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3 years ago

Factorise m2 - 10m + 24 =0

Mathematics

2 answers:

faltersainse [42]3 years ago

8 0

Answer:

(m-6) (m-4).

Step-by-step explanation:

m2-6m-4m+24

m(m-4)-6(m-4).

therefore answer is (m-6) (m-4).

Finger [1]3 years ago

4 0

Answer:

the answer is =》

$(m - 6)( m- 4)$

Step-by-step explanation:

let's solve :-

=》

${m}^{2} - 10m + 24$

=》

$m {}^{2} - 6 m - 4m + 24$

=》

$m(m - 6) - 4(m - 6)$

=》

$(m - 6)(m - 4)$

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If the complement of an angle is 25 degrees, then the measurement of the angle will be 65 degrees

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Montoya Construction needs to borrow $375,000 to build a road to install utilities in a small subdivision. It borrows the funds

MissTica

Step 1

Given;

$\begin{gathered} \text{Principal(p)= \$375000} \\ \text{First rate = 8\%=}\frac{8}{100}=0.08 \\ \text{Second rate = 20\%= }\frac{20}{100}=0.2 \\ \text{Time}=\frac{90}{365}=\frac{18}{73} \end{gathered}$

Required; To find the difference in interest between the two periods.

Step 2

State the formula for simple interest

$A=P(1+rt)$

Step 3

Find the interest when the rate is 8%

$\begin{gathered} A=375000(1+(0.08\times\frac{18}{73}) \\ A=375000(1+\frac{36}{1825}) \\ A=\text{\$}382397.26 \end{gathered}$

Therefore the interest is given as;

$A-P=382397.26-375000=\text{\$}7397.26$

Step 4

Find the interest in 1980 with a 20% rate

$\begin{gathered} A=375000(1+(0.2\times\frac{18}{73}) \\ A=\text{\$}393493.15 \end{gathered}$

The interest is given as;

$A-p=393493.15-375000=\text{\$}18493.15\text{ }$

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1 year ago

Find the lengths of the missing side . Simplify all radicals !!!<br> help mee!!!!!!

larisa86 [58]

Answer:

$e = 13\sqrt{2}$

$f = 13\sqrt{2}$

Step-by-step explanation:

The ∆ given is an isosceles ∆ with a right angle measuring 90°, and two congruent angles measuring 45° each.

Using trigonometric ratio formula, we can find the lengths of the missing side as shown below:

Finding e:

$sin(\theta) = \frac{opp}{hyp}$

$sin(\theta) = sin(45) = \frac{\sqrt{2}}{2}$

hyp = 26

opp = e = ?

Plug in the values into the formula

$\frac{\sqrt{2}}{2} = \frac{e}{26}$

Multiply both sides by 26

$\frac{\sqrt{2}}{2}*26 = \frac{e}{26}*26$

$\frac{\sqrt{2}}{2}*26 = e$

$\frac{\sqrt{2}}{1}*13 = e$

$13\sqrt{2} = e$

$e = 13\sqrt{2}$

Since side e is of the same length with side f, therefore, the length of side f = $13\sqrt{2}$

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3 years ago

Problem 4: Let F = (2z + 2)k be the flow field. Answer the following to verify the divergence theorem: a) Use definition to find

Viktor [21]

Given that you mention the divergence theorem, and that part (b) is asking you to find the downward flux through the disk $x^2+y^2\le3$ , I think it's same to assume that the hemisphere referred to in part (a) is the upper half of the sphere $x^2+y^2+z^2=3$ .

a. Let $C$ denote the hemispherical <u>c</u>ap $z=\sqrt{3-x^2-y^2}$ , parameterized by

$\vec r(u,v)=\sqrt3\cos u\sin v\,\vec\imath+\sqrt3\sin u\sin v\,\vec\jmath+\sqrt3\cos v\,\vec k$

with $0\le u\le2\pi$ and $0\le v\le\frac\pi2$ . Take the normal vector to $C$ to be

$\vec r_v\times\vec r_u=3\cos u\sin^2v\,\vec\imath+3\sin u\sin^2v\,\vec\jmath+3\sin v\cos v\,\vec k$

Then the upward flux of $\vec F=(2z+2)\,\vec k$ through $C$ is

$\displaystyle\iint_C\vec F\cdot\mathrm d\vec S=\int_0^{2\pi}\int_0^{\pi/2}((2\sqrt3\cos v+2)\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm dv\,\mathrm du$

$\displaystyle=3\int_0^{2\pi}\int_0^{\pi/2}\sin2v(\sqrt3\cos v+1)\,\mathrm dv\,\mathrm du$

$=\boxed{2(3+2\sqrt3)\pi}$

b. Let $D$ be the disk that closes off the hemisphere $C$ , parameterized by

$\vec s(u,v)=u\cos v\,\vec\imath+u\sin v\,\vec\jmath$

with $0\le u\le\sqrt3$ and $0\le v\le2\pi$ . Take the normal to $D$ to be

$\vec s_v\times\vec s_u=-u\,\vec k$

Then the downward flux of $\vec F$ through $D$ is

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt3}(2\,\vec k)\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv=-2\int_0^{2\pi}\int_0^{\sqrt3}u\,\mathrm du\,\mathrm dv$

$=\boxed{-6\pi}$

c. The net flux is then $\boxed{4\sqrt3\pi}$ .

d. By the divergence theorem, the flux of $\vec F$ across the closed hemisphere $H$ with boundary $C\cup D$ is equal to the integral of $\mathrm{div}\vec F$ over its interior:

$\displaystyle\iint_{C\cup D}\vec F\cdot\mathrm d\vec S=\iiint_H\mathrm{div}\vec F\,\mathrm dV$

We have

$\mathrm{div}\vec F=\dfrac{\partial(2z+2)}{\partial z}=2$

so the volume integral is

$2\displaystyle\iiint_H\mathrm dV$

which is 2 times the volume of the hemisphere $H$ , so that the net flux is $\boxed{4\sqrt3\pi}$ . Just to confirm, we could compute the integral in spherical coordinates:

$\displaystyle2\int_0^{\pi/2}\int_0^{2\pi}\int_0^{\sqrt3}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=4\sqrt3\pi$

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4 years ago

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dedylja [7]

Answer:

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Step-by-step explanation:

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3 years ago

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