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3 years ago

How do i solve for x?

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

4 0

The triangles are similar so ∆JKL is just an enlargement of ∆WVU. Therefore the ratios of the side lengths are the same. 10:20 = 1:2 so the side lengths for ∆JKL must be in the ratio 1:2. 15 × 2 = 6x so 30 = 6x so x = 5

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I need help with this question if you can, can you please help me need it ASAP I’m going to really appreciate it

makvit [3.9K]

Answer:

110°

Step-by-step explanation:

All the angles add to 360.

Y+W = 180

and V+X = 180

to find W,

W = 180 - Y

W = 180-70

= 110

6 0

3 years ago

really need help on these two questions for geometry, please actually answer it instead of just grabbing points and leaving :'),

egoroff_w [7]

Answer:

Step-by-step explanation:

The diagonals of a rhombus are perpendicular, so ∠AED = 90° and ∠BEC = 90°.

∠DAE = 180° - ∠AED - ∠EDA = 180° - 90° - 35° = 55°

BD is a transversal across parallel sides AD and BC, so ∠EBC = ∠EDA = 35°.

∠BCE = 180° - ∠BEC - ∠EBC = 180° - 90° - 35° = 55°

5 0

3 years ago

4x + y = 10<br> 7x + 2y = 17

Nezavi [6.7K]

Answer:

x = 3, y = - 2

Step-by-step explanation:

By substitution method,

4x + y = 10 -------> equation 1.

7x + 2y = 17 -------> equation 2.

From equation 1,

y = 10 - 4x ------> equation 3.

Substitute equation 3 in 2,

7x + 2 ( 10 - 4x ) = 17

7x + 20 - 8x = 17

7x - 8x = 17 - 20

- x = - 3

x = 3

Substitute x = 3 in equation 1,

4 ( 3 ) + y = 10

12 + y = 10

y = 10 - 12

y = - 2

Hence,

x = 3

y = - 2

6 0

2 years ago

Read 2 more answers

Find the area of the composite figure.

Umnica [9.8K]

Answer:

7168

Step-by-step explanation:

5 0

3 years ago

What is the derivative of x times squaareo rot of x+ 6?

Dafna1 [17]

Hey there, hope I can help!

$\mathrm{Apply\:the\:Product\:Rule}: \left(f\cdot g\right)^'=f^'\cdot g+f\cdot g^'$
$f=x,\:g=\sqrt{x+6} \ \textgreater \ \frac{d}{dx}\left(x\right)\sqrt{x+6}+\frac{d}{dx}\left(\sqrt{x+6}\right)x \ \textgreater \ \frac{d}{dx}\left(x\right) \ \textgreater \ 1$

$\frac{d}{dx}\left(\sqrt{x+6}\right) \ \textgreater \ \mathrm{Apply\:the\:chain\:rule}: \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx} \ \textgreater \ =\sqrt{u},\:\:u=x+6$
$\frac{d}{du}\left(\sqrt{u}\right)\frac{d}{dx}\left(x+6\right)$

$\frac{d}{du}\left(\sqrt{u}\right) \ \textgreater \ \mathrm{Apply\:radical\:rule}: \sqrt{a}=a^{\frac{1}{2}} \ \textgreater \ \frac{d}{du}\left(u^{\frac{1}{2}}\right)$
$\mathrm{Apply\:the\:Power\:Rule}: \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1} \ \textgreater \ \frac{1}{2}u^{\frac{1}{2}-1} \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{u}}$

$\frac{d}{dx}\left(x+6\right) \ \textgreater \ \mathrm{Apply\:the\:Sum/Difference\:Rule}: \left(f\pm g\right)^'=f^'\pm g^'$
$\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(6\right)$

$\frac{d}{dx}\left(x\right) \ \textgreater \ 1$
$\frac{d}{dx}\left(6\right) \ \textgreater \ 0$

$\frac{1}{2\sqrt{u}}\cdot \:1 \ \textgreater \ \mathrm{Substitute\:back}\:u=x+6 \ \textgreater \ \frac{1}{2\sqrt{x+6}}\cdot \:1 \ \textgreater \ Simplify \ \textgreater \ \frac{1}{2\sqrt{x+6}}$

$1\cdot \sqrt{x+6}+\frac{1}{2\sqrt{x+6}}x \ \textgreater \ Simplify$

$1\cdot \sqrt{x+6} \ \textgreater \ \sqrt{x+6}$
$\frac{1}{2\sqrt{x+6}}x \ \textgreater \ \frac{x}{2\sqrt{x+6}}$
$\sqrt{x+6}+\frac{x}{2\sqrt{x+6}}$

$\mathrm{Convert\:element\:to\:fraction}: \sqrt{x+6}=\frac{\sqrt{x+6}}{1} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}}{1}$

Find the LCD
$2\sqrt{x+6} \ \textgreater \ \mathrm{Adjust\:Fractions\:based\:on\:the\:LCD} \ \textgreater \ \frac{x}{2\sqrt{x+6}}+\frac{\sqrt{x+6}\cdot \:2\sqrt{x+6}}{2\sqrt{x+6}}$

$Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions$
$\frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c} \ \textgreater \ \frac{x+2\sqrt{x+6}\sqrt{x+6}}{2\sqrt{x+6}}$

$x+2\sqrt{x+6}\sqrt{x+6} \ \textgreater \ \mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c}$
$\sqrt{x+6}\sqrt{x+6}=\:\left(x+6\right)^{\frac{1}{2}+\frac{1}{2}}=\:\left(x+6\right)^1=\:x+6 \ \textgreater \ x+2\left(x+6\right)$
$\frac{x+2\left(x+6\right)}{2\sqrt{x+6}}$

$x+2\left(x+6\right) \ \textgreater \ 2\left(x+6\right) \ \textgreater \ 2\cdot \:x+2\cdot \:6 \ \textgreater \ 2x+12 \ \textgreater \ x+2x+12$
$3x+12$

Therefore the derivative of the given equation is
$\frac{3x+12}{2\sqrt{x+6}}$

Hope this helps!

8 0

3 years ago