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2 years ago

A person who is 3 m tall casts a shadow that is 6 m long. At the same time, a building

Mathematics

1 answer:

podryga [215]2 years ago

6 0

Answer:

13.5 meters

Step-by-step explanation:

25 / 2 = 13.5

because if a human is 3 meters tall and cast 2 times the length of himself then if a building casts 25 meters then it should be 2 times less than its shadow.

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HELP DUE IN 10 MINUTES!! Solve for x. −2(4x+4) - 3x - 2 = 34

Andrej [43]

Answer:

$- 2(4x + 4) - 3x - 2 = 34 \\ \\ - 8x - 8 - 3x - 2 = 34 \\ \\ - 8x - 3x = 34 + 8 + 2 \\ \\ - 11x = 34 + 8 + 2 \\ \\ - 11x = 44 \\ \\ x = - \frac{44}{11} \\ \\ x = - 4$

5 0

2 years ago

ULT ma 4. Write an expression, then evaluate. a. 1/4of the sum of 18 and 6

OlgaM077 [116]

Answer: Required expression: $\dfrac14\times(18+16)$

Result: $\dfrac{17}{2}$

Step-by-step explanation:

Given phrase: $" \dfrac14\text{ of the sum of } 18\text{ and }6. "$

Required expression: $\dfrac14\times(18+16)$ ['+' used to express sum, 'x' used in place of 'of']

Since 18+16 = 34

Then,

$\dfrac14\times(18+16)=\dfrac14\times34 \\\\=\dfrac{1}{2}\times17\ \ \text{[Divide numerator and denominator by 2]}\\\\=\dfrac{17}{2}$

Hence, $\dfrac14\times(18+16)=\dfrac{17}{2}.$

8 0

2 years ago

Can someone please help me??

dmitriy555 [2]

Answer:

I is clear that, the linear equation $5x+12=5x-7$ has no solution.

Step-by-step explanation:

Checking the first option:

$\frac{2}{3}\left(9x+6\right)=6x+4$

$6x+4=6x+4$

$\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}$

$6x+4-4=6x+4-4$

$6x=6x$

$\mathrm{Subtract\:}6x\mathrm{\:from\:both\:sides}$

$6x-6x=6x-6x$

$0=0$

$\mathrm{Both\:sides\:are\:equal}$

$\mathrm{True\:for\:all}\:x$

Checking the 2nd option:

$5x+12=5x-7$

$\mathrm{Subtract\:}5x\mathrm{\:from\:both\:sides}$

$5x+12-5x=5x-7-5x$

$\mathrm{Simplify}$

$12=-7$

$\mathrm{The\:sides\:are\:not\:equal}$

$\mathrm{No\:Solution}$

Checking the 3rd option:

$4x+7=3x+7$

$\mathrm{Subtract\:}7\mathrm{\:from\:both\:sides}$

$4x+7-7=3x+7-7$

$\mathrm{Simplify}$

$4x=3x$

$\mathrm{Subtract\:}3x\mathrm{\:from\:both\:sides}$

$4x-3x=3x-3x$

$\mathrm{Simplify}$

$x=0$

Checking the 4th option:

$-3\left(2x-5\right)=15-6x$

$\mathrm{Subtract\:}15\mathrm{\:from\:both\:sides}$

$-6x+15-15=15-6x-15$

$\mathrm{Simplify}\$

$-6x=-6x$

$\mathrm{Add\:}6x\mathrm{\:to\:both\:sides}$

$-6x+6x=-6x+6x$

$\mathrm{Simplify}$

$\mathrm{Both\:sides\:are\:equal}$

$\mathrm{True\:for\:all}\:x$

Result:

Therefore, from the above calculations it is clear that, the linear equation

$5x+12=5x-7$ has no solution.

4 0

2 years ago

The length of RT is 4x-4.Given that point S is between point R and T,

Monica [59]

Answer:

RS = 32

Step-by-step explanation:

Set-up equation like this -- 2x+6+16=4x-4

Solve for x

Then plug x in for the length of RS. You should get 32 for RS.

7 0

2 years ago

The equation giving a family of ellipsoids is u = (x^2)/(a^2) + (y^2)/(b^2) + (z^2)/(c^2) . Find the unit vector normal to each

Fynjy0 [20]

Answer:

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Step-by-step explanation:

Given equation of ellipsoids,

$u\ =\ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}$

The vector normal to the given equation of ellipsoid will be given by

$\vec{n}\ =\textrm{gradient of u}$

$=\bigtriangledown u$

$=\ (\dfrac{\partial{}}{\partial{x}}\hat{i}+ \dfrac{\partial{}}{\partial{y}}\hat{j}+ \dfrac{\partial{}}{\partial{z}}\hat{k})(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2})$

$=\ \dfrac{\partial{(\dfrac{x^2}{a^2})}}{\partial{x}}\hat{i}+\dfrac{\partial{(\dfrac{y^2}{b^2})}}{\partial{y}}\hat{j}+\dfrac{\partial{(\dfrac{z^2}{c^2})}}{\partial{z}}\hat{k}$

$=\ \dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}$

Hence, the unit normal vector can be given by,

$\hat{n}\ =\ \dfrac{\vec{n}}{\left|\vec{n}\right|}$

$=\ \dfrac{\dfrac{2x}{a^2}\hat{i}+\ \dfrac{2y}{b^2}\hat{j}+\ \dfrac{2z}{c^2}\hat{k}}{\sqrt{(\dfrac{2x}{a^2})^2+(\dfrac{2y}{b^2})^2+(\dfrac{2z}{c^2})^2}}$

$=\ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

Hence, the unit vector normal to each point of the given ellipsoid surface is

$\hat{n}\ =\ \ \dfrac{\dfrac{x}{a^2}\hat{i}+\ \dfrac{y}{b^2}\hat{j}+\ \dfrac{z}{c^2}\hat{k}}{\sqrt{(\dfrac{x}{a^2})^2+(\dfrac{y}{b^2})^2+(\dfrac{z}{c^2})^2}}$

3 0

2 years ago