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3 years ago

Find the measure of x.

Mathematics

1 answer:

Andru [333]3 years ago

5 0

Answer:

Step-by-step explanation:

Since the triangles are congruent then corresponding angles are congruent

x is the angle between line with 3 strokes and 2 strokes

The corresponding angle is therefore ∠ C

∠ C = 180° - (63 + 29)° = 180° - 92° = 88°

Then x = 88 → C

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2/3 times 1/3 please help

Ulleksa [173]

[2] \frac{x}{y} [/3]*[1] \frac{x}{y} [/3]=[2] \frac{x}{y} [/9]
2*1=2
3*3=9

5 0

3 years ago

Read 2 more answers

What is the value of x?

tigry1 [53]

Sense -4 and +2 would be 6 digits away from each other, we would then have to figure out, what would 4 times x - 2 would equal "x".

We would do, 4×2 = 8.

8 - 2 = 6.

Sense 6 is the amount of how much -4 and +2 would be from each other, this would then mean that x would then equal 8.

$\boxed{x=8}$

3 0

4 years ago

I needd helpppppppp pleaseeeee

almond37 [142]

Answer:

hi guada!

Step-by-step explanation:

well the answer i think is 35.

180-155= 25

180-120-25=35

4 0

3 years ago

Please solve this please

ale4655 [162]

Answer:

C) $\frac{2z+15}{6x-12y}$

E) $\frac{7d+5}{15d^2+14d+3}$

F) $\frac{-7a-b}{6b-4a}$

Step-by-step explanation:

One is given the following equation

$\frac{z+1}{x-2y}-\frac{2z-3}{2x-4y}+\frac{z}{3x-6y}$

In order to simplify fractions, one must convert the fractions to a common denominator. The common denominator is the least common multiple between the given denominators. Please note that the denominator is the number under the fraction bar of a fraction. In this case, the least common multiple of the denominators is ( $6x-12y$ ). Multiply the numerator and denominator of each fraction by the respective value in order to convert the fraction's denominator to the least common multiple,

$\frac{z+1}{x-2y}-\frac{2z-3}{2x-4y}+\frac{z}{3x-6y}$

$\frac{z+1}{x-2y}*\frac{6}{6}-\frac{2z-3}{2x-4y}*\frac{3}{3}+\frac{z}{3x-6y}*\frac{2}{2}$

Simplify,

$\frac{z+1}{x-2y}*\frac{6}{6}-\frac{2z-3}{2x-4y}*\frac{3}{3}+\frac{z}{3x-6y}*\frac{2}{2}$

$\frac{6z+6}{6x-12y}-\frac{6z-9}{6x-12y}+\frac{2z}{6x-12y}$

$\frac{(6z+6)-(6z-9)+(2z)}{6x-12y}$

$\frac{6z+6-6z+9+2z}{6x-12y}$

$\frac{2z+15}{6x-12y}$

In this case, one is given the problem that is as follows:

$\frac{2}{3d+1}-\frac{1}{5d+3}$

Use a similar strategy to solve this problem as used in part (c). Please note that in this case, the least common multiple of the two denominators is the product of the two denominators. In other words, the following value: ( $(3d+1)(5d+3)$ )

$\frac{2}{3d+1}-\frac{1}{5d+3}$

$\frac{2}{3d+1}*\frac{5d+3}{5d+3}-\frac{1}{5d+3}*\frac{3d+1}{3d+1}$

Simplify,

$\frac{2}{3d+1}*\frac{5d+3}{5d+3}-\frac{1}{5d+3}*\frac{3d+1}{3d+1}$

$\frac{2(5d+3)}{(3d+1)(5d+3)}-\frac{1(3d+1)}{(5d+3)(3d+1)}$

$\frac{10d+6}{(3d+1)(5d+3)}-\frac{3d+1}{(5d+3)(3d+1)}$

$\frac{(10d+6)-(3d+1)}{(3d+1)(5d+3)}$

$\frac{10d+6-3d-1}{(3d+1)(5d+3)}$

$\frac{7d+5}{(3d+1)(5d+3)}$

$\frac{7d+5}{15d^2+14d+3}$

The final problem one is given is the following:

$\frac{3a}{2a-3b}-\frac{a+b}{6b-4a}$

For this problem, one can use the same strategy to solve it as used in parts (c) and (e). The least common multiple of the two denominators is ( $6b-4a$ ). Multiply the first fraction by a certain value to attain this denomaintor,

$\frac{3a}{2a-3b}-\frac{a+b}{6b-4a}$

$\frac{3a}{2a-3b}*\frac{-2}{-2}-\frac{a+b}{6b-4a}$

Simplify,

$\frac{3a}{2a-3b}*\frac{-2}{-2}-\frac{a+b}{6b-4a}$

$\frac{-6a}{6b-4a}-\frac{a+b}{6b-4a}$

$\frac{(-6a)-(a+b)}{6b-4a}$

$\frac{-6a-a-b}{6b-4a}$

$\frac{-7a-b}{6b-4a}$

4 0

3 years ago

(1 point) Find the length traced out along the parametric curve x=cos(cos(4t))x=cos⁡(cos⁡(4t)), y=sin(cos(4t))y=sin⁡(cos⁡(4t)) a

Mazyrski [523]

The length of a curve $C$ given parametrically by $(x(t),y(t))$ over some domain $t\in[a,b]$ is

$\displaystyle\int_C\mathrm ds=\int_a^b\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

In this case,

$x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)$

$y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)$

So we have

$\displaystyle\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=16\sin^24t\sin^2(\cos4t)+16\sin^24t\cos^2(\cos4t)=16\sin^24t$

and the arc length is

$\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt$

We have

$\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4$

where $n$ is any integer; this tells us $\sin(4t)\ge0$ on the interval $\left[0,\frac\pi4\right]$ and $\sin(4t)$ on $\left[\frac\pi4,1\right]$ . So the arc length is

$=\displaystyle4\left(\int_0^{\pi/4}\sin4t\,\mathrm dt-\int_{\pi/4}^1\sin4t\,\mathrm dt\right)$

$=-\cos(4t)\bigg_0^{\pi/4}-\left(-\cos(4t)\bigg_{\pi/4}^1\right)$

$=(\cos0-\cos\pi)+(\cos4-\cos\pi)=\boxed{3+\cos4}$

7 0

3 years ago