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1 year ago

I am a factor of 18.the other factor is 9.what number am i?

Mathematics

1 answer:

Liula [17]1 year ago

3 0

The answer to this question is 2

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119.31 is 82% of what number

madam [21]

To solve
82% of x = 119.31
rewrite
x = 119.31/0.82
solve
x <span>≈ 145.5</span>

Hope this helps :)

5 0

1 year ago

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What is the sum of StartRoot negative 2 EndRoot and StartRoot negative 18 EndRoot?

Darya [45]

Answer:

$4 \sqrt{2} i$

Step-by-step explanation:

We want to find the sum of

$\sqrt{ - 2} + \sqrt{ - 18}$

We can rewrite this as

$\sqrt{ 2 \times - 1} + \sqrt{ 18 \times - 1}$

This becomes;

$\sqrt{2} \times \sqrt{ - 1} + \sqrt{18} \times \sqrt{ - 1}$

Recall that;

$i = \sqrt{ - 1}$

This implies that;

$\sqrt{2} i + 3 \sqrt{2} i$

Combine like terms:

$4 \sqrt{2} i$

3 0

1 year ago

A flat circular plate has the shape of the region x squared plus y squared less than or equals 1x2+y2≤1. the plate, including t

vredina [299]

You're looking for the extreme values of $x^2+3y^2+13x$ subject to the constraint $x^2+y^2\le1$ .

The target function has partial derivatives (set equal to 0)

$\dfrac{\partial(x^2+3y^2+13x)}{\partial x}=2x+13=0\implies x=-\dfrac{13}2$

$\dfrac{\partial(x^2+3y^2+13x)}{\partial y}=6y=0\implies y=0$

so there is only one critical point at $\left(-\dfrac{13}2,0\right)$ . But this point does not fall in the region $x^2+y^2\le1$ . There are no extreme values in the region of interest, so we check the boundary.

Parameterize the boundary of $x^2+y^2\le1$ by

$x=\cos u$

$y=\sin u$

with $0\le u$ . Then $t(x,y)$ can be considered a function of $u$ alone:

$t(x,y)=t(\cos u,\sin u)=T(u)$

$T(u)=\cos^2u+3\sin^2u+13\cos u$

$T(u)=3+13\cos u-2\cos^2u$

$T(u)$ has critical points where $T'(u)=0$ :

$T'(u)=-13\sin u+4\sin u\cos u=\sin u(4\cos u-13)=0$

$(1)\quad\sin u=0\implies u=0,u=\pi$

$(2)\quad4\cos u-13=0\implies\cos u=\dfrac{13}4$

but $|\cos u|\le1$ for all $u$ , so this case yields nothing important.

At these critical points, we have temperatures of

$T(0)=14$

$T(\pi)=-12$

so the plate is hottest at (1, 0) with a temperature of 14 (degrees?) and coldest at (-1, 0) with a temp of -12.

4 0

1 year ago

Triangle J K L is shown. Angle J K L is a right angle. An altitude is drawn from point K to point M on side L J to form a right

kompoz [17]

Answer:

$LJ=15\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the length side KJ

In the right triangle JKM

Applying the Pythagoras Theorem

$KJ^{2}=JM^{2}+KM^{2}$

we have

$JM=3\ units$

$KM=6\ units$

substitute

$KJ^{2}=3^{2}+6^{2}$

$KJ^{2}=45}$

$KJ=\sqrt{45}\ units$

simplify

$KJ=3\sqrt{5}\ units$

step 2

Find the value of cosine of angle MJK in the right triangle JKM

$cos(JKM)=JM/KJ$

substitute the values

$cos(JKM)=\frac{3}{3\sqrt{5}}$

simplify

$cos(JKM)=\frac{\sqrt{5}}{5}$ -----> equation A

step 3

Find the value of cosine of angle MJK in the right triangle JKL

$cos(JKM)=KJ/LJ$

we have

$KJ=3\sqrt{5}\ units$

$cos(JKM)=\frac{\sqrt{5}}{5}$ ----> remember equation A

substitute the values

$\frac{\sqrt{5}}{5}=\frac{3\sqrt{5}}{LJ}$

Simplify

$LJ=5(3)=15\ units$

8 0

1 year ago

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Solve the following quadratic equations by extracting square roots.Answer the questions that follow.

Vikentia [17]

Answer:

1. x=±4

2. t=±9

3. r=±10

4. x=±12

5. s=±5

Step-by-step explanation:

1. x^2 = 16

Taking square root on both sides

$\sqrt{x^2}=\sqrt{16}\\\sqrt{x^2}=\sqrt{(4)^2}\\$

x=±4

2. t^2=81

Taking square root on both sides

$\sqrt{t^2}=\sqrt{81}\\\sqrt{t^2}=\sqrt{(9)^2}$

t=±9

3. r^2-100=0

$r^{2}-100=0\\r^2 =100\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{r^2}=\sqrt{100}\\\sqrt{r^2}=\sqrt{(10)^2}$

r=±10

4. x²-144=0

x²=144

Taking square root on both sides

$\sqrt{x^2}=\sqrt{144}\\\sqrt{x^2}=\sqrt{(12)^2}$

x=±12

5. 2s²=50

$\frac{2s^2}{2} =\frac{50}{2}\\s^2=25\\Taking\ Square\ root\ on\ both\ sides\\\sqrt{s^2}=\sqrt{25}\\\sqrt{s^2}=\sqrt{(5)^2}$

s=±5 ..

4 0

1 year ago

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