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Vitek1552 [10]
2 years ago
11

Joshua traveled 5 hours from City A to City B. The distance between the cities is 260 miles. First he traveled at the constant s

peed of 40 mi/h, and then at the constant speed of 60 mi/h. How many hours did he travel at the constant speed of 60 mi/h?
Mathematics
1 answer:
ruslelena [56]2 years ago
5 0

Joshua  travelled for 3 hours at the constant speed of 60 mi/h

<h3><u>Solution:</u></h3>

Given that, Joshua traveled 5 hours from City A to City B

The distance between the cities is 260 miles.  

First he traveled at the constant speed of 40 mi/h, and then at the constant speed of 60 mi/h.  

So, let the number of hours he travelled with 40 mi/h be "n" hours and the distance travelled be "x" miles

Then, the time in which he travelled with 60 mi/h will be 5 – n hours and distance travelled will be 260 – x miles

We have to find how many hours did he travel at the constant speed of 60 mi/h

<em><u>The relation between speed and distance is given as:</u></em>

\text { distance }=\text { speed } \times \text { time. }

\begin{array}{l}{40 \mathrm{mil} / \mathrm{h} \times \mathrm{n} \text { hours }= x \text { miles }} \\\\ {\rightarrow x=40 \mathrm{n} \rightarrow(1)} \\\\ {\text { And } 60 \times(5-\mathrm{n})=260-x \text { miles }}\end{array}

\begin{array}{l}{\rightarrow 300-60 n=260-40 n(\text { from }(1))} \\\\ {\rightarrow 60 n-40 n=300-260} \\\\ {\rightarrow 20 n=40} \\\\ {\rightarrow n=2}\end{array}

So, he travelled 2 hours with speed of 40 mi/h and 5 – 2 = 3 hours with speed of 60 mi/h

Hence, he travelled for 3 hours

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Answer:

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Conclusion: If the angle bisectors of two adjacent angles are perpendicular to eaxh other, the adjacent angles are supplementary angle

Adjacent angles are when the 2 angles have a common vertex and a common arm.

if the exterior sides of 2 adjacent angles are perpendicular, then the angles are complementary angles.

Then the sum of the 2 adjacent angles is a right angle - 90°.

When 2 angles add up to 90°, they are called a pair of complementary angles.

Step-by-step explanation:

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1 year ago
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They buy the van for £8,500 + VAT 20%:

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To find the amount of deposit, subtract the total payments from total cost:

£10,200 - £6,375 = £3,825

The ratio of deposit to the payments is: 3825 : 6375

Reduce the answer by dividing both numbers by 75:

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The ratio becomes: 51/85

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In the figure, polygon ABCD is dilated by a factor of 2 to produce A′B′C′D′ with the origin as the center of dilation. Point A′
emmasim [6.3K]

You did not attach any picture to solve this problem. We cannot calculate for the value of A’ and D’ without the correct graph. However, I think I found the correct graph (see attached), please attach it next time.

So we are given that the figure is dilated by a factor of, meaning that all of its end points are multiplied by 2. By this rule, all we have to do is to simply multiply the initial coordinates of A and D by 2 to get A’ and D’, that is:

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<span>D’ = (2 * 2, -1 * 2) = (4, -2)</span>

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2 years ago
Fatima saved $15,000 toward a down payment on a house. She makes $5,375 a month. What is the maximum loan she can take on a hous
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We will use \frac{28}{36} ratio to determine how much can Fatima afford as a loan for her house.

Let gross income of Fatima = $ x

Amount earned by Fatima monthly= $ 5,375

36 % of 5,375=

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Maximum amount that can be paid as a loan monthly = $ 1935

Total money that she can pay yearly for house loan if Fatima allowable money for paying loans is $ 1935=

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(B) 28 % of $ 5375=

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7 0
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Answer:

  • See the graph attached
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Explanation:

To solve log (−5.6x + 1.3) = −1 − x graphycally, you must graph this system of equations on the same coordinate plane:

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1) To graph the equation 1 you can use these features of logarithmfunctions:

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  • Range: all real numbers (- ∞ , ∞)

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  • y-intercept:

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  • Pick some other values and build a table:

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        -1           0.8

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  • You can see such graph on the picture attached: it is the red curve.

2) Graphing the equation 2 is easier because it is a line: y = - 1 - x

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  • The graph is the blue line on the picture.

3) The solution or solutions of the equations are the intersection points of the two graphs. So, now the graph method just requires that you read the x coordinates of the intersection points. From the least to the greatest, rounded to the nearest tenth, they are:

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