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tia_tia [17]
2 years ago
14

3 T-shirts and 2 hats cost £14 2 T-shirts and 7 hats cost £15

Mathematics
2 answers:
Tcecarenko [31]2 years ago
8 0

Answer:

what is the meaning of the figure

juin [17]2 years ago
4 0

Answer:

see explanation

Step-by-step explanation:

Assuming you require the cost of a t- shirt and a hat

let t be t- shirt and h be hat , then

3t + 2h = 14 → (1)

2t + 7h = 15 → (2)

Multiplying (1) by 2 and (2) by - 3 and adding eill eliminate t

6t + 4h = 28 → (3)

- 6t - 21h = - 45 → (4)

Add (3) and (4) term by term to eliminate t

0 - 17h = - 17

- 17h = - 17 ( divide both sides by - 17 )

h = 1

Substitute h = 1 into either of the 2 equations and solve tor t

Substituting into (1)

3t + 2(1) = 14

3t + 2 = 14 ( subtract 2 from both sides )

3t = 12 ( divide both sides by 3 )

t = 4

The cost of a t- shirt is £4 and a hat is £1

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2 years ago
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Help, will give brainlist!!
Licemer1 [7]

Answer:

<u>Fred.</u>

Started hang gliding at a height of 700 ft and descends 15 feet every seconds

<u>Gene</u>

Started hang gliding at a height of 575 ft and descends 10 feet every seconds

Step-by-step explanation:

The function that models Fred's hang gliding is f(x)=-15x+700

The initial value is 700 feet. This Fred was 700 feet above see level before he starts descending.

The rate of descent is -15 ft/s. This means Fred descends 15 feet in one second.

From the table the initial height is 575 ft. This means Gene was 575 feet above sea-level at the beginning of the hang gliding.

The rate of descent is \frac{565-575}{1-0} =-10 ft/s.

This means that in every seconds, Gene descends 10 feet.

4 0
3 years ago
Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt
polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt

=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k

The first integral is trivial since (\tan t)'=\sec^2t.

The second can be done by substituting u=t^2-1:

u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C

The third can be found by integrating by parts:

u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t

\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8

\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C

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3 years ago
What is the equation of this graphed line?
Soloha48 [4]
Y=2x+2
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7 0
3 years ago
What is the slope of the line?
Simora [160]

Answer:

The slope is 0.5 or (1/2)

Step-by-step explanation:

Find the rise over run.

from -3 to -1 and 2 to 3 rises 1 runs 2 so then you divide 1/2 and get 0.5.

8 0
2 years ago
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