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3 years ago

Find the angles in this quadilateral

Mathematics

1 answer:

ch4aika [34]3 years ago

7 0

Answer:

Therefore a= 90°,b=54°, x=54°, y= 162°

Step-by-step explanation:

a=90°

a:b=5:3

5+3= 8

5/8 x A = 90

A is the sum of the angles of a and b divided in the ratio 5:3

5A/8 = 90

cross multiply

5A= 90 X8 = 720

5A=720

A= 720/5= 144°

b= 3/8 x 144 = 3x144/8 = 432/8 = 54

a= 90

b= 54

x:y is in the ratio of 1:3

the Sum of angles in a Quadrilateral is 360°

if the sum of a and b is 144°

then the reamining angles is 360-144= 216°

then x:y=1:3

1+3=4

x= 1/4 x 216= 54°

y= 3/4 x 216= 162°

Therefore a= 90°,b=54°, x=54°, y= 162°

we can see that b = x = 54°

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Find the value of x

lbvjy [14]

Answer:

x = 14.5

Step-by-step explanation:

Exterior angle thm

<C = <D

7x - 2 = 9x - 31

7x - 9x = -31 + 2

-2x = -29

x = 14.5

5 0

3 years ago

137.5 is 125% of what number

Lana71 [14]

Percent means parts out of 100
125%=125/100=1.25

'of' menas multily

'is' means equals

137.5 is 125% of what
137.5=1.25 times what
divide both sides by 1.25
108=what

the number is 108

3 0

4 years ago

Answer with explanation<br> I will and rate on the feedback.<br> Will mark brainliest too

yarga [219]

Answer: 6.6

Step-by-step explanation:

We have two triangles. Find the hypothenuse of the first triangle to make it easier to solve the second triangle.

First triangle:

a = 6

b = 8

c = ?

Use pythagorean's theorem

$c^2=a^2+b^2\\c=\sqrt{a^2+b^2}$

$c=\sqrt{(6)^2+(8)^2}\\ c=\sqrt{36+64}\\ c=\sqrt{100}\\ c=10$

This hypothenuse is valid for both triangles. Having said this, we already have 2 sides of the second triangle; c and a. We need to find b.

$c^2=a^2+b^2\\b^2=c^2-a^2\\b=\sqrt{c^2-a^2}$

$b=\sqrt{(12)^2-(10)^2}\\ b=\sqrt{144-100}\\ b=\sqrt{44}\\ b=6.6$

6 0

3 years ago

AP Calculus Help Please!

ivanzaharov [21]

4. By the fundamental theorem of calculus,

$\displaystyle\frac{\mathrm dg(x)}{\mathrm dx}=\frac{\mathrm d}{\mathrm dx}\int_{-2}^xf(t)\,\mathrm dt=f(x)$

$g(x)$ is increasing on those intervals where $g'(x)=\dfrac{\mathrm dg(x)}{\mathrm dx}>0$ . So you have

$g'(x)=f(x)=\begin{cases}3&\text{for }-3\le x$

which is clearly positive for $x\in[-3,0)$ , and in the second interval you have

$-x+3>0\implies x$

Together, this means $g'(x)>0$ for all $x\in[-3,3)$ .

5. When $0\le x\le6$ , $f(x)$ reduces to $-x+3$ , so you have

$g(x)=\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt$
$g(x)=6+\left(3t-\dfrac12t^2\right)\bigg|_{t=0}^{t=x}$
$g(x)=6+3x-\dfrac12x^2$

4 0

4 years ago

What is the length of the altitude of the equilateral triangle below

Setler79 [48]

Using Sine.

Sine = Opposite / Hypotenuse

sin 60° = a / (6√3) But sin60° = √3/2

√3/2 = a/(6√3)

Cross multiply

2*a = 6√3 * √3

2a = 6 * √(3*3)

2a = 6*3

a = 6*3/2

a = 3*3 = 9

Option E.

Hope this helps.

3 0

3 years ago