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Greeley [361]
3 years ago
8

Can someone help me in number 3 I’m not shure how to do it

Mathematics
1 answer:
Rudik [331]3 years ago
4 0

well (a) hope this helps


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James has $325 in his savings account. He spends $28 a week. How many, w, weeks until he has $157 left in his account?​
otez555 [7]

Answer:

6 weeks

Step-by-step explanation:

Total savings= $325

For every week, his total savings decrease by $28.

Let's form an expression for the amount of money left in his account after w weeks.

325 -28w= 157

28w= 325 -157

28w= 168

<em>Divide by 28 on both sides:</em>

w= 168 ÷28

w= 6

Thus, there are 6 weeks until he has $157.

3 0
2 years ago
Roberto rowed 20 miles downstream in 2.5 hours. The trip back, however, took him 5 hours. Find the rate that Roberto rows in sti
almond37 [142]

Answer:

<em>Roberto's speed in still water is 6 miles/hour and the river speed is 2 miles/hour</em>

Step-by-step explanation:

<u>Relative Speed</u>

When a body is moving at a constant speed v, the distance traveled in a time t is:

d=v.t

When Roberto rows downstream, his speed in still water is added to the speed of the water, making it easier to travel the required distance.

When Roberto rows upstream, his speed in still water is affected by the speed of the water, both are subtracted and the required distance is covered in more time.

Let's call

x = Roberto's rowing speed in still water

y = Speed of the river current

The speed when rowing downstream is x+y, thus the distance traveled is

d=(x+y).t_1

Where t1=2.5 hours. Substituting values:

20=(x+y)*2.5

Rearranging, we find the downstream equation:

2.5x+2.5y=20\qquad[1]

The speed when rowing upstream is x-y, and the distance traveled is

d=(x-y).t_2

Where t2=5 hours. Substituting values:

20=(x-y)*5

Rearranging, we find the upstream equation:

5x-5y=20\qquad[2]

Multiplying [1] by 2:

5x+5y=40

Adding this equation to [2]:

10x=60

Solving:

x=60/10=6

Dividing [2] by 5:

x-y=4

Solving for y

y=x-4=6-4=2

Thus, Roberto's speed in still water is 6 miles/hour and the river speed is 2 miles/hour

3 0
3 years ago
Please help me solve this problem ASAP
DiKsa [7]

\bold{\huge{\blue{\underline{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • <u>The </u><u>right </u><u>angled </u><u>below </u><u>is </u><u>formed </u><u>by </u><u>3</u><u> </u><u>squares </u><u>A</u><u>, </u><u> </u><u>B </u><u>and </u><u>C</u>
  • <u>The </u><u>area </u><u>of </u><u>square </u><u>B</u><u> </u><u>has </u><u>an </u><u>area </u><u>of </u><u>1</u><u>4</u><u>4</u><u> </u><u>inches </u><u>²</u>
  • <u>The </u><u>area </u><u>of </u><u>square </u><u>C </u><u>has </u><u>an </u><u>of </u><u>1</u><u>6</u><u>9</u><u> </u><u>inches </u><u>²</u>

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>area </u><u>of </u><u>square </u><u>A</u><u>? </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u><u> </u></h3>

The right angled triangle is formed by 3 squares

<u>We </u><u>have</u><u>, </u>

  • Area of square B is 144 inches²
  • Area of square C is 169 inches²

<u>We </u><u>know </u><u>that</u><u>, </u>

\bold{ Area \: of \: square =  Side × Side }

Let the side of square B be x

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 144 =  x × x }

\sf{ 144 =  x² }

\sf{ x = √144}

\bold{\red{ x = 12\: inches }}

Thus, The dimension of square B is 12 inches

<h3><u>Now, </u></h3>

Area of square C = 169 inches

Let the side of square C be y

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ 169 =  y × y }

\sf{ 169 =  y² }

\sf{ y = √169}

\bold{\green{ y = 13\: inches }}

Thus, The dimension of square C is 13 inches.

<h3><u>Now, </u></h3>

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3><u>Therefore</u><u>, </u><u>By </u><u>using </u><u>Pythagoras </u><u>theorem</u><u>, </u></h3>

  • <u>The </u><u>sum </u><u>of </u><u>squares </u><u>of </u><u>base </u><u>and </u><u>perpendicular </u><u>height </u><u>equal </u><u>to </u><u>the </u><u>square </u><u>of </u><u>hypotenuse </u>

<u>That </u><u>is</u><u>, </u>

\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}

<u>Here</u><u>, </u>

  • Base = x = 12 inches
  • Perpendicular = z
  • Hypotenuse = y = 13 inches

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ (z)² + (x)² = (y)² }

\sf{ (z)² + (12)² = (169)² }

\sf{ (z)² + 144 = 169}

\sf{ (z)² = 169 - 144 }

\sf{ (z)² = 25}

\bold{\blue{ z = 5 }}

Thus, The dimensions of square A is 5 inches

<h3><u>Therefore</u><u>,</u></h3>

Area of square

\sf{ = Side × Side }

\sf{ = 5 × 5  }

\bold{\orange{ = 25\: inches }}

Hence, The area of square A is 25 inches.

6 0
2 years ago
GIVING BRAILEST<br> Describe a situation that the equation could represent g+3<br> _____ =15<br> 6
krek1111 [17]

Answer:

g=153

Step-by-step explanation:

i hope this helped...

6 0
2 years ago
Using standard algorithm what does 82 times 18 equal
VashaNatasha [74]
82 times 18 would add up to 1,476. Hope this helped. 
6 0
3 years ago
Read 2 more answers
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