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3 years ago

Choose the correct solution for the given equation x^2-6x=40

Mathematics

2 answers:

Sindrei [870]3 years ago

8 0

Answer:

x=10,-4

Step-by-step explanation:

Move all of the terms to the left and then set x to zero . Then set each factor equal to zero .

shutvik [7]3 years ago

7 0

Answer:

10,-4

Step-by-step explanation:

not sure where the options are but if you were to solve this equation first bring everything to one side.

x^2 - 6x - 40 = 0

factor it

(x-10)(x+4) = 0

set each part to 0

x-10 = 0 and x+4 = 0

solutions are 10 and -4

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Undersea explorers discover a second sentient race living on Earth. This species is scientifically much more advanced but cannot

Pavel [41]

Answer:

46 cm

Step-by-step explanation:

Let p represent the length in cm of 1 bap'ai; let k represent the length in cm of 1 bok'ai. Then we have ...

12p +2k = 100

10p +10k = 100

Subtracting the second equation from 5 times the first, we get ...

5(12p +2k) -(10p +10k) = 5(100) -(100)

50p = 400

p = 8 . . . . cm

Then the second equation tells us ...

10(8) +10k = 100

10k = 20

k = 2 . . . . cm

Then 5p+3k = 5(8) +3(2) = 46 cm.

The distance 5 bap'ai and 3 bok'ai is 46 cm.

8 0

3 years ago

Okay so imugwy and wat is 100*3/6

jonny [76]

Answer:

what is your questions mate I mm didn't understand ¯\_(ツ)_/¯

Step-by-step explanation:

,When a percent amount is multiplied to another number, the operation produces a value that equals the given percent of the original number. ... Multiplying a number by 100 percent is a just variation of the multiplicative identity and will result in the value being unchanged.

4 0

2 years ago

What is the value of the expression (81 1/4) ^4

Mama L [17]

the answer is 43580627.44 according to my calculations!

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3 years ago

Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an

Leya [2.2K]

Parameterize the lateral face $T_1$ of the cylinder by

$\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v$

where $0\le u\le2\pi$ and $0\le v\le3$ , and parameterize the disks $T_2,T_3$ as

$\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)$
$\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)$

where $0\le r\le2$ and $0\le\theta\le2\pi$ .

The integral along the surface of the cylinder (with outward/positive orientation) is then

$\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS$
$=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr$
$=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr$
$=136\pi$

7 0

3 years ago

Question 1

vivado [14]

Answer:

For Lin's answer

Step-by-step explanation:

When you have a triangle, you can flip it along a side and join that side with the original triangle, so in this case the triangle has been flipped along the longest side and that longest side is now common in both triangles. Now since these are the same triangle the area remains the same.

Now the two triangles form a quadrilateral, which we can prove is a parallelogram by finding out that the opposite sides of the parallelogram are equal since the two triangles are the same(congruent), and they are also parallel as the alternate interior angles of quadrilateral are the same. So the quadrilaral is a paralllelogram, therefore the area of a parallelogram is bh which id 7 * 4 = 7*2=28 sq units.

Since we already established that the triangles in the parallelogram are the same, therefore their areas are also the same, and that the area of the parallelogram is 28 sq units, we can say that A(Q)+A(Q)=28 sq units, therefore 2A(Q)=28 sq units, therefore A(Q)=14 sq units, where A(Q), is the area of triangle Q.

7 0

3 years ago

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