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3 years ago

Please Help.!!!!!!!!!!

Mathematics

1 answer:

frosja888 [35]3 years ago

5 0

Answer:

um i think its i dont know

Step-by-step explanation:

sorry but i dont know this

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Solve the following by substitution : -<br><br>x + y = 10<br>x - y = 5

koban [17]

$\begin{cases} x + y = 10 \\ x - y = 5 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x - y = 5 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x - (10-x) = 5 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x - 10+x = 5 \end{cases}\\ \\$

$\begin{cases} y = 10 -x\\ 2x - 10 = 5 \end{cases}\\ \\\begin{cases} y = 10 -x\\ 2x = 5 +10 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ 2x =15 \ \ /:2 \end{cases}\\ \\ \begin{cases} y = 10 -x\\ x =\frac{15}{2} \end{cases}\\ \\$

$\begin{cases} y = 10 - 7\frac{1}{2}\\ x =7\frac{1}{2} \end{cases}\\ \\ \begin{cases} y = 2\frac{1}{2}\\ x = 8\frac{1}{2} \end{cases}\\ \\$

7 0

3 years ago

Read 2 more answers

Compare 9 × (11 – 4) and 3 × (11 – 4). A. 9 × (11 – 4) is three times as much as 3 × (11 – 4). B. 9 × (11 – 4) is nine times as

liubo4ka [24]

Hello there!

So there are two separate problems: 9 * (11-4) and 3 * (11-4). To find the value for each expression, use order of operations. We'll solve what's in the parenthesis before multiplying that value by the other number. 11-4 is 7. 7 * 9 is 63. 11-4 is 7. 7 * 3 is 21. You can divide 63 by 21 to get 3. This means that 9 * (11 - 4) is three times as much as 3 *(11 - 4). Even just by looking at the problem, the expression in the parenthesis is the same thing, but the front numbers are different. 9 is three times as much as 3. The answer is A.

3 0

3 years ago

Read 2 more answers

Use cylindrical coordinates. find the volume of the solid that is enclosed by the cone z = x2 + y2 and the sphere x2 + y2 + z2 =

sashaice [31]

Let $R$ be the solid. Then the volume is

$\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$

which follows from the facts that

$\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=\zeta\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm d\zeta$
(by computing the Jacobian)

and

$z=x^2+y^2=r^2\implies z+z^2=72\implies z=-9\text{ or }z=8$
(we take the positive solution, since it's clear that $R$ lies above the $x$ - $y$ plane)
$r^2+z^2=72\implies z=\pm\sqrt{72-r^2}$
(again, taking the positive root for the same reason)
$z=r^2\implies 8=r^2\implies r=\sqrt8$

$\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}r(\sqrt{72-r^2}-r^2)\,\mathrm dr$
$=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr$
$=\dfrac{32(27\sqrt2-35)\pi}3$

7 0

3 years ago

Anyone wanna help with this?????

madreJ [45]

The answer is c hope it’s right

4 0

3 years ago

Given that (x+2) is a factor of x^3+8, find the other quadratic factor.

Semmy [17]

If you know the formular a^3+b^3=(a+b)(a^2-ab+b^2), you can solve this problem.
8 is 2 cubed, so x^3+2^3=(x+2)(x^2-2x+4)
so the other quadratic factor is x^2-2x+4

3 0

3 years ago