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vovangra [49]
3 years ago
8

The balance of Susu's savings account can be represented by the variable b. The inequality describing her balance is b greater-t

han 30 dollars. Which could be a solution to the inequality? $15 $20 $30 $35
Mathematics
2 answers:
Sphinxa [80]3 years ago
8 0

Answer:

$35

Step-by-step explanation:

b is greater than 30

35 is the only value greater than 30

kompoz [17]3 years ago
3 0
$35 it is there only value larger than 30
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balu736 [363]

Answer:

Option A. one rectangle and two triangles

Option E. one triangle and one trapezoid

Step-by-step explanation:

step 1

we know that

The area of the polygon can be decomposed into one rectangle and two triangles

see the attached figure N 1

therefore

Te area of the composite figure is equal to the area of one rectangle plus the area of two triangles

so

A=(8)(4)+2[\frac{1}{2}((8)(4)]=32+32=64\ yd^2

step 2

we know that

The area of the polygon can be decomposed into one triangle and one trapezoid

see the attached figure N 2

therefore

Te area of the composite figure is equal to the area of one triangle plus the area of one trapezoid

so

A=\frac{1}{2}(8)(4)+\frac{1}{2}((4+8)(8)=16+48=64\ yd^2

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You decide to enter in a rowing competition. To train, you go to the boat house and begin rolling down stream. You row for the s
fredd [130]

Answer:

Time spent rowing down stream =100\ seconds

Speed of boat in still water =14\ ms^{-1}

Step-by-step explanation:

Let speed of boat in still water be = x\ ms^{-1}

Speed of current = 10\ ms^{-1}

Speed of boat down stream = \textrm{Speed of boat in still water +Speed of current}= (x+10)\ ms^{-1}

Distance rowed down stream = 2400 m

Time spent rowing down stream = \frac{Distance}{Speed}=\frac{2400}{x+10}\ s = \frac{Distance}{Speed}=\frac{2400}{x+10}\ s

Speed of boat up stream = \textrm{Speed of boat in still water -Speed of current}= (x-10)\ ms^{-1}

Distance rowed up stream = \frac{1}{6} \textrm{ of distance rowed downstream}=\frac{1}{6}\times 2400 = 400\ m

Time spent rowing up stream = \frac{Distance}{Speed}=\frac{400}{x-10}\ s

We know that,

\textrm{Time spent rowing down stream =Time spent rowing up stream}

So,

\frac{2400}{x+10}=\frac{400}{x-10}

Cross multiplying

2400(x-10)=400(x+10)

Dividing both sides by 400

\frac{2400(x-10)}{400}=\frac{400(x+10)}{400}

6(x-10)=x+10

6x-60=x+10

Adding 60 to both sides.

6x-60+60=x+10+60

6x=x+70

Subtracting both sides by x

6x-x=x+70-x

5x=70

Dividing both sides by 5.

\frac{5x}{5}=\frac{70}{5}

∴ x=14

Speed of boat in still water =14\ ms^-1

Time spent rowing down stream =\frac{2400}{14+10}=\frac{2400}{24}=100\ s

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