P(A|B) = 0.2

P(A|B) = P(A) = 0.2

Step-by-step explanation:

P(A) is the probability that the selected student plays soccer.

Then:

$P(A)=\dfrac{40}{200}=0.2$

P(B) is the probability that the selected student plays basketball.

Then:

$P(B)=\dfrac{50}{200}=0.25$

P(A and B) is the probability that the selected student plays soccer and basketball:

$P(A\&B)=\dfrac{10}{200}=0.05$

P(A|B) is the probability that the student plays soccer given that he plays basketball. In this case, as it is given that he plays basketball only 10 out of 50 plays soccer:

$P(A|B)=\dfrac{P(A\&B)}{P(B)}=\dfrac{10}{50}=0.2$

P(A | B) is equal to P(A), because the proportion of students that play soccer is equal between the total group of students and within the group that plays basketball. We could assume that the probability of a student playing soccer is independent of the event that he plays basketball.

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2 years ago

A ball is thrown upward and outward from a height of 77 feet. the height of the ball, f(x), in feet, can be modeled by f left

Oksi-84 [34.3K]

From the given function modeling the height of the ball:
f(x)=-0.2x^2+1.4x+7
A] The maximum height of the ball will be given by:
At max height f'(x)=0
from f(x),
f'(x)=-0.4x+1.4
solving for x we get:
-0.4x=-1.4
x=3.5ft
thus the maximum height would be:
f(3.5)=-0.2(3.5)^2+1.4(3.5)+7
f(3.5)=9.45 ft

b]
How far from where the ball was thrown did this occur:
from (a), we see that at maximum height f'(x)=0
f'(x)=-0.4x+1.4
solving for x we get:
-0.4x=-1.4
x=3.5ft
This implies that it occurred 3.5 ft from where the ball was thrown.

c] How far does the ball travel horizontally?
f(x)=-0.2x^2+1.4x+7
evaluationg the expression when f(x)=0 we get:
0=-0.2x^2+1.4x+7
Using quadratic equation formula:
x=-3.37386 or x=10.3739
We leave out the negative and take the positive answer. Hence the answer 10.3739 ft horizontally.

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2 years ago