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Ganezh [65]
3 years ago
12

Question up there. ^ |​

Mathematics
1 answer:
wel3 years ago
8 0

Answer: (5,12) hope this helped

Step-by-step explanation: can i get brainliest

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A car traveling at 88 km an hour over takes a bus traveling at 64 km an hour if the bus has a 1.5 hour Headstart how far from th
PIT_PIT [208]

Answer: 352 miles

Step-by-step explanation:

At the point where the car overtook the bus, the car and the bus would have travelled the same distance.

Let x represent the distance covered by bus and the car before the car overtook the bus.

Let t represent the time taken by the bus to travel x miles.

The bus was traveling at 64 km an hour.

Distance = speed × time

Distance travelled by the bus in t hours would be

x = 64 × t = 64t

if the bus has a 1.5 hour Headstart, it means that the car started moving 1.5 hours after the bus has moved.

Therefore, time spent by the car would be t - 15 hours.

The car traveling at 88 km an hour

Distance covered by the car in t - 1.5 hours would be

x = 88(t - 1.5) = 88t - 132

Since the distance covered is equal, then

88t - 132 = 64t

88t - 64t = 132

24t = 132

t = 132/24

t = 5.5 hours

Therefore, the distance from the starting point when the car overtook the bus would be

x = 64 × 5.5 = 352 miles

3 0
3 years ago
Which of these absolute values is the greatest? a. |140| b. |-104| c. |104| d. |-20
Wewaii [24]

Answer: a

/////////////////

5 0
2 years ago
Read 2 more answers
1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a
katovenus [111]

1. Let a and b be coefficients such that

\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}

Combining the fractions on the right gives

\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}

\implies 1 = (2a+b)x + 3a

\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23

so that

\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}

2. a. The given ODE is separable as

x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}

Using the result of part (1), integrating both sides gives

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C

Given that y = 1 when x = 1, we find

\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)

so the particular solution to the ODE is

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)

We can solve this explicitly for y :

\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)

\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|

\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|

\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}

2. b. When x = 9, we get

y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}

8 0
3 years ago
12x -3 divided by -5
Vesna [10]

Answer: 1.8x

Step-by-step explanation:

4 0
4 years ago
A vendor sells hotdogs and bags of potato chips .A customer buys 4 hot dogs and 4 bags of chips for $17.00 another customer buys
mario62 [17]
Let use h= hotdogs price and c=chips price
we know that
4h +4c= 17
5h+3c=17.75
t use the elimination method we will multiply the first equation by 5 and the second one by (-4)
20h +20c=85
-20h -12c= -71
8c=14, we will divide by 8 both side and c=1.75, so a bag of potato chips cost $ 1.75
now we will substitute the c in the first equation
4h+7=17     therefore  4h=17-7,    4h=10,     h=2.50

a bag of potato chips is 1.75
a hot dog costs 2.50
7 0
3 years ago
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