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pickupchik [31]
3 years ago
13

The slope of the line through the points is 2. Which statement describes how the slope relates to the height of the water is the

pool?

Mathematics
1 answer:
SOVA2 [1]3 years ago
6 0

Answer:

The height of the water increases 2 inches per minute. Or A

Step-by-step explanation:

Edge

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In the number 9,999.999, how does the 9 in the tenths place compare to the 9 in the place to its right
Aleks04 [339]
It’s ten times the value of the 9 in the tenths place.
8 0
3 years ago
Factor the expression.<br> 20b + 35 =<br> (Factor completely.)
Airida [17]

Answer:

I think it would be 5(4b+7)

Step-by-step explanation:

I divided both terms by 5.

6 0
3 years ago
Which equation represents a line which is perpendicular to the line 5x + 2y = 12?
o-na [289]

Answer:

<h2>          y = ²/₅ x - 3</h2>

Step-by-step explanation:

Changing to slope-intercept form:

5x + 2y = 12         {subtract 5x from both sides}

2y = -5x + 12         {divide both sides by 2}

y = -⁵/₂ x + 6

y=m₁x+b₁   ⊥   y=m₂x+b₂   ⇔    m₁×m₂ = -1

{Two lines are perpendicular if the product of theirs slopes is equal -1}

y =-⁵/₂ x + 1    ⇒     m₁ =  -⁵/₂

-⁵/₂× m₂ = -1      ⇒   m₂ = ²/₅

So, any line perpendicular to 5x + 2y = 12 must have slope m =²/₅

7 0
3 years ago
A pair of equations is shown below.
PtichkaEL [24]

Answer:

a

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Tim has to cover 3 tanks completley with paint. ​
bija089 [108]

<em>answer:</em>

<h3><em>Total </em><em>surface </em><em>area </em><em>of </em><em>3</em><em> </em><em>tanks=</em><em>3</em><em>4</em><em>.</em><em>3</em><em> </em><em>m^</em><em>2</em></h3><h3><em>7</em><em> </em><em>tins </em><em>of </em><em>paint </em><em>are </em><em>required</em><em>.</em></h3>

<em>Solution</em><em>,</em>

<em>diameter </em><em>of </em><em>tank=</em><em>1</em><em>.</em><em>4</em><em> </em><em>m</em>

<em>Height </em><em>of </em><em>tank=</em><em>1</em><em>.</em><em>9</em><em> </em><em>m</em>

<em>Given </em><em>tank </em><em>has </em><em>top </em><em>and </em><em>bottom </em>

<em>Now,</em><em> </em><em>The </em><em>total </em><em>surface </em><em>area </em><em>of </em><em>cylinder </em><em>tank </em><em>is </em><em>given </em><em>by:</em>

<em>tsa = 2\pi \:  {r}^{2}  + (2\pi \: r \: ) \times h</em>

<em>r(</em><em> </em><em>radius </em><em>of </em><em>tank)</em><em>=</em>

<em>\frac{diameter}{2}  =  \frac{1.4}{2}  = 0.7 \: m</em>

<em>h(</em><em>height </em><em>of </em><em>tank)</em><em>=</em><em>1</em><em>.</em><em>9</em><em> </em><em>m</em>

<em>TSA </em><em>of </em><em>1</em><em> </em><em>tank</em>

<em>=</em>

<em>2 \times 3.14 \times  {(0.7)}^{2}  + 2 \times 3.14 \times 0.7 \times 1.9 \\  = 11.435 \:  {m}^{2}</em>

<em>TSA </em><em>Of </em><em>3</em><em> </em><em>tanks</em>

<em>=</em>

<em>3 \times 11.435 \:  {m}^{2}  \\  = 34.3 \:  {m}^{2}</em>

<em>Given </em><em>a </em><em>tin </em><em>of </em><em>plate </em><em>covers </em><em>5</em><em>m</em><em>^</em><em>2</em><em> </em><em>area.</em>

<em>No.of </em><em>tins </em><em>required:</em>

<em>1</em><em> </em><em>tin</em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>5</em><em>m</em><em>^</em><em>2</em>

<em>x </em><em>tin</em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>3</em><em>4</em><em>.</em><em>3</em><em> </em><em>m^</em><em>2</em>

<em>5x = 34.3(cross \: multiplication) \\ or \: x =  \frac{34.3}{5}  \\ x = 6.86 \\ x = 7 \: tins</em>

<em>there</em><em>fore</em><em> </em><em>7</em><em> </em><em>tins </em><em>are </em><em>required</em><em>.</em>

<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em>

8 0
3 years ago
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