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o-na [289]
4 years ago
10

6 × 2/5= please help me I'm new here

Mathematics
2 answers:
STatiana [176]4 years ago
7 0
2.4. Hop this helped
svetoff [14.1K]4 years ago
4 0
The answer is 2.4 c:

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Complete the equation so that it has solutions of –5 and 7. x2 + x +
zubka84 [21]

Answer:

x² - 2x - 35

Step-by-step explanation:

x₁ = -5

x₂ = 7

-5 * 7 = -35

5 - 7 = -2

x² - 2x - 35 = 0

(x - 7)(x + 5) = 0

x - 7 = 0

x₁ = 7

x + 5 = 0

x₂ = -5

7 0
4 years ago
What is the answer for number 2 and 4 thank you
DaniilM [7]

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5 0
4 years ago
Q10.<br> Prove algebraically that the recurring decimal 0.178 can be written as the fraction 59/330
bearhunter [10]

Answer:

bbv

Step-by-step explanation:

5 0
3 years ago
Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect
Tasya [4]
4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1

defines an ellipse centered at (0,0) with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the x-axis. So if cross sections are taken perpendicular to the x-axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at x=x_0 will have a base of length

\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}
\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a \sqrt2:1 ratio with either of the remaining legs. So if the hypotenuse is \dfrac43\sqrt{9-{x_0}^2}, then either leg will have length \dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}.

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is \dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}.

Finally, x_0 can be chosen from any value in -3\le x_0\le3. We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)

and so the volume would be

\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx
=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}
=16

6 0
3 years ago
Divisibility rules for 2, 5, and 10
Liono4ka [1.6K]

Answer:

B is divisible by 5

Step-by-step explanation:

5 divisibility rule is that the number has to end in 5 or 0. Only B ends in 0, so b is the answer.

4 0
3 years ago
Read 2 more answers
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