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strojnjashka [21]
2 years ago
15

Can you please help me. If you help me i will give you brainliest ( question 6)

Mathematics
1 answer:
quester [9]2 years ago
8 0
Similar can be described as same angles and proportional, however is this proportional? Think about it, 5:10 as 8:14?
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Given: 5x + 3 &gt; 4x + 7.<br> Choose the graph of the solution set.
Mkey [24]

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The first one

Step-by-step explanation:

We simplify this inequality to x>4. This means it is moving past the four to the right with an open dot

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If we deposit $2,200 in an account that pays 4% per year, find the amount of simple
RUDIKE [14]

Step-by-step explanation:

Principal(P) =$2200

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3 years ago
What is another way to express five to the second power? <br>​
Zina [86]

Answer: Squared because it equals to the 5 squared

Step-by-step explanation: Its because 5 to the second power is the same as 5 squared Hope this helped :)

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99 POINT QUESTION, PLUS BRAINLIEST!!!
Elden [556K]
We draw region ABC. Lines that connect y = 0 and y = x³ are vertical so:
(i) prependicular to the axis x - disc method;
(ii) parallel to the axis y - shell method;
(iii) parallel to the line x = 18 - shell method.

Limits of integration for x are easy x₁ = 0 and x₂ = 9.
Now, we have all information, so we could calculate volume.

(i)

V=\pi\cdot\int\limits_a^bf^2(x)\, dx\qquad\implies \qquad a=0\qquad b=9\qquad f(x)=x^3


V=\pi\cdot\int\limits_0^9(x^3)^2\, dx=\pi\cdot\int\limits_0^9x^6\, dx=\pi\cdot\left[\dfrac{x^7}{7}\right]_0^9=\pi\cdot\left(\dfrac{9^7}{7}-\dfrac{0^7}{7}\right)=\dfrac{9^7}{7}\pi=\\\\\\=\boxed{\dfrac{4782969}{7}\pi}

Answer B. or D.

(ii)

V=2\pi\cdot\int\limits_a^bx\cdot f(x)\, dx


V=2\pi\cdot\int\limits_0^{9}(x\cdot x^3)\, dx=2\pi\cdot\int\limits_0^{9}x^4\, dx=&#10;2\pi\cdot\left[\dfrac{x^5}{5}\right]_0^9=2\pi\cdot\left(\dfrac{9^5}{5}-\dfrac{0^5}{5}\right)=\\\\\\=2\pi\cdot\dfrac{9^5}{5}=\boxed{\dfrac{118098}{5}\pi}

So we know that the correct answer is D.

(iii)
Line x = h

V=2\pi\cdot\int\limits_a^b(h-x)\cdot f(x)\, dx\qquad\implies\qquad h=18


V=2\pi\cdot\int\limits_0^9\big((18-x)\cdot x^3\big)\, dx=2\pi\cdot\int\limits_0^9(18x^3-x^4)\, dx=\\\\\\=2\pi\cdot\left(\int\limits_0^918x^3\, dx-\int\limits_0^9x^4\, dx\right)=2\pi\cdot\left(18\int\limits_0^9x^3\, dx-\int\limits_0^9x^4\, dx\right)=\\\\\\=2\pi\cdot\left(18\left[\dfrac{x^4}{4}\right]_0^9-\left[\dfrac{x^5}{5}\right]_0^9\right)=2\pi\cdot\Biggl(18\biggl(\dfrac{9^4}{4}-\dfrac{0^4}{4}\biggr)-\biggl(\dfrac{9^5}{5}-\dfrac{0^5}{5}\biggr)\Biggr)=\\\\\\

=2\pi\cdot\left(18\cdot\dfrac{9^4}{4}-\dfrac{9^5}{5}\right)=2\pi\cdot\dfrac{177147}{10}=\boxed{\dfrac{177147\pi}{5}}

Answer D. just as before.

6 0
3 years ago
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