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Greeley [361]
3 years ago
15

Solve: 0.5a+2.54=0.8a+4.94.

Mathematics
2 answers:
Kisachek [45]3 years ago
7 0

Answer:

0.5a+2.54=0.8a+4.94

0.5a-0.8a=4.94-2.54

-0.3a=2.40

a=-2.40/0.3=-8

gregori [183]3 years ago
7 0

Answer:

0.5a+2.54=0.8a+4.94

0.5a-0.8a=4.94-2.54

-0.3a=2.40

a=-2.40/0.3=-8

Step-by-step explanation:

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Zina goes shopping. She buys in the same store two scarves and three T-shirts for
Umnica [9.8K]

Answer:

The price of T-shirts before the sale = 42 euros

The price of scarves before the sale = -77/2 euros

Step-by-step explanation:

The cost of the items Zina buys from the store initially are;

Two scarves and three T-shirts for 49 euros

The reduction in the price of scarves a week later during sales = 1.5 euros

The reduction in the price of T-shirts a week later during sales = 2 euros

The items Zina bought for 40 euros at the reduced price of the sales opportunity are;

Four scarves and five T-shirts for 40 euros

Let 'x' represent the initial price of a scarf and let 'y' represent the initial price of a T-shirt, we  have;

2·x + 3·y = 49...(1)

4·(x - 1.5) + 5·(y - 2) = 40...(2)

Expanding equation (2) gives;

4·x - 6 + 5·y - 10 = 40

4·x + 5·y = 40 + 6 + 10 = 56

∴ 4·x + 5·y = 56...(3)

By multiplying equation (1) by 2 and subtracting the result from equation (3), we get;

4·x + 5·y - 2 × (2·x + 3·y) = 56 - 2 × 49 = -63

-y = -42

y = 42

The price of T-shirts before the sale = 42 euros

x = (49 - 3 × 42)/2 = -77/2

x = -77/2

The price of scarves before the sale = -77/2 euros

(However, if the had bought the 4 scarves and 5 T-shirts during sales at 70 euros, we get;

The price of T-shirts before the sale = 12 euros

The price of scarves before the sale = 6.5 euros.

The allowable total price for the reduced 4 scarves and 5 T-shirts is between 66 and about 82 for both initial prices to be +ve)

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Sam raked 3 bags of leaves in 16 minutes. if he continues to work at the same rate about how long will it take him to rake 5 bag
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3 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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99 POINT QUESTION, PLUS BRAINLIEST!!!
VladimirAG [237]
First, we have to convert our function (of x) into a function of y (we revolve the curve around the y-axis). So:


y=100-x^2\\\\x^2=100-y\qquad\bold{(1)}\\\\\boxed{x=\sqrt{100-y}}\qquad\bold{(2)} \\\\\\0\leq x\leq10\\\\y=100-0^2=100\qquad\wedge\qquad y=100-10^2=100-100=0\\\\\boxed{0\leq y\leq100}

And the derivative of x:

x'=\left(\sqrt{100-y}\right)'=\Big((100-y)^\frac{1}{2}\Big)'=\dfrac{1}{2}(100-y)^{-\frac{1}{2}}\cdot(100-y)'=\\\\\\=\dfrac{1}{2\sqrt{100-y}}\cdot(-1)=\boxed{-\dfrac{1}{2\sqrt{100-y}}}\qquad\bold{(3)}

Now, we can calculate the area of the surface:

A=2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\left(-\dfrac{1}{2\sqrt{100-y}}\right)^2}\,\,dy=\\\\\\= 2\pi\int\limits_0^{100}\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=(\star)

We could calculate this integral (not very hard, but long), or use (1), (2) and (3) to get:

(\star)=2\pi\int\limits_0^{100}1\cdot\sqrt{100-y}\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\left|\begin{array}{c}1=\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\end{array}\right|= \\\\\\= 2\pi\int\limits_0^{100}\dfrac{-2\sqrt{100-y}}{-2\sqrt{100-y}}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\,\,dy=\\\\\\ 2\pi\int\limits_0^{100}-2\sqrt{100-y}\cdot\sqrt{100-y}\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\dfrac{dy}{-2\sqrt{100-y}}=\\\\\\

=2\pi\int\limits_0^{100}-2\big(100-y\big)\cdot\sqrt{1+\dfrac{1}{4(100-y)}}\cdot\left(-\dfrac{1}{2\sqrt{100-y}}\, dy\right)\stackrel{\bold{(1)}\bold{(2)}\bold{(3)}}{=}\\\\\\= \left|\begin{array}{c}x=\sqrt{100-y}\\\\x^2=100-y\\\\dx=-\dfrac{1}{2\sqrt{100-y}}\, \,dy\\\\a=0\implies a'=\sqrt{100-0}=10\\\\b=100\implies b'=\sqrt{100-100}=0\end{array}\right|=\\\\\\= 2\pi\int\limits_{10}^0-2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=(\text{swap limits})=\\\\\\

=2\pi\int\limits_0^{10}2x^2\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4}\cdot\sqrt{1+\dfrac{1}{4x^2}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^4}{4x^2}}\,\,dx= 4\pi\int\limits_0^{10}\sqrt{x^4+\dfrac{x^2}{4}}\,\,dx=\\\\\\= 4\pi\int\limits_0^{10}\sqrt{\dfrac{x^2}{4}\left(4x^2+1\right)}\,\,dx= 4\pi\int\limits_0^{10}\dfrac{x}{2}\sqrt{4x^2+1}\,\,dx=\\\\\\=\boxed{2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx}

Calculate indefinite integral:

\int x\sqrt{4x^2+1}\,dx=\int\sqrt{4x^2+1}\cdot x\,dx=\left|\begin{array}{c}t=4x^2+1\\\\dt=8x\,dx\\\\\dfrac{dt}{8}=x\,dx\end{array}\right|=\int\sqrt{t}\cdot\dfrac{dt}{8}=\\\\\\=\dfrac{1}{8}\int t^\frac{1}{2}\,dt=\dfrac{1}{8}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}=\dfrac{1}{8}\cdot\dfrac{t^\frac{3}{2}}{\frac{3}{2}}=\dfrac{2}{8\cdot3}\cdot t^\frac{3}{2}=\boxed{\dfrac{1}{12}\left(4x^2+1\right)^\frac{3}{2}}

And the area:

A=2\pi\int\limits_0^{10}x\sqrt{4x^2+1}\,dx=2\pi\cdot\dfrac{1}{12}\bigg[\left(4x^2+1\right)^\frac{3}{2}\bigg]_0^{10}=\\\\\\= \dfrac{\pi}{6}\left[\big(4\cdot10^2+1\big)^\frac{3}{2}-\big(4\cdot0^2+1\big)^\frac{3}{2}\right]=\dfrac{\pi}{6}\Big(\big401^\frac{3}{2}-1^\frac{3}{2}\Big)=\boxed{\dfrac{401^\frac{3}{2}-1}{6}\pi}

Answer D.
6 0
4 years ago
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That will be one-third.
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